Bohr Model and Quantum Theory Lecture-2. Bohr Atom The Planetary Model of the Atom

Bohr Model and Quantum Theory

Lecture-2

Bohr Atom

The Planetary Model of the Atom

Bohr’s Model

Nucleus

Electron

Orbit

Energy Levels

Bohr’s Model

Photons

Max Planck in 1900 stated that the light emitted by a hot object (black body radiation) is given off in discrete units or quanta. The higher the frequency of the light, the greater the energy per quantum.

Max Planck (1858-1947)

(a) and (b) represent two waves that are traveling at the same speed.

In (a) the wave has long wavelength and low frequency

In (b) the wave has shorter wavelength and higher frequency

Frequency

9-2. Photons

The system shown here detects people with fevers on the basis of their infrared emissions, with red indicating skin temperatures above normal. In this way people with illnesses that may be infectious can be easily identified in public places.

Photons

All the quanta associated with a particular frequency of light have the same energy. The equation is E = hf where E = energy, h = Planck's constant (6.63 x 10-34 J s), and f = frequency.

Electrons can have only certain discrete energies, not energies in between.

The Photoelectron Effect

The photoelectric effect is the emission of electrons from a metal surface when light shines on it. The discovery of the photoelectric effect could not be explained by the electromagnetic theory of light. Albert Einstein developed the quantum theory of light in 1905.

What is light?

Light exhibits either wave characteristics or particle (photon) characteristics, but never both at the same time. The wave theory of light and the quantum theory of light are both needed to explain the nature of light and therefore complement each other.

Bohr Model (1913)

Assumptions

1) Only certain set of allowable circular orbits for an electron in an atom

2) An electron can only move from one orbit to another. It can not stop in between. So discrete quanta of energy involved in the transition in accord with Planck (E = h)

3) Allowable orbits have unique properties particularly that the angular momentum is quantized.

2

nh mvr

Bohr Model (1913)

• Equations derived from Bohr’s Assumption

• Radius of the orbit

r1

r2

r3

n=1 n=2

n=3

rn n 2a o

Z22

2

o me4

h a

n = orbit number

Z = atomic number

h = Planck’s constantm = mass of electrone = charge on electron

Bohr Model (1913)

For H:

For He+ (also 1 electron)

r1 12ao

1 ao Called Bohr radius

r2 22ao

1 4ao

r1 12ao

2

12ao Smaller value for the radius.

This makes sense because of the larger charge in the center

For H and any 1 electron system:

n = 1 called ground staten = 2 called first excited staten = 3 called second excited stateetc.

Bohr Model (1913)

Which of the following has the smallest radius?

A) First excited state of H

B) Second excited state of He+

C) First excited state of Li+2

D) Ground state of Li+2

E) Second excited state of H

Bohr Model (1913)

Which of the following has the smallest radius?

A) First excited state of H

B) Second excited state of He+

C) First excited state of Li+2

D) Ground state of Li+2

E) Second excited state of H

r2 22ao

1 4ao

r3 32ao

2

92ao

r2 22ao

3

43ao

r1 12ao

3

1

3ao

r3 32ao

1 9ao

Calculate the radius of 5th orbit of the hydrogen atom.

Problem

22

22

n Zme4

hn r

n=5h= 6.62 x 10-34 J secm=9.109x10-31kge=1.602x10-19Cπ=3.14Z=1

r5= 13.225x10-10m

9-9. The Bohr Model

Electron orbits are identified by a quantum number n, and each orbit corresponds to a specific energy level of the atom. An atom having the lowest possible energy is in its ground state; an atom that has absorbed energy is in an excited state.

Bohr Model (1913)

• Energy of the Electron

E n AZ2

n 2

constant, A = 2.18 x 10-18 J

E1 A12

12

A

E 2 A12

22

1

4A

E 3 A12

32

1

9A

What’s happening to the energy of the orbit as the orbit number increases?

Energy is becoming less negative, therefore it is increasing. The value approaches 0.

E 0 (have formed ion) Completely removed the electron from the atom.

Bohr Model (1913)

E1 A

E 2 A

4

E = E 2 - E1 A

4 ( A)=+

3

4A

+ sign shows that energy was absorbed.

xJ) = 1.64 x 10-18 J

What is E when electron moves from n = 2 to n = 1?

E = E1 - E 2 A (A

4)=-

3

4A

xJ) = - 1.64 x 10-18 J

So: Ephoton = |E|transition = h = h(c/)

h = Planck’s constant = 6.62 x 10-34 J sec

c = speed of light = 3.00 x 108 m/sec

When E is positive, the photon is absorbed

When E is negative, the photon is emitted

E = - A1

n f2

1

n i2

A green line of wavelength 4.86x107 m is observed in the emission spectrum of hydrogen.a) Calculate the energy of one photon of this green light.b) Calculate the energy loses by the one mole of H atoms.

a)

b)

Problem

SolutionWe know the wavelength of the light, and we calculate its frequency so that we can then calculate the energy of each photon.

DeBroglie Postulate (1924)

Said if light can behave as matter, i.e. as a particle, then matter can behave as a wave. That is, it moves in wavelike motion.

So, every moving mass has a wavelength () associated with it.

where h = Planck’s constant

v = velocity

m = mass

hmv

What is the in nm associated with a ping pong ball (m = 2.5 g) traveling at 35.0 mph.

A) 1.69 x 10-32 B) 1.7 x 10-32

C) 1.69 x 10-22 D) 1.7 x 10-23

hmv

6.62x10 34 kgm2

s2s

2.5g 1kg1000g

35.0mi.hr

1.609kmmi

1000mkm

1hr3600s

1.69 x10 32m1.7 x10 23nm

(a)Calculate the wavelength in meters of an electron traveling at 1.24 x107 m/s. The mass of an electron is 9.11x 10-28 g. (b) Calculate the wavelength of a baseball of mass 149g traveling at 92.5 mph. Recall that 1 J = 1 kgm2/s2.

Problem

b)m= 149g= 0.149kg

Heisenberg Uncertainty Principle

To explain the problem of trying to locate a subatomic particle (electron) that behaves as a wave

Anything that you do to locate the particle, changes the wave properties

He said: It is impossible to know simultaneously both the momentum(p) and the position(x) of a particle with certainty