Question

The threshold frequency of a metal corresponds to the wavelength of $xnm$. In two separate experiments $A$ and $B$, incident radiations of wavelengths $\frac{1}{2}xnm$ and $\frac{1}{4}xnm$ respectively are used. The ratio of kinetic energy of the released electrons in experiment $B$ to that in experiment $A$ is:

• A. $\frac{1}{3}$
• B. 2
• C. 4
• D. 3
• E. $\frac{1}{2}$

Answer 1

The correct option is D 3

In experiment A, incident radiations of 1 wavelength used $=\frac{1}{2}X$

In experiment B, incident radiations of wavelength used $=\frac{1}{4}X$

The ratio of kinetic energy of the released electrons in experiment 'B' to that in experiment 'A':
$\frac{(KE{)}_B}{(KE{)}_A}=\frac{\frac{1}{{\lambda }_B}-\frac{1}{{\lambda }_0}}{\frac{1}{{\lambda }_A}-\frac{1}{{\lambda }_0}}[lets{\lambda }_0=x]$

$\frac{(KE{)}_B}{(KE{)}_A}=\frac{\frac{4}{x}-\frac{1}{x}}{\frac{2}{x}-\frac{1}{x}}=3$