AP Physics B– Atomic Physics and Quantum Effects– Multiple Choice Practice Questions

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The essential points you need to remember in atomic physics and quantum effects (especially to make you strong in working out multiple choice questions in the stipulated time) were given in the post dated 23^rd February 2008. Some typical multiple choice questions (with solution) were discussed in the posts dated 25^th February 2008 and 1^st March 2008. Two free response questions (with model answers) in this section were discussed in the post dated 8^th March 2008. You can access all these posts by clicking on the label ‘atomic physics’ below this post.

Today we will discuss some more multiple choice questions in this section:

(1) The photoelectric work function of a metal surface is Φ. The thermionic work function of the same metal surface will be

(a) less than Φ

(b) equal to Φ

(c) greater than Φ

(d) dependent on the method of heating used

(e) independent of the photoelectric work function

The difference between photoelectric emission and thermionic emission is in the method of supplying energy to the electrons in the metal to make them come out of the metal. The minimum energy required for the electrons to escape from the surface of a given metal is a constant. This means that the photoelectric work function of a given metal surface is equal to its thermionic work function [Option (b)].

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(2) The adjoining figure shows four energy levels of an atom. The wave lengths emitted due to transitions 1, 2, 3 and 4 (fig.) are λ₁, λ₂, λ₃ and λ₄ and the corresponding frequencies are ν₁, ν₂, ν₃ and ν₄ respectively. Then

(a) λ₁ = λ₂ + λ₃ + λ₄

(b) λ₄ = λ₁ + λ₂ + λ₃

(c) ν₁ = ν₂ + ν₃ + ν₄

(d) ν₄ = ν₁ + ν₂ + ν₃

(e) λ₄ = λ₁ λ₂ λ₃ /(λ₁ + λ₂ + λ₃)

From Bohr’s frequency condition we have

hν₄ = hν₁ + hν₂ + hν₃ where h is Planck’s constant.

Therefore ν₄ = ν₁ + ν₂ + ν₃

(3) In the above question the emitted wave lengths are related as

(a) λ₁ = λ₄ – (λ₂ + λ₃)

(b) λ₄ = λ₁ – (λ₂ + λ₃)

(c) 1/λ₁ = 1/(λ₂ + λ₃ + λ₄)

(d) 1/λ₄ = 1/(λ₁ + λ₂ + λ₃)

(e) λ₄ = λ₁λ₂λ₃ /(λ₁λ₂ + λ₂λ₃ + λ₃λ₁)

As shown above, ν₄ = ν₁ + ν₂ + ν₃

Or, c/λ₄ = c/λ₁ + c/λ₂ + c/λ₃ where c is the speed of light.

Therefore, 1/λ₄ = 1/λ₁ + 1/λ₂ + 1/λ₃ = (λ₂λ₃ + λ₃λ₁ + λ₁λ₂) /λ₁λ₂λ₃ from which

λ₄ = λ₁λ₂λ₃ /(λ₁λ₂ + λ₂λ₃ + λ₃λ₁)

(4) What is the change in the angular momentum of the electron in the hydrogen atom when it undergoes a transition to emit the H_α line in the Balmer series? (Planck’s constant = h).

(a) h/2π

(b) h/π

(c) 2h/π

(d) 3h/2π

(e) zero

When the electron is in the orbit of quantum number n its angular momentum is nh/2π. The H_α line results due to the transition from the 3^rd orbit to the 2^nd orbit. The angular momentum therefore changes from 3h/2π to 2h/2π so that the change in the angular momentum is h/2π.

(5) In an X-ray tube the accelerating voltage is V volt. If the electronic charge is e coulomb, Planck’s constant is h joule second and the speed of light in free space is c ms^–1, the range of wave lengths possible in the continuous X-ray spectrum is

(a) zero to infinity

(b) zero to hc/eV

(c) zero to eV/hc

(d) hc/eV to infinity

(e) eV/hc to infinity

The minimum wave length λ_min of the X-rays produced is given by

hc/λ_min = eV

We have equated the entire energy eV of the electron to the energy hν (= hc/λ) of the photon. This gives λ_min = hc/eV.

All wave lengths above this value are possible so that the correct option is (d).

You will find some useful questions in this section here

AP Physics B– Atomic Physics and Quantum Effects – Equations to be Remembered

The topics included under atomic physics and quantum effects are the following:

(i) Photons, the photoelectric effect, Compton scattering, x-rays

(ii) Atomic energy levels

(iii) Wave-particle duality

Remember the following:

(1) Energy of a photon, E = hν = hc/λ where ‘h’ is Planck’s constant, ν is the frequency of light, ‘c’ is the speed of light and λ is the wave length. [Note that the speed and the wave length of light depend on the medium where as the frequency is independent of the medium].

Since the energy of a photon of wave length 1000 Angstrom is very nearly 12.4 electron volt, in the case of any photon, the product of energy in eV and the wave length in Ǻ is 12400. This can be used to calculate the energy or wave length of a photon.

