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Modern Physics
Q1. What is matter wave? Or What is Debroglie hypothesis?
Ans: 1. According to debroglie hypothesis, any moving particle is associatedwith a wave.
2. The waves associated with a particle is called matter wave/debrogliewave.
3.Wavelength associated with the matter wave is called debrogliewavelength. λ = Or λ =
m= mass of the particle v= velocity of the particle p=mv= momentum of the particle.
Q2. Derive the expression for de-broglie wavelgth.
Ans:
mvh h
p
E = hν=hc/ hv/ –-------1 E = mc2 = mv2 ---------------------2
1=2 hv/ = mv2.
Cancel v. h/ = mv since mv = p.
h/ = p
= h/p
Q3. Expression for debroglie wavelgth in different forms
.
Q4. Write the properties of Matter waves.
Q5: Describe the DavissionGermer expriment for the proof of wavenature of matter Or Describe the DavissionGermer expriment for the exprimental evidenceof matter waves.
Ans: 1. Experimental arrangement consist of filament, Ni crystal, detector.2. Electron beam (electrons) are generated from the filament by passing high currentthrough it. When filament gets heated up, it emits electrons.3. Emitted electrons are accelerated by potential (voltage) applied between filament andthe anode.4. Accelerated electron beam is allowed to fall on the Ni crystal. 5. These electrons are scattered by Ni crystal in all the directions.
6. The detector D measures the number of electrons scattered by the crystal in differentdirections/angles (090 degrees).7. Experiment was performed for different accelerated potentials .8. At 54V accelearting voltage, intensity graph have shown below. It is cleared that at 50degrees, a intense hump (intense peak) is observed. This indicates that electrons arescattered more. This behaviour is not observed for other accelerating voltages.
9. If you calculate wave length associated with electrons at 54 V
= 1.66 Å------------------------1
10. From below figure, we can measure angle of incidence of electron beamon the Ni crystal planes and it is found that 65 degrees.
Or
11. Substitute above values in braggs law of x-ray diffraction
According to braggs law 2dsinθ=nλ
d=0.91 Å, n=1, θ=65o
λ= 1.65 Å ------------------------2
12. from the values of wavelength obtianed by equation 1 (De-broglie) and 2 (braggs) are same/agreed well.
Ф
Ф
At 54 V
Intensity Peak /hump
Ф=50O
13. Therefore, this experiment gave a evidence that electrons exhibit diffraction (wave nature). This implies the existance of matter waves.
Q6: Describe the G P Thomson's expriment for the proof of wave natureof matter Or Describe the G P Thomson's expriment for the exprimental evidence ofmatter waves.
Ans: 1. Experimental arrangment consist of Filament (F), Anode (A), Photograhic plate (P), Thin gold foil (G) (Poly crystalline) and Metal block (B).
2. Electrons are produced from the heated filament (F) and accelerated through highpotentail given to the anode (A).
3. Electron beam passes through a fine hole in metal block (B) and falls on the goldfoil of thickness 0.1 μm. The electron are passing through foils are received on thephotographic plate (P).
4. Metals are poly crystalline in which grains are oriented randomaly and somegrains always satisfies braggs law with respect to the incident electron beam angleand produce the braggs reflection.
5. A concentric ring pattern is produced on the photographic plate like as shown inbelow.
6. X-ray diffraction pattern of powder (poly crystalline) samples is shown in below.
Form above two figures, the diffraction pattern produced by electron beam is similarto the x-ray diffraction pattern produced using x-rays.
Therefore, this experiment provides the evidence for the wave nature of electron.Which implies the existance/evidence of matter waves.
Q7. Explain the terms (a) Wave packet (b) Group velocity (c) Phasevelocity.
Ans:
(a) Wavepacket: It is representation of matter wave associated with aparticle. Which is represnted bleow
Wave packet is the resultant of superpostion of large number of harmonicwaves slightly differ in frequency
(b) Group velocity (Vg): The velocity with which wave packet advances(moving) in the medium is called group velocity. Vg =
dω
dk
(c) Phase velocity (Vp): The individual waves forming the wave packetpropogate at a velocity known as the phase velocity (Vp)
Vp= ω/k
Q8. What is Heisenberge uncertainity principle and write itsapplications
Ans: Principle: It is impossible to measure both the position and momentum of aparticle simulataneously and precisely.
Δp Δx≥
h4 π
Other forms ΔE Δt≥h
4 π
Application 1
Application 2
Application 3 Particle in a box : Let us consider a particle confined to a box of length l.The uncertainity Δx in the position is l
Δx .ΔP ≈ h ( h cut/h bar)
ΔP=h/l ( Δx=l )
E= P2/2m
E= (ΔP)2/2m
E= h2/2ml
This Energy result agrees with the result obtained from the schrodingerwave equation .
Q9. Derive the Schrodinger’s time independent waveequation.
Ans:
Q 10 Write physical significance of wave function.
Ans:
1. It gives a statistical relationship between the particle and wave.
2.
3.
4.
The probability of finding a particle within a volume dv
P= ∫ ΨΨ* dv= ∫ |Ψ|2dv
dv=dxdydz, Ψ2 Gives the probability finding a particle.
When the particle definately exist in a volume
P= ∫ ΨΨ* dv=1 (Normalization condition )
When the particle does not exist in a volume
P= ∫ ΨΨ* dv=0
Q11. Derive Schrodinger Equation for a particle in a one dimensional "box" and find its energy (eigen ) values, wave (eigen) functions and probability.
Q.12 Explain the formation of energy bands in solids
Q.13 Distingush between insulator, conductor, and semiconductor.
1. Conductors
2. Semiconductors
3. Insulator
Q. 14 Explain in detail the comparison among Maxwell Boltzmen (M-B), Fermi Dirac (F-D) and Bose-Einstein (B-E) statistics
Q . 15 Define the tunneling phenomenon and find the transmission and reflection coefficient of the partcile (electron) in case of rectangular potential barrier.
OrExplain the tunneling phenomenon in case of rectangular potential barrier.
Ans: Defination : Quantum tunnelling or tunneling refers to the quantum mechanical phenomenon where a particle tunnels through a barrier that it classically could not not possible like as shown in the figure.
Rectangular potential barrier:
V(x)=V0
V=0 V=0
X=0 X=L
Region - I Region -II Region -III
The-end