Phuong Phap Giai Cac Bai Tap Ly 12

7/28/2019 Phuong Phap Giai Cac Bai Tap Ly 12

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PHNG PHP GII BI TP VT L 12.

CHNG I : DAO NG C HC

Dng 1: i cng v dao ng iu hoa1) Phng trinh dao ng: x = Acos(t + ) (m,cm,mm)

Trong x: li hay lch khi v tr cn bng (m,cm,mm)A: (A>0) bin hay li cc i (m,cm,mm): tn s gc hay tc gc (rad/s)t + : pha dao ng thi gian t (rad) : pha ban u (rad)

2) Chu ky, tn s:

a. Chu k dao ng iu ha: T =2

=N

tt: thi gian (s) ; T: chu k (s)

b. Tn s f =1

T=

2

3) Vn tc, gia tc:a. Vn tc: v = -Asin(t + )

vmax = A khi x = 0 (ti VTCB) v = 0 khi x = A (ti v tr bin)

b. Gia tc: a = 2Acos (t + ) = 2x amax = 2A khi x = A (ti v tr bin) a = 0 khi x = 0 (ti VTCB)

4) Lin h gia x, v, A: A2 = x2 +2

2

v

.

Lin h : a = - 2x Lin h a v v : 122

2

42

2

=+ A

v

A

a

5) Cac h qu:+ Qu o dao ng iu ha l 2A

+ Thi gian ngn nht i t bin ny n bin kia lT

2

+ Thi gian ngn nht i t VTCB ra VT bin hoc ngc li lT

4+ Qung ng vt i c trong mt chu k l 4A

Dng 2: Tinh chu k con lc lo xo theo c tinh cu to1) Cng thc tnh tn s gc, chu k v tn s dao ng ca con lc l xo:

+ Tn s gc: = km

vi

k: ocng cua loxo (N/m)

m : khoi lng cua vat nang (kg)

+ Chu k: T = 2m

k=N

t=2

g

l*l : gin ra ca l xo (m)

* N: s ln dao ng trong thi gian t

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+ Tn s: f =1 k

2 m

2) Chu k con lc l xo v khi lng ca vt nngGi T1 v T2 l chu k ca con lc khi ln lt treo vt m1 v m2 vo l xo c cng kChu k con lc khi treo c m1 v m2: m = m1 + m2 l T2 =

2

1T + 22T .

3) Chu k con lc v cng k ca l xo.Gi T1 v T2 l chu k ca con lc l xo khi vt nng m ln lt mc vo l xo k1 v l xo k2 cng tng ng v chu k ca con lc khi mc phi hp hai l xo k1 v k2:

a- Khi k 1 ni tip k2 th1 2

1 1 1

k k k= + v T2 = 21T +

2

2T .

b- Khi k1 song song k2 th k = k1 + k2 v 2 2 21 2

1 1 1

T T T= + .

Ch : cng ca l xo t l nghch vi chiu di t nhin ca n.

Dng 3: Chiu di lo xo1) Con lc l xo thng ng:+ Gi lo :chiu di t nhin ca l xo (m)

l: dn ca l xo v tr cn bng: l =mg

k(m)

+ Chiu di l xo VTCB: lcb = lo + l+ Chiu di ca l xo khi vt li x:

l = lcb + x khi chiu dng hng xung.l = lcb x khi chiu dng hng ln.

+ Chiu di cc i ca l xo: lmax = lcb + A+ Chiu di cc tiu ca l xo: lmin = lcb A

h qu:

max min

cb

max min

2

A2

+ = =

l ll

l l

2) Con lc nm ngang:S dng cc cng thc v chiu di ca con lc l xo thng ng nhng vi l = 0

* bin dng ca l xo thng ng:mg

lk

= 2l

Tg

=

* bin dng ca l xo nm trn mt phng nghing c gc nghing :

sinmglk

= 2sin

lTg

=

Mt l xo c cng k, chiu di lc ct thnh cc l xo c cng k1, k2, v chiu di tng ng l l1,

l2, th c: kl = k1l1 = k2l2 =

Dng 4: Lc n hi ca lo xo1) Con lc l xo thng ng:

a- Lc n hi do l xo tc dng ln vt ni c li x:Fh = kl + x khi chn chiu dng hng xung

Trang 2

O (VTCB)x

o

cb

o


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hay Fh = kl x khi chn chiu dng hng lnb- Lc n hi cc i:Fh max = k(l + A) ; Fh max : (N) ; l (m) ; A(m)c- Lc n hi cc tiu:

Fh min = 0 khi A l (vt VT l xo c chiu di t nhin)Fh min = k(l- A) khi A < l (vt VT l xo c chiu di cc tiu)Fh min : ( lc ko v) n v (N)

2) Con lc nm ngang:

S dng cc cng thc v lc n hi ca con lc l xo thng ng nhng vi l = 0*Lc n hi, lc hi phc:

a. Lc n hi:

( )

( ) ( ) neu

0 neu l A

hM

h hm

hm

F k l A

F k l x F k l A l A

F

= +

= + = > =

b. Lc hi phc:0

hpM

hphpm

F kAF kx

F

==

=hay

2

0

hpM

hp

hpm

F m AF ma

F

== =

lc hi phc lun hng v v tr cn bng.c . Fh v tr thp nht: Fh = k (l0 + A ).d. Fh v tr cao nht: Fh = k /l0 A/.e. Lc hi phc hay lc phc hi (l lc gy dao ng cho vt) l lc a vt v v tr cn bng (l hp lcca cc lc tc dng ln vt xt phng dao ng), lun hng v VTCB. F = - Kx. Vi x l ly ca vt.

+ Fmax = KA (vt VTB).+ Fmin = 0 (vt qua VTCB).

Dng 5: Nng lng dao ng ca con lc lo xo

Th nng: Wt =1

2kx2 * Wt : th nng (J) ; x : li (m)

ng nng: W =

1

2mv2 * W : ng n ng (J) ; v : vn tc (m/s)

C nng ca con lc l xo: W = Wt + W = Wt max = W max =1

2kA2 =

1

2m2A2 = const .

W : c nng (nng l ng) (J) A : bi n (m); m: khi lng (kg)

Ch : ng nng v th nng bin thin iu ha cng chu k T =T

2hoc cng tn s f = 2f

Dng 6: Vit phng trnh dao ng iu hoa

+ Tm = 2 f =T

2=

m

k

+ Tm A: s dng cng thc A2 = x2 +2

2

v

hoc cc cng thc khc nh :

+ cho: cho x ng vi v A = .)( 22

vx + Nu v = vmax x = 0 A = .max

v

+ cho: chiu di qu o CD A=2

CD.

+ Cho lc FMAX = KA. A=K

FMAX .

+ Cho lmax v lmin A =2

minllMAX .

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+ Cho c nng hoc ng nng cc i hoc th nng cc i A =k

E2.Vi E = Emax =Etmax =

2

2

1KA .

+ Cho lCB,lmax hoc lCB, lmax A = lmax lCB hoc A = lCB lmin.

+ Tm : T iu kin kch thch ban u: t = 0,o

o

x x

v v

= = , gii phng trnh lng gic tm . Th chon gi

tr k=0Ch : a phng lng gic v dng

* sin a = sinb

+=

+=

2

2

kba

kbak=0,1,2..

* cosa = cosb a = b+ k2 ( k= 0,1,2.)+ Lu :

- Vt i theo chiu dng th v > 0

sin < 0; i theo chiu m th v 0- Cc trng hp c bit:

- Gc thi gian l lc vt qua VTCB theo chiu dng th =-/2.( khi t = 0, x = 0, v > 0 = -2

(rad) )

- Gc thi gian l lc vt qua VTCB theo chiu m th = /2 (khi t = 0, x = 0, v < 0 =2

(rad) )

- Gc thi gian l lc vt VTB dng th =0 (khi t = 0, x = A ;v = 0 = 0. )- Gc thi gian l lc vt VTB m th = (khi t = 0, x = A , v = 0 = (rad) ) .

Mt s trng hp khc ca :

khi t = 0, x = 2

A

, v = 0 = - 3

(rad)khi t = 0, x = -

2

A, v = 0 =

3

(rad)

Dng 7: Tinh thi gian vt chuyn ng t v tri x1 n x2:B1: V ng trn tm O, bn knh A. v trc Ox thng ng hng ln v trc

vung gc vi Ox ti O.B2: xc nh v tr tng ng ca vt chuyn ng trn u.

Nu vt dao ng iu ha chuyn ng cng chiu dng th chn v tr ca vtchuyn ng trn u bn phi trc Ox.

Nu vt dao ng iu ha chuyn ng ngc chiu dng th chn v tr cavt chuyn ng trn u bn tri trc Ox.

B3: Xc nh gc qutGi s: Khi vt dao ng iu ha x1 th vt chuyn ng trn u M

Khi vt dao ng iu ha x2 th vt chuyn ng trn u NGc qut l = MON (theo chiu ngc kim ng h)S dng cc kin thc hnh hc tm gi tr ca (rad)

B4: Xc nh thi gian chuyn ng

t

=

vi l tn s gc ca dao ng iu ha (rad/s)

Ch : Thi gian ngn nht vt i Trang 4

x

OM

N


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+ t x = 0 n x = A/2 (hoc ngc li) l T/12+ t x = 0 n x = - A/2 (hoc ngc li) l T/12+ t x = A/2 n x = A (hoc ngc li) l T/6+ t x = - A/2 n x = - A (hoc ngc li) l T/6

Ch :

Gi O l trung im ca qu o CD v M l trung im ca OD; thi gian i t O n M l12

OM

Tt =

thi gian i t M n D l6

MDTt = .

T v tr cn bng 0x= ra v tr 2

2x A= mt khong thi gian

8

Tt= .

T v tr cn bng 0x= ra v tr 3

2x A= mt khong thi gian

6

Tt= .