(2) Momentum of photon, p = hν/c = h/λ = E/c

(3) Einstein’s photoelectric equation relates the energy of incident photon to the maximum kinetic energy (K_max) of the photo electron and the work function (Ф) of the photosensitive surface and can be written as

hν = K_max + Ф

Or, hν = K_max + hν₀ where ν₀ is the threshold frequency (minimum frequency of light to initiate photo electron emission from the surface).

The above equation can be written as K_max = h(ν – ν₀) = hc[(1/λ) – (1/λ₀)] where λ₀ is the threshold wave length (maximum wave length of light to initiate photo electron emission from the surface).

The cut-off or stopping potential is the minimum negative potential to be applied on the anode (plate) so that the photoelectric current becomes zero (stops).

In terms of the stopping potential V_s, Einstein’s equation can be written as

eV_s = h(ν – ν₀), since K_max = eV_s.

(4) de Broglie wave length is l = h/p = h/mv where λ is the wave length associated with a particle of momentum ‘p’.

[In the case of an electron accelerated by a small voltageV (so that relativistic mass increase is negligible), the wave length in Angstrom is very nearly √(150/V)]

(5) The minimum wave length (λ) of X-rays produced by an X-ray tube is inversely proportional to the anode voltage V.

Since the entire energy of the impinging electron is converted in to the energy of the X-ray photon of minimum wave length, the energy of the photon must be V electron volt. Therefore, the minimum wave length (λ_min ) in Angstrom of X-rays produced by an X-ray tube operating with anode voltage V volt must be given by

λ_min×V = 12400 [See (1) above].

(6) In Compton effect, the change (dλ) in the wave length of the scattered x-ray photon is given by

dλ = (h/mc) (1– cosφ) where ‘h’ is Planck’s constant, ‘m’ is the mass of the electron, ‘c’ is the speed of light in free space and ‘φ’ is the angle of scattering.

The above equation shows that the maximum change in wave length of the scattered X-ray photon is 2h/mc which happens when the photon is turned back (φ = 180º).

(7) In a hydrogen like atom of atomic number Z, the energy (E_n) of the electron in the n^th orbit is given by

E_n = – 13.6 Z²/n² electron volt. Here n = 1,2,3,4 etc.

Note that the energy is inversely proportional to the square of the quantum number n.

The above energy is the total energy of the electron and is made of negative potential energy and positive kinetic energy. The potential energy value is twice the kinetic energy value (as in the case of planetary motion) and hence the total energy is negative.

(8) Momentum of the electron in the n^th orbit is nh/2π.

(9) The Bohr radius (r₀) is the radius of the innermost orbit in a hydrogen atom and is given by

r₀ = h²ε₀/πme² = 0.53 Ǻ (nearly)

(10) The radius (r) of the n^th orbit in a a hydrogen like atom is given by

r = n²r₀/Z

Note that the orbital radius is directly proportional to the square of the quantum number n.

(11) The orbital period T is related to the orbital radius r as

T² α r³, as in planetary motion.

(12) When an electron undergoes a transition from an orbit of energy E₂ to an orbit of energy E₁, the frequency of the radiation emitted is given by Bohr’s frequency condition,

ν = (E₂ – E₁)/h

Since the emitted photon has linear momentum p = hν/c = h/λ = E/c, the atom receives an equal and opposite recoil momentum.

[When an atom is excited by absorbing a photon of energy hν, the electron in the atom undergoes transition from lower energy E₁ to higher energy E₂ and the same frequency condition (Bohr’s) holds].

(13) Rydberg’s relation for the wave number`ν (number of waver per metre) of the spectral line emitted by a hydrogen atom is

`ν = 1/ λ = R[(1/n₁²) – (1/n₂²)] where n₁ and n₂ are the quantum numbers of the inner and outer orbits.

(14) The important spectral series in the case of hydrogen atom are Lyman, Balmer, Paschen, Brackett and Pfund series. Of these, the Balmer series contains spectral lines in the visible region.

Wave numbers of the spectral lines in the Lyman series are obtained by putting n₁ =1 and n₂ = 2,3,4,5…etc. in the Rydberg relation (since the spectral lines in this series arise due to transitions from outer orbits to the innermost orbit).

For Balmer series, n₁ = 2 and n₂ = 3,4,5,6.…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the second orbit).

For Paschen series, n₁ = 3 and n₂ = 4,5,6.…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the third orbit).

For Brackett series, n₁ = 4 and n₂ = 5,6,7..…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the fourth orbit).

For Pfund series, n₁ = 5 and n₂ = 6,7,8..…etc. (since the spectral lines in this series arise due to transitions from outer orbits to the fifth orbit).

In the next post, I’ll discuss questions from this section. Meanwhile, find some useful multiple choice questions here.