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 ${}^{\prime }{B}^{\prime }$ to that in experiment ${}^{\prime }{A}^{\prime }$ is

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

Answer 1

Answer: (D)

In experiment $A$, incident radiations of 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 ${\lambda }_0=x$]

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