Chuyn ng t O n D l chuyn ng chm dn u( 0;av a v < ), chuyn ng t D n O l

chuyn ng nhanh dn u( 0;av a v > )Vn tc cc i khi qua v tr cn bng (li bng khng), bng khng khi bin (li cc i).

Dng 8: Tinh quang ng vt i c t thi im t1 n t2:B1: Xc nh trng thi chuyn ng ca vt ti thi im t1 v t2.

thi im t1: x1 = ?; v1 > 0 hay v1 < 0 thi im t2: x2 = ?; v2 > 0 hay v2 < 0

B2: Tnh qung nga- Quang ng vt i c t thi im t1 n khi qua v tr x1 ln cui cng trong khong thi gian t t1

n t2:

+ Tnh 2 1t t

T

= a Phn tch a = n + b, vi n l phn nguyn

+ S1 = N.4A

b- Tnh qung ng S2 vt i c t thi im vt i qua v tr x1 ln cui cng n v tr x2:+ cn c vo v tr ca x1, x2 v chiu ca v1, v2 xc nh qu trnh chuyn ng ca vt. m tbng hnh v.+ da vo hnh v tnh S2.

c- Vy qung ng vt i t thi im t1 n t2 l: S = S1 + S2

d- Ch : Qung ng:

Neu th4

Neu th 22

Neu th 4

Tt s A

Tt s A

t T s A

= =

= =

= =

suy ra

Neu th 4

Neu th 44

Neu th 4 2

2

t nT s n A

Tt nT s n A A

Tt nT s n A A

= = = + = + = + = +

Dng 9: Tinh vn tc trung bnh+ Xc nh thi gian chuyn ng (c th p dng dng 6)+ Xc nh qung ng i c (c th p dng dng 7)

+ Tnh vn tc trung bnh:S

vt

= .

Dng 10: Chu k con lc n v phng trnh

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1) Cng thc tnh tn s gc, chu k v tn s dao ng ca con lc n:

+ Tn s gc: =l

gvi

l

2g: gia toc trong trng tai ni treo con lac (m/s )

: chieu dai cua con lac n (m)

+ Chu k: T = 2l

g

+ Tn s: f = 1

2 l

g

2) Chu ky dao ng iu ha ca con lc n khi thay i chiu di:Gi T1 v T2 l chu k ca con lc c chiu di l1 vl2

+ Con lc c chiu di l 1 2= +l l l th chu k dao ng l: T2 =2

1T + 22T .

+ Con lc c chiu di l l = l1 l2 th chu k dao ng l: T2 =2

1T 22T .

3) Chu ki con lc n thay i theo nhit :

oT . t

T 2 = vi o o oT =T' - T

t t ' t

= nhit tng th chu k tng v ngc lai

Trong : Chieu dai bien oi theo nhiet o : l = lo(1 +t). l h s n di (K-1) .

T l chu k ca con lc nhit to.T l chu k ca con lc nhit to.

4) Chu ki con lc n thay i theo cao so vi mt t:

T h

T R

= vi T = T T T lun ln hn T

Trong : T l chu k ca con lc mt tT l chu k ca con lc cao h so vi mt t.R l bn knh Tri t. R = 6400km

5) Thi gian chy nhanh, chm ca ng h qu lc trong khong thi gian :T = T T > 0 : ng chy chmT = T T < 0 : ng h chy nhanh

Khong thi gian nhanh, chm: t = T

T

.

Trong : T l chu k ca ng h qu lc khi chy ng, T = 2s. l khong thi gian ang xt

6) Chu ky dao ng iu ha ca con lc n khi chu thm tc dng ca ngoi lc khng i:

T = 2g'l vi : chieu dai con lac n

g':gia toc trong trng bieu kienl

ViF

g' gm

= +r r

vi F : ngoi lc khng i tc dng ln con lc

S dng cc cng thc cng vect tm g

+ Nu F c phng nm ngang ( F g ) th g2 = g2 +2

F

m

.

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Khi , ti VTCB, con lc lch so vi phng thng ng 1 gc : tg =F

P.

+ Nu F thng ng hng ln ( F g ) th g = g F

m g < g

+ Nu F thng ng hng xung ( F g ) th g = g +F

m g > g

Cc dng ngoi lc:+ Lc in trng: F = q E F = q.ENu q > 0 th F cng phng, cng chiu vi ENu q < 0 th F cng phng, ngc chiu vi E

+ Lc qun tnh: F = maF ngc chieu a

F ma

=Ch : chuyn ng thng nhanh dn u a cng chiu vi v

chuyn ng thng chm dn u a ngc chiu vi vc . Phng trnh dao ng:

s = S0 cos(t + ) hoc = 0cos(t + ) vi s = l, S0 = 0l v 100

v = s = - S0sin(t + ) = l0cos(t + +2 )

a = v = -2S0cos(t + ) = -2l0cos(t + ) = -2s = -2lLu : S0 ng vai tr nh A cn s ng vai tr nh x

3. H thc c lp: a = -2s = -2l * 2 2 20

( )v

S s

= + ;2

2 2

0

v

gl = +

4. C nng: 2 2 2 2 2 2 0 0 0 0

1 1 1 1

2 2 2 2tmg

E E E m S S mgl m ll

= + = = = =

Vi 2

1

2E mv= (1 os ).tE mgl c =

Dng 11: Nng lng, vn tc v lc cng dy ca con lc n1) Nng lng dao ng ca con lc n:

- ong nang : W =2

1mv2.

-The nang : Wt = = mgl(1 - cos) =2

1mgl2.

- C nang : W = W + Wt = mgl(1 - cos) +2

1mv 2

Vn tc ca con lc ti v tr dy treo hp vi phng thng ng mt gc

v = ( ) l o2g cos cos .

2) Lc cng dy ca con lc ti v tr dy treo hp vi phng thng ng mt gc T = mg(3cos 2coso) .

Dng 12: Tng hp dao ng . Dao ng tt dn , dao ng cng bc , cng hng

I . TNG HP DAO NG lch pha gia hai dao ng cng tn s: x1 = A1cos(t + 1) v x2 = A2cos(t + 2)

+ lch pha gia dao ng x1 so vi x2: = 2 1Nu > 0 1 > 2 th x1 nhanh pha hn x2.Nu < 0 1 < 2 th x1 chm pha hn x2.

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+ Cc gi tr c bit ca lch pha: = 2k vi k= 0 hai dao ng cng pha = (2k+1) vi k Z hai dao ng ngc pha

= (2k + 1)2

vi k Z hai dao ng vung pha

Dao ng tng hp: x = Asicos(t + )

+ Bin dao ng tng hp: A

2

=

2

1

A+

2

2

A+ 2A1A2cos(2 1)Ch : A1 A2 A A1 + A2

Amax = A1 + A2 khi x1 cng pha vi x2Amin = A1 A2khi x1 ngc pha vi x2

+ Pha ban u: tan2211

211

CosACosA

SinASinA

+

+=

II. DAO NG TT DN DAO NG CNG BC - CNG HNG

1. Mt con lc l xo dao ng tt dn vi bin A, h s ma st

.

* Qung ng vt i c n lc dng li l:2 2 2

2 2

kA AS

mg g

= =

* gim bin sau mi chu k l:2

4 4mg gA

k

= =

* S dao ng thc hin c:2

4 4

A Ak AN

A mg g

= = =

* Thi gian vt dao ng n lc dng li:

.

4 2

AkT At N T

mg g

= = = (Nu coi dao ng tt dn c tnh tun hon vi chu k2

T

= )

3. Hin tng cng hng xy ra khi: f = f0 hay = 0 hay T = T0Vi f, , T v f0, 0, T0 l tn s, tn s gc, chu k ca lc cng bc v ca h dao ng.

CHNG II. SNG C V SNG M

I. SNG C HC

1. Bc sng: = vT = v/f

Trong : : Bc sng; T (s): Chu k ca sng; f (Hz): Tn s ca sng

v: Tc truyn sng (c n v tng ng vi n v ca )

2. Phng trnh sng

Ti im O: uO = Acos(t + )

Ti im M cch O mt on x trn phng truyn sng.

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T

x

O

O

x

M

x


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* Sng truyn theo chiu dng ca trc Ox th uM = AMcos(t + -x

v ) = AMcos(t + - 2

x

)

* Sng truyn theo chiu m ca trc Ox th uM = AMcos(t + +x

v ) = AMcos(t + + 2

x

)

3. lch pha gia hai im cch ngun mt khong x1, x2

1 2 1 22x x x x

v

= =

Nu 2 im nm trn mt phng truyn sng v cch nhau mt khong x th:

2x x

v

= =

Lu :n v ca x, x1, x2, v v phi tng ng vi nhau

Nu 2 im nm trn mt phng truyn sng v cch nhau mt khong d th:

2d d

v

= =

a. Nhng im dao ng cng pha: 2d d

v

= = = 2k d = k (k Z). im gn nht dao

ng cng pha c: d = .

b. Nhng im dao ng ngc pha: 2d d

v

= = = (2k + 1) d = (2k + 1)/2 (k Z). im

gn nht dao ng ngc pha c: d = /2.

c. Nhng im dao ng vung pha: 2d d

v

= = = (2k + 1)/2 d = (2k + 1)/4

(k Z). im gn nht dao ng vung pha c: d = /4. - C n gn li th c (n 1) bc sng: L = (n 1)4. Trong hin tng truyn sng trn si dy, dy c kch thch dao ng bi nam chm in vi tn s

dng in l f th tn s dao ng ca dy l 2f.

II. SNG DNG

1. Mt s ch

* u c nh hoc u dao ng nh l nt sng.

* u t do l bng sng

* Hai im i xng vi nhau qua nt sng lun dao ng ngc pha.

* Hai im i xng vi nhau qua bng sng lun dao ng cng pha.

