Question

The threshold frequency of a metal corresponds to the wavelength of x nm. In two separate experiments A and B, incident radiations of wavelengths 1/2 x nm and 1/4 x nm respectively are used. The ratio of kinetic energy of the released electrons in experiment B to that in experiment A is:
(A) 1/3
(B) 2
(C) 4
(D) 3
(E) 1/2

Answer 1

Hint: Here the wavelengths of the radiations are given and we know that the kinetic energy is inversely proportional to the wavelength of the radiation. And we use this relation in this question.

Complete step by step solution:
Let \[{{\lambda }_{0}}\]= x
In experiment A, wavelength of incident radiations (\[{{\lambda }_{A}}\]) = \[\dfrac{1}{2}X\]nm
In experiment B, wavelength of incident radiations (\[{{\lambda }_{B}}\])= \[\dfrac{1}{4}X\]nm
And we know that
 \[h\nu =h{{\nu }_{0}}+KE\]
By rearranging
\[KE=h{{\nu }_{0}}-h\nu \]
And we know that
\[\nu =\dfrac{c}{\lambda }\]
By putting this value in above equation for A:
\[K{{E}_{A}}=\dfrac{hc}{{{\lambda }_{A}}}-\dfrac{hc}{{{\lambda }_{0}}}\]
And now we will write the KE for B:
\[K{{E}_{B}}=\dfrac{hc}{{{\lambda }_{B}}}-\dfrac{hc}{{{\lambda }_{0}}}\]
The ratio of kinetic energy of the released electrons in experiment 'B' to that in experiment 'A':
\[\dfrac{{{(KE)}_{B}}}{{{(KE)}_{A}}}=\dfrac{\dfrac{1}{{{\lambda }_{B}}}-\dfrac{1}{{{\lambda }_{0}}}}{\dfrac{1}{{{\lambda }_{A}}}-\dfrac{1}{{{\lambda }_{0}}}}\]
And putting given data in above ratio:
\[\dfrac{{{(KE)}_{B}}}{{{(KE)}_{A}}}=\dfrac{\dfrac{4}{x}-\dfrac{1}{x}}{\dfrac{2}{x}-\dfrac{1}{x}}\]
By solving this equation
\[\dfrac{{{(KE)}_{B}}}{{{(KE)}_{A}}}\]= (3 / 1) = 3

So, from the above derivation we can say that the correct answer is option “D”

Note: So, here the KE of does not depend on the unit of wavelength. And if the frequencies are given then also we can find the ratio of kinetic energies.