Question

Two separate monochromatic light beams $\text{A}$ and $\text{B}$ of the same intensity (energy per unit area per unit time) are falling normally on a unit area of a metallic surface. Their wavelength is ${\lambda }_A$ and ${\lambda }_B$ respectively. Assuming that all the incident light is used in ejecting the photoelectrons, the ratio of the number of photoelectrons ejected by beam $\text{A}$ to that beam $\text{B}$ is-

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

Answer 1

The correct option is A $\left(\frac{{\lambda }_A}{{\lambda }_B}\right)$

The number of photo electron depends on the number of photons,

$\because I=Nh\nu$

Where, $Nh\nu =\text{Amount of energy pass through the unit area per unit time}N=\text{No. of photons unit area per unit time}$

$\therefore I=Nh\nu =\frac{Nhc}{\lambda }$

$\Rightarrow N=\frac{I\lambda }{hc}$

$\Rightarrow N\propto \lambda$

$\therefore \frac{N_A}{N_B}=\frac{{\lambda }_A}{{\lambda }_B}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.