* Cc im trn dy u dao ng vi bin khng i nng lng khng truyn i

* Khong thi gian gia hai ln si dy cng ngang (cc phn t i qua VTCB) l na chu k.

2. iu kin c sng dng trn si dy di l:

* Hai u l nt sng: *( )2

l k k N

=

S bng sng = s b sng = k

S nt sng = k + 1

* Mt u l nt sng cn mt u l bng sng: (2 1) ( )4

l k k N

= +

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S b sng nguyn = k

S bng sng = s nt sng = k + 1

3. Phng trnh sng dng trn si dy CB (vi u C c nh hoc dao ng nh l nt sng)

* u B c nh (nt sng):

Phng trnh sng ti v sng phn x ti B: os2Bu Ac ft = v ' os2 os(2 )Bu Ac ft Ac ft = =

Phng trnh sng ti v sng phn x ti M cch B mt khong d l:

os(2 2 )M

du Ac ft

= + v ' os(2 2 )M

du Ac ft

=

Phng trnh sng dng ti M: 'M M Mu u u= +

2 os(2 ) os(2 ) 2 sin(2 ) os(2 )2 2 2M

d du Ac c ft A c ft

= + =

Bin dao ng ca phn t ti M: 2 os(2 ) 2 sin(2 )2M

d dA A c A

= + =

* u B t do (bng sng):

Phng trnh sng ti v sng phn x ti B: ' os2B Bu u Ac ft = =Phng trnh sng ti v sng phn x ti M cch B mt khong d l:

os(2 2 )M

du Ac ft

= + v ' os(2 2 )M

du Ac ft

=

Phng trnh sng dng ti M: 'M M Mu u u= +

2 os(2 ) os(2 )M

du Ac c ft

=

Bin dao ng ca phn t ti M: 2 cos(2 )Md

A A

=

Lu : *Vi x l khong cch t M n u nt sng th bin : 2 sin(2 )Mx

A A

=

* Vi x l khong cch t M n u bng sng th bin : 2 cos(2 )Md

A A

=

III. GIAO THOA SNG

Giao thoa ca hai sng pht ra t hai ngun sng kt hp S1, S2 cch nhau mt khong l:

Xt im M cch hai ngun ln lt d1, d2

Phng trnh sng ti 2 ngun 1 1Acos(2 )u ft = + v 2 2Acos(2 )u ft = +

Phng trnh sng ti M do hai sng t hai ngun truyn ti:

11 1

Acos(2 2 )M

du ft

= + v 22 2Acos(2 2 )M

du ft

= +

Phng trnh giao thoa sng ti M: uM= u1M+ u2M

1 2 1 2 1 22 os os 22 2M

d d d d u Ac c ft

+ + = + +

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Bin dao ng ti M: 1 22 os2M

d dA A c

= +

vi 1 2 =

Ch : * S cc i: (k Z)2 2

l lk

+ < < + +

* S cc tiu:1 1

(k Z)2 2 2 2

l lk

+ < < + +

1. Hai ngun dao ng cng pha ( 1 2 0 = = )

* im dao ng cc i: d1 d2 = k (kZ)

S ng hoc s im (khng tnh hai ngun):l l

k

< ZC th u sm pha hn i l2

=> =2

(rad)

- Nu ZL < ZCth u tr pha hn i l2

=> = -

2

(rad)

- BI TP P DNG

Bi 4: Cho on mch nh hnh v: R C LBiu thc cng dng in trong mch: A M N B

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i = 2 2 cos(100t +6

) (A); R = 50, L =

3(H) v C =

35

103

(F)

1. Tnh ZAN, ZMB v ZAB;2. Vit biu thc hiu in th tc thi uAM, uNB v uAB.

Bi 5: Cho on mch in xoay chiu nh hnh v: L R CHai u on mch AB ta duy tr mt hiu in th: A M N B

uAB = 200 2 cos(100t) (V)

R = 100, C =

410 (F), bit cng sut tiu th ca on mch l P = 100W.

a. Tnh tng tr ca on mch v h s t cm L ca cun dy.b. Vit biu thc cng dng in trong mch;c. Vit biu thc hiu in th uMB hai u on mch.

Bi 6: Cho on mch in xoay chiu nh hnh v:Hai u on mch AB ta duy tr mt hiu in th xoay chiu: R M Ro, L N C

u = 200 6 cos 100t (V) A B

Cho bit R = 100, Ro = 50, L =2

3(H) v C =

3

104

(F)

a. Tnh tng tr ca on mch, cng dng in hiu dng trong mch.b. Vit biu thc cng dng in qua mch.c. Vit biu thc hiu in th hai on mch uMN v uMB.d. hiu in th v cng dng in trong on mch cng pha th t in phi c in dung l

bao nhiu?Bi 7: Cho on mch in xoay chiu nh hnh v:Hai u on mch AB ta duy tr mt hiu in th xoay chiu: R M L N C

u = 60 2 cos 100t (V) A B

Cho bit R = 30, L =2

4,0(H) v C =

8

10 3(F)

a. Vit biu thc cng dng in qua mch.

b. Vit biu thc hiu in th hai on mch uAN v uMB.c. Mc vo hai im M v N mt ampre k c in tr khng ng k th s ch ca ampre k l baonhiu?Bi 8: Cho on mch in xoay chiu nh hnh v: R L

Cun dy thun cm c t cm L =

3(H) A B

in tr thun R = 100, cng dng in trong mch c dng: i = 2cos(100t +6

) (A)

1. Tnh tng tr on mch;2. Vit biu thc cng dng in trong mch.3.Tnh hiu in th hai u in tr R v hai u cun cm L;

4. Vit biu thc hiu in th hai u in tr R v hai u cun cm L.

DNG 3: CNG SUT DNG XOAY CHIU*Biu thc tnh cng sut dng xoay chiu: P = UIcos = RI2.

* H s cng sut: k = cos =Z

R

Mt s bi ton lin quan n tm i lng cng sut tiu th trn on mch khng phnnhnh RLC c cc tr:

Bi ton 1: Tm L, C cng sut t gi tr cc i.

Trang 14


15/34


Phng php: Vit biu thc cng sut P = RI2 = 2CL

2

2

2

2

)ZZ(R

RU

Z

RU

+= ;

Khi : P -> Pmax Z -> Zmin = R ZL = ZC: Xy ra hin tng cng hng in.T ta suy ra gi tr L, C cn tm.

=> Pmax =R

U 2

Bi ton 2: Tm R cng sut tiu th trn on mch RLC t gi tr cc i:

Phng php: Vit biu thc cng sut P = RI2 = yU

)R

ZZ(R

U

Z

RU 2

2CL

2

2

2

=

+

=;

Khi : P -> Pmax y -> ymin

S dng bt ng thc Cauchy: y = R + CL2

CL ZZ2R

ZZ

ymin = CL ZZ2 R = CL ZZ

Khi cng sut tiu th cc i ca mch l: Pmax =CL

22

min

2

ZZ

U

R2

U

y

U

==

Bi 9: Cho mch in xoay chiu nh hnh v:in tr thun R = 100 3 ; t in c in dung C. R CDuy tr hiu in th hai u on mch: A Bu = 200 2 cos100t (V) th cng dng in hiu dngtrong mch l 1A.

1. Xc nh gi tr in dung C ca t in;2. Vit biu thc tc thi cng dng in qua mch v hiu in th tc thi hai u mi dng c

inj;3. Tnh cng sut tiu th ca on mch.

Bi 10: Mt on mch in xoay chiu RLC c in tr R = 50, C =

410.2(F), L =

1(H). Hiu in

th t vo hai u on mch c dng: u = 100 2 cos100t (V) .1. Vit biu thc cng dng in tc thi qua mch;2. Vit biu thc hiu in th hai u mi dng c in.3. Tnh cng sut tiu th ca on mch v h s cng sut ca on mch.4. Gi nguyn cun cm v in tr, thay t in c in dung C bng t in c in dung C th

cng sut tiu th ca on mch t gi tr cc i. Xc nh gi tr C v cng sut cc i .

Bi 11: Cho on mch RLC ni tip, in tr R = 50, cun dy thun cm c t cm L =

3 H. Biu

thc cng dng in qua mch l i = 2 2 cos(100t) (A) v nhanh pha hn hiu in th hai u on

mch l3

(rad).

1. Tnh in dung C ca t in;2. Tnh cng sut tiu th ca on mch v h s cng sut ca on mch;3. Vit biu thc tc thi hiu in th hai u mi dng c in v hai u on mch.4. Gi nguyn t in v cun dy, thay i in tr R bng in tr R th cng sut tiu th trn

on mch t gi tr cc i. Tnh gi tr R v cng sut cc i .

Trang 15


16/34


Bi 12: Mt on mch in xoay chiu gm mt cun dy thun cm c t cm L =

1H, mt t in c

in dung C =

4

103

F v mt bin tr R mc ni tip vi nhau. Hai u on mch ta duy tr mt hiu in

th xoay chiu u = 120 2 cos100t (V) .1. iu chnh bin tr cng sut tiu th trn on mch c gi tr 84,84W 260 W. Tnh gi tr

tng ng ca in tr R.

2. Xc nh in tr R cng sut tiu th trn on mch t gi tr cc i. Tnh cng sut cc iny.Bi 13: Mt mch in xoay chiu khng phn nhnh RLC c in tr thun R = 100 , cun dy c t

cm L = 0,636H

2

H v t in c in dung C. Hiu in th hiu dng hai u on mch l 200V v tn

s 50Hz.

1.Bit hiu in th hai u on mch nhanh pha hn cng dng in trong mch l4

, tnh gi

tr in dung C ca t in.2. Tnh cng sut tiu th ca on mch.

3. Ly pha ban u ca hiu in th hai u on mch l 4

(rad), vit biu thc cng dngin trong mch v biu thc hiu in th hai u mi dng c.Bi 14: Cho mt on mch in RLC c R = 100, mt t in c in dung C = 31,8F, cun dy thuncm c t cm L c th thay i c. Hai u on mch ta duy tr mt hiu in th xoay chiu: u = 2002 cos100t (V) .

