Question

Two separate monochromatic light beams $A$ and $B$ of the same intensity are falling normally on a unit area of a metallic surface. Their wavelengths are ${\lambda }_A$ and ${\lambda }_B$, respectively. Assuming that all the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons from beam $A$ to that from $B$ is:

• A. $\frac{{\lambda }_A}{{\lambda }_B}$
• B. $\frac{{\lambda }_B}{{\lambda }_A}$
• C. ${\left(\frac{{\lambda }_B}{{\lambda }_A}\right)}^2$
• D. ${\left(\frac{{\lambda }_A}{{\lambda }_B}\right)}^2$

Answer 1

The correct option is A $\frac{{\lambda }_A}{{\lambda }_B}$

Energy of each beam

$E_A=n_A\times \frac{hc}{{\lambda }_A}$

$E_B=n_B\times \frac{hc}{{\lambda }_B}$

The light have the same intensity and energy (for ejection to occur).

Hence,
$E_A=E_B$

or, $n_A\frac{hc}{{\lambda }_A}=n_B\frac{hc}{{\lambda }_B}$

or, $\frac{n_A}{n_B}=\frac{{\lambda }_A}{{\lambda }_B}$