The essen tial 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 23labe l ‘atomic physics’ below this post.

^{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
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

*λ*

_{1},

*λ*

_{2},

*λ*

_{3}and

*λ*

_{4}and the corresponding frequencies are

*ν*

_{1},

*ν*

_{2},

*ν*

_{3}

*and*

*ν*

_{4 }respectively. Then

(a)

*λ*_{1 }=*λ*_{2 }+*λ*_{3}+*λ*_{4}
(b)

*λ*_{4 }=*λ*_{1 }+*λ*_{2}+*λ*_{3}
(c)

*ν*_{1}=*ν*_{2}+*ν*_{3}*+**ν*_{4 }
(d)

*ν*_{4}=*ν*_{1}+*ν*_{2}*+**ν*_{3 }
(e)

*λ*_{4 }=*λ*_{1}*λ*_{2}*λ*_{3}/(*λ*_{1 }+*λ*_{2}+*λ*_{3})
From Bohr’s frequency condition we have

*hν*

_{4}=

*hν*

_{1}+

*hν*

_{2}+

*hν*

_{3 }where

*h*is Planck’s constant.

Therefore

*ν*_{4}=*ν*_{1}+*ν*_{2}*+**ν*_{3}**(3)**In the above question the emitted wave lengths are related as

(a)

*λ*_{1 }=*λ*_{4}– (*λ*_{2 }+*λ*_{3})
(b)

*λ*_{4 }=*λ*_{1}– (*λ*_{2 }+*λ*_{3})
(c) 1/

*λ*_{1}= 1/(*λ*_{2 }+*λ*_{3}+*λ*_{4})
(d) 1/

*λ*_{4 }= 1/(*λ*_{1 }+*λ*_{2}+*λ*_{3})
(e)

*λ*_{4 }=*λ*_{1}*λ*_{2}*λ*_{3}/(*λ*_{1}*λ*_{2}*+**λ*_{2}*λ*_{3}+*λ*_{3}*λ*_{1})
As shown above,

*ν*_{4}=*ν*_{1}+*ν*_{2}*+**ν*_{3}
Or,

*c/**λ*_{4}=*c/**λ*_{1}*+**c/**λ*_{2}*+**c/**λ*_{3}where*c*is the speed of light.
Therefore, 1

*/**λ*_{4}= 1*/**λ*_{1}*+ 1**/**λ*_{2}*+ 1**/**λ*_{3}= (*λ*_{2}*λ*_{3}+*λ*_{3}*λ*_{1}+*λ*_{1}*λ*_{2}) /*λ*_{1}*λ*_{2}*λ*_{3}from which*λ*

_{4 }=

*λ*

_{1}

*λ*

_{2}

*λ*

_{3}/(

*λ*

_{1}

*λ*

_{2}

*+*

*λ*

_{2}

*λ*

_{3}+

*λ*

_{3}

*λ*

_{1})

**(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) 2

*h/*π
(d) 3

*h/*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 3*h/*2π to 2*h/*2π so that the change in the angular momentum is*h/*2π.
(5) In an X-ray tube the accelerating volta ge 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

## No comments:

## Post a Comment