1. Xc nh gi tr t cm L ca cun dy h s cng sut tiu th on mch c gi tr ln nht.Tnh cng sut tiu th ca on mch trong trng hp ny.

2. Xc nh gi tr t cm ca cun dy cng sut tiu th ca on mch l 100W. Vit biuthc cng dng in qua mch trong trng hp ny.Bi 15: Mt cun cm c in tr thun r = 10, t cm L = 0,159H mc ni tip vi mt bin tr R v

mt t in c in dung CV bin thin. Hai u on mch duy tr mt hiu in th xoay chiu u =200cos100t (V) .

1. Cho CV = C1 = F1000

. cng sut tiu th trn on mch t gi tr cc i phi cho bin tr

c gi tr l bao nhiu? Tnh cng sut cc i y v vit biu thc cng dng in trong mch trongtrng hp ny.

2. Cho R = R2 = 10. hiu in th hai u cun cm t gi tr cc i phi iu chnh cho CV cgi tr l bao nhiu? Tnh hiu in th cc i y. Vit biu thc hiu in th hai u cun cm trong trnghp ny.

DNG 4: CC BI TON LIN QUAN N MY BIN TH

Cc cng thc lin quan n my bin th:

* Ch khng ti:2

1

2

1

n

n

U

U=

* Ch c ti: Cun th cp ni vi ti tiu th in nng:

U1I1 U2I2 =>1

2

2

1

I

I

U

U

Trang 16


17/34


* Cng sut hao ph trn ng dy ti in2

22

U

PRRIP ==

* gim th trn ng dy ti in: U = IR

* Hiu sut my bin th: H =11

22

IU

IU

* Hiu sut ti in: H =P

PP

BI TP P DNG

Bi 1: Mt my bin th, cun s cp c 1100 vng, cun th cp c 50 vng. Cun th cp c mc vomch in gm in tr thun R, cun dy c t cm L v t in c in dung C mc ni tip vi nhau.Bit tn s ca dng in l 50Hz. Hiu in th hai u cun s cp l 220V, cng dng in hiu dng

qua cun s cp l 0,032A =44

2A, cng sut tiu th ca mch th cp l 5W, in dung ca t in l C =

212F =15

104

F . Tnh gi tr in tr R v t cm L ca cun dy, bit hiu sut ca my bng 1.

Bi gii:

*Hiu in th hai u cun th cp: U2 =1

2

nn U1 = 10V.

* Cng dng in trong mch cun th cp: I2 =2

1

U

UI1 =

2

2A.

* in tr trong mch cun th cp: P = R 22

22

I

PRI ==> = 10;

*Tng tr ca mch th cp: Z2 =2

2

I

U= 10 2

* Gii ra ta tm c hai gi tr ca L l: 0,08H v 0,16H.Bi 2: Mt my pht in cung cch mch ngoi mt cng sut P 1 = 2MW, hiu in th gia hai cc camy pht l U1 = 2000V.

1. Tnh cng dng in hiu dng do my cung cp, bit hiu in th cng pha vi cng dng in.

2. Dng in c a vo cun s cp ca mt my bin th c hiu sut H = 97,5%. Cun s cp cN1 = 160 vng, cun th cp c N2= 1200 vng. Dng in cun th cp c dn n ni tiu th bng dydn c in tr R = 10. Tnh hiu in th, cng sut ni tiu th v hiu sut ti in.

Bi gii:

1. Cng dng in hiu dng do my pht cung cp: I1 =1

1

U

P= 1000A.

2. Hiu in th, cng sut, hiu sut ti in:

+ Hiu in th gia hai u cun th cp: U2=1

2

NN U1 = 15000V.

+ Cng dng in trong cun th cp: I2 = H2

11

U

IU= 130A;

+ gim in th trn ng dy: U = I2R = 1300V;+ Hiu in th ni tiu th: U3 = U2 - U = 13700V;+ Cng sut dng in ti ni tiu th: P3 = U3I3 = 1781000W.

Trang 17


18/34


+ Hiu sut ti in: Ht =1

3

P

P= 89%

Bi 3: Mt my bin th, cun s cp c 6250 vng v cun th cp c 1250 vng. Bit hiu sut ca mybin th l 96%. My nhn cng sut t 10kW cun s cp.

1. Tnh hiu in th cun th cp, bit hiu in th hai u cun s cp l 1000V (bit hiu sutkhng nh hng n hiu in th).

2. Tnh cng sut nhn c cun th cp v cng hiu dng trong mch th cp. Bit h s

cng sut ca mch th cp l 0,8.3. Bit h s t cm tng cng ca mch th cp l 0,2H. Tm in tr ca mch th cp. Cho bit tn

s dng in l 50Hz.Bi gii:

1. Tnh hiu in th hai u cun th cp: U2=1

2

N

NU1 = 200V

2. Tnh cng sut tiu th cun th cp v cng dng in cun th cp.+ Cng sut mch th cp: P2 = H.P1 = 9600W.

+ Mt khc ta c: P2 = U2I2cos2 => I2 =22

2

cosU

P

= 60A.

3. Tm in tr mch th cp: Ta c k = 2L

2 ZR

RZR

+= => R = 83,7.

BI TP T GII

Bi 4: Mt trm pht in truyn i mt cng sut P 1 = 100kW trn dy in c in tr R = 8 . Bit hiuin th t trm pht in chuyn i l U1 = 1000V.

1. Tnh hiu sut ti in.2. Tnh li cu trn nu trm pht in c ni vi my bin th c h s bin th k = 0,1. Coi hiu

sut ca my bin th l H = 1 (cho biu thc ca h s bin th l k =2

1

n

n).

Bi 5: Mt my bin th, cun s cp c n 1 = 1000 vng, cun th cp c n2 = 100 vng. Cun th cp cmc vo mt mch in gm in tr thun R = 12 , cun dy thun cm c t cm l L=0,016H v tin c in dung l C = 320F . Bit tn s ca dng in l f = 50Hz, hiu in th hai u cun s cp lU1 = 117V. Hiu sut ca my bin th l Hm = 0,95 v gi thit rng hiu sut ny ch nh hng n cng

dng in. Tnh cng dng in hiu dng trong cun s cp. Ly

1= 0,32.

Bi 6: Cho on mch in xoay chiu nh hnh v: R L

Cun dy thun cm c t cm L =

3 (H) A B

in tr thun R = 100, cng dng in trong mch c dng: i = 2cos(100 + 6

) (A)1. Tnh tng tr on mch;2. Vit biu thc cng dng in trong mch.3.Tnh hiu in th hai u in tr R v hai u cun cm L;4. Vit biu thc hiu in th hai u in tr R v hai u cun cm L.

Lu :1. on mch RLC c R thay i:

Trang 18


19/34


* Khi R=ZL-ZC th2 2

ax 2 2M L C

U U

Z Z R= =

P

* Khi R=R1 hoc R=R2 th P c cng gi tr. Ta c2

2

1 2 1 2; ( )

L C

UR R R R Z Z+ = =

P

V khi 1 2R R R= th2

ax

1 22M

U

R R=P

* Trng hp cun dy c in tr R0 (hnh v)Khi

2 2

0 ax

02 2( )L C M

L C

U UR Z Z R

Z Z R R= = =

+P

Khi2 2

2 2

0 ax2 2

00 0

( )2( )2 ( ) 2

L C RM

L C

U UR R Z Z

R RR Z Z R= + = =

++ +P

2. on mch RLC c L thay i:

* Khi2

1L

C= th IMax URmax; PMax cn ULCMinLu : L v C mc lin tip nhau

* Khi

2 2

CL

C

R Z

Z Z

+= th

2 2

axC

LM

U R ZU R

+= v

2 2 2 2 2 2

ax ax ax; 0LM R C LM C LMU U U U U U U U = + + =

* Vi L = L1 hoc L = L2 th UL c cng gi tr th ULmax khi1 2

1 2

1 2

21 1 1 1( )

2L L L

L LL

Z Z Z L L= + =

+

* Khi2 24

2C C

L

Z R ZZ

+ += th ax 2 2

2 R

4RLM

C C

UU

R Z Z=

+ Lu : R v L mc lin tip nhau

3. on mch RLC c C thay i:

* Khi2

1C

L= th IMax URmax; PMax cn ULCMinLu : L v C mc lin tip nhau

* Khi

2 2

LC

L

R ZZ Z

+= th

2 2

axL

CM

U R ZU R+= v

2 2 2 2 2 2

ax ax ax; 0CM R L CM L CM U U U U U U U U = + + =

* Khi C = C1 hoc C = C2 th UC c cng gi tr th UCmax khi1 2

1 21 1 1 1

( )2 2C C C

C CC

Z Z Z

+= + =

* Khi2 24

2L L

C

Z R ZZ

+ += th ax 2 2

2 R

4RCM

L L

UU

R Z Z=

+ Lu : R v C mc lin tip nhau

4. Mch RLC c thay i:

* Khi1

LC= th IMax URmax; PMax cn ULCMinLu : L v C mc lin tip nhau

* Khi 21 1

2

C L RC

=

th ax

2 2

2 .

4LM

U LU

R LC R C=

* Khi21

2

L R

L C= th ax

2 2

2 .

