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Albert Einstein

Thursday, March 12, 2009

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

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 23rd February 2008. Some typical multiple choice questions (with solution) were discussed in the posts dated 25th February 2008 and 1st March 2008. Two free response questions (with model answers) in this section were discussed in the post dated 8th 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)].
(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
4 = 1 +2 +3 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 Hα line in the Balmer series? (Planck’s constant = h).
(a) h/
(b) h/π
(c) 2h/π
(d) 3h/
(e) zero
When the electron is in the orbit of quantum number n its angular momentum is nh/2π. The Hα line results due to the transition from the 3rd orbit to the 2nd 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 λmin of the X-rays produced is given by
hc/λmin = eV
We have equated the entire energy eV of the electron to the energy (= hc/λ) of the photon. This gives λmin = 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

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