4CM

U LU

R LC R C=

* Vi = 1 hoc = 2 th I hoc P hoc URc cng mt gi tr th IMax hoc PMax hoc URMax khi

1 2 = tn s 1 2f f f=

Trang 19

A

CR L,R0


20/34


5. Hai on mch AM gm R1L1C1 ni tip v on mch MB gm R2L2C2 ni tip mc ni tip vi nhau cUAB = UAM + UMB uAB; uAMv uMBcng pha tanuAB = tanuAM= tanuMB6. Hai on mch R1L1C1 v R2L2C2 cng u hoc cng i c pha lch nhau

Vi 1 111

tanL CZ Z

R

= v 2 22

2

tan L CZ Z

R

= (gi s 1 > 2)

C 1 2 = 1 2

1 2

tan tantan

1 tan tan

=

+

Trng hp c bit = /2 (vung pha nhau) th tan1tan2 = -1.VD: * Mch in hnh 1 c uAB v uAM lch pha nhau

y 2 on mch AB v AM c cng i v uAB chm pha hn uAM

AMAB = tan tan

tan1 tan tan

=

+AM AB

AM AB

Nu uAB vung pha vi uAM th tan tan =-1 1L CLAM ABZ ZZ

R R

=

* Mch in hnh 2: Khi C = C1 v C = C2 (gi s C1 > C2) th i1 v i2 lch pha nhau y hai on mch RLC1 v RLC2 c cng uAB

Gi 1 v 2 l lch pha ca uAB so vi i1 v i2th c 1 > 21 - 2 = Nu I1 = I2 th 1 = -2 = /2

Nu I1 I2 th tnh1 2

1 2

tan tantan

1 tan tan

=

+

-----------------------------------------------

CHNG V. SNG NH SNG

DNG 1: GIAO THOA NH SNG N SC

+ Cng thc tnh khong vn: i =a

D;

+ Bc sng ca nh sng n sc dng lm th nghim:D

ai= ;

+ V tr vn sng: x = ki = ka

D

- Nu k = 0: Ta c vn sng trung tm;

- Nu k = 1: Ta c vn sng bc 1;- Nu k = 2: Ta c vn sng bc 2

+ V tr vn ti: x = (k + 0,5)i = (k + 0,5)a

D

- Nu k = 0: vn ti th nht;- Nu k = 1: Vn ti th hai.Lu : Khi gii cc bi tp v giao thoa sng nh sng, cc i lng D,a,i,x phi cng n v.

BI TP P DNG

Trang 20

R L CMA

Hnh 1

R L CMA

Hnh 2


21/34


Bi 1: Hai khi Yong S1, S2 cch nhau 1mm, ngun sng n sc dng lm th nghim c bc sng =0,6mTnh khong cch gia hai vn sng hoc hai vn ti k nhau, bit rng khong cch t hai khe n mn l2m.Bi 2: Trong th nghim Yong v giao thoa nh sng, khong cch gia hai khe l 1mm, khong cch t haikhe n mn l 1m. Bc sng nh sng dng lm th nghim l =0,6m.

1. Tnh hiu quang l t hai ngun S1 v S2 n im M trn mn cch vn trung tm 1,5cm.2. Tnh khong vn (khong cch gia hai vn sng hoc hai vn ti k nhau).

Bi 3: Hai khe Young cch nhau 0,5mm. Ngun sng cch u cc khe pht ra bc x n sc c bc sng=0,5m. Vn giao thoa hng c trn mn E cch cc khe 2m. Tm khong vn i?Bi 4: Quan st giao thoa trn mn E, ngi ta o c khong cch gia hai vn sng lin tip l 1,5mm.Khong cch gia hai khe n mn l 2m, khong cch gia hai khe l 1mm. Xc nh bc sng ca bc xn sc dng lm th nghim.Bi 5:Ngi ta m c trn mn 12 vn sng tri di trn b rng l 13,2mm. Tnh khong vn ca hintng giao thoa.Bi 6: Trong th nghim Young v hin tng giao thoa nh sng, khong cch gia hai khe l 0,9mm,khong cch t hai khe n mn hng E l 2m. Khong cch t vn sng th nht n vn sng th 11 cngbn so vi vn trung tm l 15mm. Tnh bc sng ca nh sng n sc dng lm th nghim.Bi 7: Trong th nghim Young v hin tng giao thoa nh sng, bc sng dng lm th nghim l

=0,5m, khong cch gia hai khe l 1mm.a. Tm khong cch gia hai khe n mn trn mn ti v tr cch vn trung tm 2,5mm ta c vn

sng bc 5.b. ti l vn sng bc 2 th phi di mn E mt on l bao nhiu? Theo chiu no?

Bi 8: Trong th nghim Young v hin tng giao thoa nh sng, khong cch gia hai khe sng l 0,3mmkhong cch t hai khe n mn l 1m v khong vn o c l 2mm.

1. Tm bc sng dng lm th nghim.2. Xc nh v tr ca vn sng bc 5.3. Xc nh khong cch t vn sng bc 5 v vn ti th 3 nm hai bn so vi vn trung tm.

Bi 9: Trong th nghim Young v hin tng giao thoa nh sng, khong cch gia hai khe l 2mm, khongcch t hai khe n mn l 1m, bc sng ca nh sng n sc dng lm th nghim =0,5m.

1. Tnh khong vn i ca hin tng giao thoa nh sng.2. Xc nh v tr ca vn sng bc hai v vn ti th nm. V tnh khong cch gia chng trong hai

trng hp:a. Hai vn cng bn so vi vn trung tm.b. Hai vn hai pha so vi vn trung tm.

DNG 2: GIAO THOA TRNG - S VN GIAO THOA

* Khong vn ca bc x n sc: i =a

D;

* Xc nh b rng na giao thoa trng: l n = 2k.

+Nu l = (k+0,5)i: Vn ngoi cng l vn ti th k + 1, s=2

L

+ Nu l = ki: th vn ngoi cng l vn bc k, s vn sng v vn ti quan st c trn giao thoatrng l:

- S vn sng l: n = 2k+1;- S vn ti l : vn sng, vn ti quan st c trn giao thoa trng l:

+ S vn ti l: n = 2(k+1);+ S vn sng l : n = 2k + 1.

Trang 21


22/34


Lu : S vn sng trn giao thoa trng l s l, s vn ti trn giao thoa trng l s chn;* Xc nh s vn sng, vn ti trong vng giao thoa (trng giao thoa) c b rng L (i xng qua vn trungtm)

+ S vn sng (l s l): 2 12SL

Ni

= +

+ S vn ti (l s chn): 2 0,52tL

Ni

= +Trong [x] l phn nguyn ca x. V d: [6] = 6; [5,05] = 5; [7,99] = 7

* Xc nh s vn sng, vn ti gia hai im M, N c to x1, x2 (gi s x1 < x2)+ Vn sng: x1 < ki < x2+ Vn ti: x1 < (k+0,5)i < x2

S gi tr k Z l s vn sng (vn ti) cn tmLu : M v N cng pha vi vn trung tm th x1 v x2 cng du.

M v N khc pha vi vn trung tm th x1 v x2 khc du.* Xc nh khong vn itrong khong c b rng L. Bit trong khong L c n vn sng.

+ Nu 2 u l hai vn sng th:1

Li

n=

-

+ Nu 2 u l hai vn ti th: Lin

=

+ Nu mt u l vn sng cn mt u l vn ti th:0,5

Li

n=

-

* S trng nhau ca cc bc x 1, 2 ... (khong vn tng ng l i1, i2 ...)+ Trng nhau ca vn sng: xs = k1i1 = k2i2 = ... k11 = k22 = ...+ Trng nhau ca vn ti: xt = (k1 + 0,5)i1 = (k2 + 0,5)i2 = ... (k1 + 0,5)1 = (k2 + 0,5)2 = ...

Lu : V tr c mu cng mu vi vn sng trung tm l v tr trng nhau ca tt c cc vn sng ca cc bcx.

BI TP P DNG

Bi 10: Trong th nghim Young v hin tng giao thoa nh sng, hai khe cch nhau 0,5mm vi nh sngn sc dng lm th nghim c bc sng = 0,5m v quan st hin tng giao thoa trn mn E cch haikhe 2m.

1. Ti cc im M1 v M2 cch vn trung tm ln lt 7mm v 10mm thu c vn g? Bc (th) my?2. Bit chiu rng ca vng giao thoa trng trn mn l 26mm, tnh s vn sng, vn ti quan st

c?

3. Nu thc hin giao thoa trong nc c chit sut n =3

4th c hin tng g xy ra? Tnh khong

vn trong trng hp ny?

Bi 11: Trong th nghim Young v hin tng giao thoa nh sng, hai khe cch nhau 1,5mm, hai khe sngcch mn l 3m, ngi ta o c khong cch gia vn sng bc 2 n vn sng bc 5 cng nm v mt phaso vi vn trung tm la 3mm.

1. Tnh bc sng ca bc x dng lm th nghim;2. Tnh khong cch gia vn sng bc 3 v vn ti th 8 nm v hai pha so vi vn trung tm.3. Tm s vn sng v vn ti quan st c trn giao thoa trng c b rng 11mm.

Bi 12: Trong th nghim Young v hin tng giao thoa nh sng, hai khe cch nhau 2mm, hai khe sngcch mn l 3m, bc x n sc dng lm th nghim c bc sng = 0,5m.

1. Xc nh s vn sng, vn ti trn b rng 3cm ca giao thoa trng;

Trang 22


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2. Thay nh sng n sc bng nh sng n sc = 0,6m th khong vn tng hay gim, tm svn sng, vn ti trn giao thoa trng 3cm nh trn.

DNG 3: GIAO THOA NG THI HAI BC XXC NH V TR TRNG NHAU CA HAI VN SNG, HAI VN TI

* Khong vn: i =a

D; i =

a

D'

* V tr vn sng ca hai bc x: xs () = k1i = k1a

D; xs (') = k2i = k2

a

D';

Hai vn sng trng nhau khi: xs () = xs (')

=> 2212

1 k'

ki

'ik

'

i

'i

k

k

===>

==

Lu : + k1, k2Z;+ Da vo iu kin bi ton (gii hn giao thoa trng) gii hn k1, k2.

BI TP P DNG

Bi 13: Trong th nghim Young v hin tng giao thoa nh sng, ngun sng ng thi pht ra hai bc xn sc c bc sng 1 = 0,5m v 2 = 0,6m. Hai khe cch nhau 1,5mm v cch mn 1,5m. Xc nh v trca vn sng bc 4 ca hai bc x ny (nm cng mt pha so vi vn trung tm). Khong cch gia hai vnny l bao nhiu?Bi 14: Trong th nghim Young v hin tng giao thoa nh sng, khong cch gia hai khe l 1,2mm,khong bc sng ca bc x n sc dng lm th nghim 1 = 0,6m. Trn mn ngi ta m c 16 vnsng tri di trn b rng 18mm.

1. Tnh khong cch t hai khe n mn.2. Thay nh sng n sc c bc sng 2 th vng quan st trn ngi ta m c 21 vn sng. Tnh

2.3. Ti v tr cch vn trung tm 6cm ta thu c vn g, bc (th) my ca hai bc x n sc trn?

Bi 15: Trong th nghim Young v hin tng giao thoa nh sng, ngun sng ng thi pht ra hai bc x1 = 0,6m v 2. Trn mn ngi ta thy vn ti th nm ca h vn ng vi bc x 1 trung vi vn sngbc 5 ca bc x 2. Tnh bc sng 2 dng lm th nghim.Bi 16: Trong th nghim Young v hin tng giao thoa nh sng, bc x dng lm th nghim c bc sng= 0,55m, khong vn hng c trn mn l i.

1. Khi thay bc x n sc c bc sng bng bc x n sc c bc sng , ngi ta thy khongvn i = 1,2i. Tnh .

2. Nu chiu ng thi hai bc x n sc trn, xc nh cc v tr m vn sng ca hai bc x trngnhau.Bi 17: Trong th nghim Young v hin tng giao thoa nh sng, bc x n sc dng lm th nghim c

bc sng = 0,6m. Hai khe cch nhau 2mm, hai khe cch mn 2m. Tnh s vn sng, vn ti quan st ctrn giao thoa trng rng 25,8mm.Bi 18: Trong th nghim Young v hin tng giao thoa nh sng, hai khe cch nhau 2mm, hai khe cch mn2m.

1. Ngi ta chiu ti hai khe ng thi hai bc x n sc c bc sng ln lt l 1 = 0,45m2 = 0,5m. Xc nh nhng v tr m hai vn sng ca hai bc x trng nhau.

2. Chiu ti hai khe mt thnh phn n sc th ba c bc sng 3 = 0,6m. nh v tr m c ba vntrng nhau trn mn.

Trang 23


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Bi 19: Trong th nghim Young v hin tng giao thoa nh sng, khong cch gia hai khe l 1mm, khongcch gia hai khe n mn l 2m. nh sng n sc dng lm th nghim c bc sng 1=0,66m. Bit brng ca giao thoa trng l 13,2mm.

1. Tnh khong vn, s vn sng v vn ti quan st c trn giao thoa trng.2. Nu chiu ng thi hai bc x 1,2 th vn sng bc 3 ca bc x 2 trng vi vn sng th hai ca

bc x 1. Tnh 2 .Bi 20: Trong th nghim Young v hin tng giao thoa nh sng, khong cch gia hai khe l 2mm, bc x

n sc dng lm th nghim c bc sng 1 = 0,6m v trn mn, vn sng th 5 cch vn trung tm l3mm.

1. Tnh khong cch t hai khe n mn.2. Tnh khong cch t vn ti th 3 n vn sng th 4 nm hai bn so vi vn trung tm.3. Nu chiu vo hai khe ng thi hai bc x 1 v 2 th ta thy vn sng bc 5 ca bc x 1 trng

vi vn ti th 5 ca bc x 2 . Tm 2? Bc x 2 nm trong vng no ca thang sng in t?Bi 21: Trong th nghim Young v hin tng giao thoa nh sng, hai khe cch nhau 2mm, hai khe cch mn2m, bc x n sc dng lm th nghim c bc sng 1 = 0,6m.

1. Tnh khong vn i1.2. Ngi ta ng thi chiu hai bc x n sc 1 v 2= 0,4m th vn sng bc 3 ca bc x 1 trng

vi vn no ca bc x 2?3. Tnh khong cch t vn sng bc 2 ca bc x 1 n vn ti th 6 ca bc x 2, bit rng hai vnny cng nm mt pha so vi vn trung tm.Bi 22: Trong th nghim Young v hin tng giao thoa nh sng, khong cch gia hai khe sng l 1mm,khong cch t hai khe n mn l 2m.

1. Chiu vo hai khe sng bc x n sc c bc sng 1 = 0,5m. Tnh khong cch gia vn sngbc 3 v vn sng bc 5 nm hai bn so vi vn trung tm.

2. Nu chiu vo hai khe bc x n sc c bc sng 2 th ti im M cch vn trung tm 4,8mm cvn sng bc 4, tnh bc sng 2 ?

3. Nu chiu ng thi hai bc x n sc c bc sng 1, 2 th trn mn c nhng v tr no trngnhau ca cc vn sng ca hai bc x, bit b rng ca giao thoa trng l 24mm.Bi 23: Trong th nghim Young v hin tng giao thoa nh sng, hai khe cch nhau 2mm, trn mn giaothoa xut hin vn sng bc 5 cch vn trung tm 4mm. Bit khong cch t hai khe n mn l 3m.

1. Tnh bc sng 1 dng lm th nghim.2. Nu chiu vo hai khe hai bc x c bc sng 2 th ngi ta o c trong khong 4,5mm c 6

vn sng lin tip. Tm bc sng 2 v b rng quang ph bc 1 ng vi bc x 2 .Bi 24: Trong th nghim Young v hin tng giao thoa nh sng, hai khe cch nhau 1mm, hai khe cch mn1m.

1. Ban u dng bc x n sc c bc sng 1 th khong cch gia vn sng bc 3 n vn ti th 5 cng mt pha so vi vn trung tm l 1,5mm. Tm bc sng 1 .

2. Dng ng thi hai bc x n sc c bc sng 1, 2 th vn ti th hai ca bc x 1 trng v

vn sng bc 3 ca bc x 2 .Tnh 2.3. Xc nh v tr trng nhau hai vn sng ca bc x nm cng mt bn gn vn trung tm nht.

Bi 25: Trong th nghim Young v hin tng giao thoa nh sng, hai khe cch nhau 2mm v hai khe cchmn 2m.

1. Vi nh sng n sc c bc sng 1 = 0,45m. Tnh khong cch gia vn sng bc 4 v vnsng bc 5 nm v hai pha so vi vn trung tm.

2. Vi snh sng n sc c bc sng 2. Bit b rng 5 khong vn lin tip l 30mm. Tnh bcsng 2. Ti v tr cch vn trung tm 9mm l vn sng hay vn ti? Bc (th) my?

Trang 24


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3. ng thi chiu vo hai khe hai bc x 1, 2 tm v tr gn nhau nht so vi vn trung tm m vnsng ca hai bc x trng nhau.

DNG 4: GIAO THOA VI NH SNG TRNG

* Khong vn: i =a

D;

* V tr vn sng ca bc x: xs () = ki = kaD ;

*V tr vn ti ca bc x: : xt () = ki = ka

D;

* nh sng trng c min bc sng: 0,38m 0,75mLu : + Nhiu khi cho min bc sng ca nh sng trng: 0,4m 0,76m.

PHNG PHP GII MT S DNG TON THNG GP

Bi ton 1 : Tm s bc x cho vn sng ti v tr cch vn trung tm x.Phng php:

+ Ti v tr x cho vn sng: x = ka

D => =kD

ax

+ Dc vo min bc sng ca nh sng trng: 0,38m 0,75m

=> 0,38m kD

ax 0,75m =>

D38,0

axk

D75,0

ax , kZ

Tm k t h bt phng trnh trn, c bao nhiu k th c by nhiu bc x cho vn sng ti v tr .

Thay gi tr k vo biu thc =kD

ax, ta tm c bc sng ca cc bc x.

Bi ton 2 : Tm s bc x cho vn ti (b tt) ti v tr cch vn trung tm x.Phng php:

+ Ti v tr x cho vn sng: x = (k + 0,5) aD

=> = D)5,0k(ax

+

+ Dc vo min bc sng ca nh sng trng: 0,38m 0,75m

=> 0,38m D)5,0k(

ax

+ 0,75m => 5,0

D38,0

axk5,0

D75,0

ax , kZ

Tm k t h bt phng trnh trn, c bao nhiu k th c by nhiu bc x cho vn ti (b tt) ti v tr.

Thay gi tr k vo biu thc =D)5,0k(

ax

+, ta tm c bc sng ca cc bc x.

Bi ton 3: Tm b rng quang ph bc k ca nh sng trng:

xk= xk() - xk(t) = kaD ( - t) = kx1

- Khong cch di nht v ngn nht gia vn sng v vn ti cng bc k:

[k ( 0,5) ]

Min t

Dx k

a =

ax[k ( 0,5) ]

M t

Dx k

a = + Khi vn sng v vn ti nm khc pha i vi vn trung tm.

ax[k ( 0,5) ]M t

Dx k

a = Khi vn sng v vn ti nm cng pha i vi vn trung tm

Trang 25


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BI TP P DNG

Bi 26: Trong th nghim Young v hin tng giao thoa nh sng, khong cch gia hai khe l 2mm, khongcch t hai khe n mn l 3,2m. Trn mn ngi ta xc nh c v tr ca vn sng bc 14 cch vn trungtm 11,2mm.

1. Tnh bc sng ca bc x n sc dng lm th nghim.2. Th nghim vi nh sng trng, th ti im M cch vn trung tm 6,72mm c nhng bc x no b

tt?Bi 27: Trong th nghim Young v hin tng giao thoa nh sng, khong cch t hai khe n mn l 2m.Bc x n sc dng lm th nghim c bc sng =0,5m, trn mn ngi ta o c khong cch t vnsng th hai n vn ti th 4 (nm cng mt pha so vi vn trung tm) l 0,75mm.

1. Tnh khong vn i v khong cch gia hai khe sng.2. Th nghim vi nh sng trng, th ti im cch vn trung tm 4mm c vn sng ca nhng bc x

n sc no?Bi 28: Trong th nghim Young v hin tng giao thoa nh sng trng, hai khe cch nhau 2mm. khongcch t hai khe n mn l 1,6m. Bit bc sng ca nh sng trng bin thin lin tc t 0,4 m n 0,75mTnh b rng ca quang ph bc nht v quang ph bc 3 ca hin tng giao thoa (ch xt n cc vn nm cng mt pha so vi vn trung tm).

Bi 29: Trong th nghim Young v hin tng giao thoa nh sng trng, khong cch gia hai khe l 2mmkhong cch gia hai khe n mn l 1,6m. Hy xc nh bc sng ca cc bc x b tt ti v tr cch vntrung tm 3,5mm.Bi 30: Trong th nghim Young v hin tng giao thoa nh sng trng, khong cch gia hai khe l 0,5mm,khong cch t hai khe n mn l 2m, cho bit = 0,4m, t = 0,75m.

1. Tnh b rng quang ph bc 1 v quang ph bc 3.2. Xc nh cc bc sng ca cc bc x b tt ti v tr cch vn trung tm 7,2mm.

Bi 31: Trong th nghim Young v hin tng giao thoa nh sng, ngun sng pht ra cc bc x c minbc sng 0,4m n 0,8m.Hai khe cch nhau mt on 1mm v cch mn 1,5m. Ti im M cch vntrung tm 3mm c bao nhiu bc x n sc c cng cc i. Tm bc sng ca cc bc x n sctrn?

Bi 32: Trong th nghim Young v hin tng giao thoa nh sng, khong cch gia hai khe l 2mmkhong cch t hai khe n mn l 2m.

1. Tin hnh th nghim ng thi vi hai bc x n sc c bc sng 1 = 0,5m, 2 = 0,6m. Xcnh nhng v tr m vn sng ca hai bc x trng nhau.

2. Tin hnh th nghim vi nh sng trng (c bc sng t 0,4m n 0,75m).a. Tnh b rng quang ph bc nht v quang ph bc hai ca hin tng giao thoa;b. Xc nh bc sng ca cc bc x b tt ti v tr M cch vn trung tm 3,3mm.

Bi 33: Trong th nghim Young v hin tng giao thoa nh sng, hai khe sng cch nhau 0,2mm, hai khecch mn 1m.

1. Bit khong cch gia 10 vn sng k nhau l 2,7mm. Tnh bc sng ca bc x n sc do ngunsng pht ra.

2. Thay nh sng n sc bng nh sng trng, ti im cch vn trung tm 2,7mm c bao nhiu bcx b tt? xc nh bc sng ca cc bc x trn.Bi 34: Trong th nghim Young v hin tng giao thoa nh sng, hai khe cch nhau 1mm, hai khe cch mn2m.1. Tin hnh th nghim vi bc x n sc c bc sng = 0,656m. Tnh khong vn thu c trn mn.

2. Tin hnh th nghim vi bc x mu lc, bit b rng ca 10 khong vn lin tip l 1,09cm. Tnhbc sng ca bc x mu lc dng lm th nghim.

3. Tin hnh th nghim vi nh sng trng (c bc sng t 0,4m n 0,7m). Xc nh nhng bcx b tt ti v tr cch vn trung tm 1,2cm.

Trang 26


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DNG 5: GIAO THOA TRN LNG LNG KNH FRESNEL

* H lng knh Fresnel gm hai lng knh hon tonging ht nhau, chung cnh y v c gc chit quang A nh.

* Ngun sng S qua hai lng knh cho 2 nh S 1, S2

ng vai tr l hai ngun sng kt hp.a = S1S2 = 2SO = 2(n 1)A.SO

BI TP P DNGBi 35: Mt lng lng knh Fresnel gm hai lng knh ging nhau, cc y st nhau. Lng knh c gc chitquang A v chit sut n. t mt khe sng S song song vi cnh ca lng lng knh. Khong cch t khe Sn lng lng knh l d1. mn E cch lng lng knh l d2.

1. Chng minh rng lng lng knh Fresnel tng ng vi h giao thoa khe Young . V trnggiao thoa v tnh b rng giao thoa trng trn mn E.

2. Hy thit lp h thc xc nh hiu quang trnh ti mt im M trn mn E cch tm O ca mn mt

on x theo A, n, d1 v d2.Tnh khong vn trong trng hp A = 3.10-3rad, n = 1,5, d1 = 50cm, d2 = 1m v = 0,55m.Bi gii:

Bi 36: 1. Chng minh lng lng knh Fresnel tng ng vi h giao thoa khe Young:Chm tia sng xut pht t ngun sng S, sau khi khc x qua lng lng knh th b tch thnh hai

Hai chm tia sng ny coi nh xut pht t S1, S2 l hai nh o ca S. Hai ngun o S1, S2 v cc chm tiasng do chng pht ra i xng nhau qua mt phng cha y ca lng knh. S 1 v S2 ng vai tr l haingun sng kt hp nn s gy ra hin tng giao thoa.

Vy, lng lng knh Fresnel c tc dng tng ng vi khe Young.* B rng giao thoa trng trn mn E:+ Gc lch tia sng qua lng knh: = (n 1)A (vi A rt nh)

+ Khong ch gia hai ngun S1, S2: a = S1S2 = 2d1tan =2d1(n 1)A.=> B rng ca giao thoa trng: l = a

1

2

d

d= 2d2(n 1)A

2. Thit lp h thc tnh hiu quang l ti mt im M trn mn.T cng thc tnh hiu quang l trong hin tng giao thoa:

=D

ax, vi D = d1 + d2 => =

21

2

dd

x)1n(Ad2

+

*Khong vn i c tnh bi cng thc: i =)1n(Ad2

)dd(

a

D

2

21

+=

Thay cc gi tr t gi thit bi ton, ta tm c khong vn i = 0,55mmBi 37: Mt lng lng knh Fresnel c gc chit quang A = 1 o lm bng thu tinh c chit sut n =1,5. Mtngun sng S n sc t trn mt phng y chung ca hai lng knh v cch y lng knh mt on d =25cm.

1. Xem rng nh S1, S2 ca S to bi hai lng knh c v tr c dch i so vi S theo phng vunggc vi mt phng y chung , hy tnh khong cch a = S1S2, cho 1= 3.10-4rad.

2. t mt mn quan st E vung gc vi mt phng y chung ca hai lng knh, cch lng knh mton 2m, ngi ta quan st thy cc vn giao thoa. Tnh khong vn v s vn quan st c trn mn E, bitbc sng nh sng ca ngun sng S dng lm th nghim l = 0,5m. Nu ngun sng S dch chuyn ra

Trang 27

I

S1 A

a S O C

S2 A


28/34


xa lng knh theo phng vung gc vi mn E th khong vn v s vn quan st c trn mn E thay ith no?

----------------------------------------

CHNG VI. LNG T NH SNG

1. Nng lng mt lng t nh sng (ht phtn)2hchf mce

l= = =

Trong h = 6,625.10-34 Js l hng s Plng.

c = 3.108m/s l vn tc nh sng trong chn khng.

f, l tn s, bc sng ca nh sng (ca bc x).

m l khi lng ca phtn

2. Tia Rnghen (tia X)

Bc sng nh nht ca tia Rnghen

Min

hcE

l =

Trong 22

0 2 2

mvmvE e U= = + l ng nng ca electron khi p vo i catt (i m cc)

U l hiu in th gia ant v catt

v l vn tc electron khi p vo i catt

v0 l vn tc ca electron khi ri catt (thng v0 = 0)

m = 9,1.10-31 kg l khi lng electron

3. Hin tng quang in*Cng thc Anhxtanh2

0 ax

2M

mvhchf Ae

l= = = +

Trong 0

hcA

l= l cng thot ca kim loi dng lm catt

0 l gii hn quang in ca kim loi dng lm catt

v0Max l vn tc ban u ca electron quang in khi thot khi catt

f, l tn s, bc sng ca nh sng kch thch

* dng quang in trit tiu th UAK Uh (Uh < 0), Uh gi l hiu in th hm2

0 ax

2M

h

mveU =

Lu : Trong mt s bi ton ngi ta ly Uh > 0 th l ln.

* Xt vt c lp v in, c in th cc i VMax v khong cch cc i dMax m electron chuyn ng trong

in trng cn c cng E c tnh theo cng thc:

Trang 28


29/34


2

ax 0 ax ax

1

2M M Me V mv e Ed = =

* Vi U l hiu in th gia ant v catt, vA l vn tc cc i ca electron khi p vo ant, vK= v0Max l

vn tc ban u cc i ca electron khi ri catt th:

2 21 1

2 2A Ke U mv mv= -

* Hiu sut lng t (hiu sut quang in) :0

nHn

=

Vi n v n0 l s electron quang in bt khi catt v s phtn p vo catt trong cng mt khong thi

gian t.

Cng sut ca ngun bc x: 0 0 0n n hf n hc

pt t t

e

l= = =

Cng dng quang in bo ho:bh

n eqI

t t= =

bh bh bhI I hf I hc

H p e p e p e

e

l = = =

* Bn knh qu o ca electron khi chuyn ng vi vn tc v trong t trng u B

, = ( ,B)

sin

mvR v

e Ba

a=

r ur

Xt electron va ri khi catt th v = v0Max

Khi sin 1mv

v B Re B

a^ = =r ur

Lu : Hin tng quang in xy ra khi c chiu ng thi nhiu bc x th khi tnh cc i lng: Vn

tc ban u cc i v0Max, hiu in th hm Uh, in th cc i VMax, u c tnh ng vi bc x c

Min (hoc fMax)

4. Tin Bo - Quang ph nguyn t Hir

* Tin Bo

mn m n

mn

hchf E E e

l= = = -

* Bn knh qu o dng th n ca electron trong nguyn t hir:

rn = n2

r0Vi r0 =5,3.10-11m l bn knh Bo ( qu o K)

* Nng lng electron trong nguyn t hir:

2

13,6( )

nE eV

n=- Vi n N*.

* S mc nng lng

- Dy Laiman: Nm trong vng t ngoi

ng vi e chuyn t qu o bn ngoi v qu o K

Trang 29

hfmn

hfmn

nhn phtn pht phtE

m

En

Em

> En


30/34


Lu : Vch di nht LKkhi e chuyn t L K

Vch ngn nht Kkhi e chuyn t K.

- Dy Banme: Mt phn nm trong vng t ngoi, mt phn nm trong vng nh sng nhn thy

ng vi e chuyn t qu o bn ngoi v qu o L

Vng nh sng nhn thy c 4 vch:

Vch H ng vi e: M L

Vch lam H ng vi e: N L

Vch chm Hng vi e: O L

Vch tm H ng vi e: P L

Lu : Vch di nht ML (Vch H)

Vch ngn nht L khi e chuyn t L.

- Dy Pasen: Nm trong vng hng ngoi

ng vi e chuyn t qu o bn ngoi v qu o

M

Lu : Vch di nht NM khi e chuyn t N M.

Vch ngn nht M khi e chuyn t M.

Mi lin h gia cc bc sng v tn s ca cc

vch quang ph ca nguyn t hir:

13 12 23

1 1 1

= + v f13 = f12 +f23 (nh cng

vct)

--------------------------------------------

CHNG VII. HT NHN NGUYN T

1. Hin tng phng x

* S nguyn t cht phng x cn li sau thi gian t

0 0.2 .

t

tTN N N e l-

-= =

* S ht nguyn t b phn r bng s ht nhn con c to thnh v bng s ht ( hoc e- hoc e+) c to

thnh:

0 0(1 )tN N N N e l-D = - = -

* Khi lng cht phng x cn li sau thi gian t

0 0.2 .

t

tTm m m e l-

-= =

Trong : N0, m0 l s nguyn t, khi lng cht phng x ban u

T l chu k bn r

Trang 30

Laiman

K

M

NO

L

P

Banme

Pasen

H

H

H

H

n=1

n=2

n=3

n=4n=5

n=6


31/34


2 0,693ln

T Tl = = l hng s phng x

v T khng ph thuc vo cc tc ng bn ngoi m ch ph thuc bn cht bn trong ca chtphng x.

* Khi lng cht b phng x sau thi gian t

0 0(1 )tm m m m e l-D = - = -

* Phn trm cht phng x b phn r:0

1 tm

em

l-D = -

Phn trm cht phng x cn li:0

2t

tTm

em

l-

-= =

* Khi lng cht mi c to thnh sau thi gian t

1 0 11 1 0

(1 ) (1 )t t

A A

A N ANm A e m e

N N Al l- -D= = - = -

Trong : A, A1 l s khi ca cht phng x ban u v ca cht mi c to thnh

NA = 6,022.10-23 mol-1 l s Avgar.

Lu : Trng hp phng x +, - th A = A1 m1 = m

* phng x H

L i lng c trng cho tnh phng x mnh hay yu ca mt lng cht phng x, o bng s phn r

trong 1 giy.

0 0.2 .

t

tTH H H e Nl l-

-= = =

H0 = N0 l phng x ban u.

n v: Becren (Bq); 1Bq = 1 phn r/giy

Curi (Ci); 1 Ci = 3,7.1010 Bq

Lu : Khi tnh phng x H, H0 (Bq) th chu k phng x T phi i ra n v giy(s).

2. H thc Anhxtanh, ht khi, nng lng lin kt

* H thc Anhxtanh gia khi lng v nng lng

Vt c khi lng m th c nng lng ngh E = m.c2

Vi c = 3.108 m/s l vn tc nh sng trong chn khng.

* ht khi ca ht nhn AZX

m = m0 m

Trong m0 = Zmp + Nmn = Zmp + (A-Z)mn l khi lng cc nucln.m l khi lng ht nhn X.

* Nng lng lin kt E = m.c2 = (m0-m)c2

* Nng lng lin kt ring (l nng lng lin kt tnh cho 1 nucln):E

AD

Lu : Nng lng lin kt ring cng ln th ht nhn cng bn vng.

3. Phn ng ht nhn

Trang 31


32/34


* Phng trnh phn ng: 31 2 41 2 3 41 2 3 4

AA A A

Z Z Z ZX X X X+ +

Trong s cc ht ny c th l ht s cp nh nucln, eletrn, phtn ...

Trng hp c bit l s phng x: X1 X2 + X3X1 l ht nhn m, X2 l ht nhn con, X3 l ht hoc

* Cc nh lut bo ton

+ Bo ton s nucln (s khi): A1 + A2 = A3 + A4+ Bo ton in tch (nguyn t s): Z1 + Z2 = Z3 + Z4

+ Bo ton ng lng:1 2 3 4 1 1 2 2 4 3 4 4

m m m mp p p p hay v v v v+ = + + = +

+ Bo ton nng lng:1 2 3 4X X X X

K K E K K+ +D = +

Trong : E l nng lng phn ng ht nhn

21

2X x xK m v= l ng nng chuyn ng ca ht X

Lu : - Khng c nh lut bo ton khi lng.

- Mi quan h gia ng lng pX v ng nng KX ca ht X l:2

2X X Xp m K=- Khi tnh vn tc v hay ng nng K thng p dng quy tc hnh bnh hnh

V d:1 2

p p p= + bit 1 2,p p=

uur uur

2 2 2

1 2 1 22p p p p p cosj= + +

hay 2 2 21 1 2 2 1 2 1 2( ) ( ) ( ) 2mv m v m v m m v v cosj= + +

hay 1 1 2 2 1 2 1 22mK m K m K m m K K cosj= + +

Tng t khi bit 1 1

,p p=uur ur

hoc 2 2

,p p=uur ur

Trng hp c bit:1 2

p p^ 2 2 21 2p p p= +

Tng t khi1

p p^ hoc2

p p^

v = 0 (p = 0) p1 = p21 1 2 2

2 2 1 1

K v m A

K v m A= =

Tng t v1 = 0 hoc v2 = 0.

* Nng lng phn ng ht nhn: E = (M0 - M)c2

Trong :1 20 X X

M m m= + l tng khi lng cc ht nhn trc phn ng.

3 4X XM m m= + l tng khi lng cc ht nhn sau phn ng.

Lu : - Nu M0 > M th phn ng to nng lng E di dng ng nng ca cc ht X3, X4 hoc phtn .

Cc ht sinh ra c ht khi ln hn nn bn vng hn.

Nu M0 < M th phn ng thu nng lng |E| di dng ng nng ca cc ht X1, X2 hoc phtn .

Cc ht sinh ra c ht khi nh hn nn km bn vng.

* Trong phn ng ht nhn 31 2 41 2 3 41 2 3 4

AA A A

Z Z Z ZX X X X+ +

Cc ht nhn X1, X2, X3, X4 c:

Trang 32

p

1puur

2puur


33/34


Nng lng lin kt ring tng ng l1

,2

,3

,4

.

Nng lng lin kt tng ng l E1, E2, E3, E4

ht khi tng ng l m1, m2, m3, m4

Nng lng ca phn ng ht nhn

E = A3 3 +A4 4 - A1 1 - A2 2

E = E3 + E4 E1 E2E = (m3 + m4 - m1 - m2)c2

* Quy tc dch chuyn ca s phng x

+ Phng x ( 42He ):4 4

2 2

A A

Z ZX He Y-

- +

So vi ht nhn m, ht nhn con li 2 trong bng tun hon v c s khi gim 4 n v.

+ Phng x - ( 10e- ): 01 1

A A

Z ZX e Y

- + +

So vi ht nhn m, ht nhn con tin 1 trong bng tun hon v c cng s khi.

Thc cht ca phng x - l mt ht ntrn bin thnh mt ht prtn, mt ht electrn v mt ht

ntrin:n p e v- + +

Lu : - Bn cht (thc cht) ca tia phng x - l ht electrn (e-)

- Ht ntrin (v) khng mang in, khng khi lng (hoc rt nh) chuyn ng vi vn tc ca

nh sng v hu nh khng tng tc vi vt cht.

+ Phng x + ( 10e+ ): 01 1

A A

Z ZX e Y

+ - +

So vi ht nhn m, ht nhn con li 1 trong bng tun hon v c cng s khi.

Thc cht ca phng x + l mt ht prtn bin thnh mt ht ntrn, mt ht pzitrn v mt ht

ntrin:p n e v+ + +

Lu : Bn cht (thc cht) ca tia phng x + l ht pzitrn (e+)

+ Phng x (ht phtn)

Ht nhn con sinh ra trng thi kch thch c mc nng lng E1 chuyn xung mc nng lng E2 ng

thi phng ra mt phtn c nng lng

1 2

hchf E E e

l= = = -

Lu : Trong phng x khng c s bin i ht nhn phng x thng i km theo phng x v .4. Cc hng s v n v thng s dng

* S Avgar: NA = 6,022.1023 mol-1

* n v nng lng: 1eV = 1,6.10-19 J; 1MeV = 1,6.10-13 J

* n v khi lng nguyn t (n v Cacbon): 1u = 1,66055.10 -27kg = 931 MeV/c2

* in tch nguyn t: |e| = 1,6.10-19 C

* Khi lng prtn: mp = 1,0073u

* Khi lng ntrn: mn = 1,0087u Trang 33


34/34


* Khi lng electrn: me = 9,1.10-31kg = 0,0005u

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