Question

Light rays of wavelengths $6000\mathring{A}$ and of Photon intensity $39.6{\text{W m}}^2$ is incident on a metal surface. If only one percent of photons incident on the surface, emit photo electrons, then the number of electrons emitted per second per unit area from the surface will be-
[Use $h=6.64\times {10}^{-34}\text{J s}$]

• A. $12\times {10}^{18}$
• B. $12\times {10}^{17}$
• C. $12\times {10}^{16}$
• D. $12\times {10}^{15}$

Answer 1

The correct option is B $12\times {10}^{17}$

Given,

$\lambda =6000\mathring{A};P=39.6{\text{W m}}^2$

The power of the emitted radiation is,

$P=\frac{E}{t}=\frac{n}{t}\frac{hc}{\lambda }$

$\because I=\frac{P}{A}\Rightarrow IA=\frac{n}{t}\frac{hc}{\lambda }$

Thus, the number of photons incident per sec per unit area is,

$\frac{n}{At}=\frac{I\lambda }{hc}=\frac{39.6\times 6000\times {10}^{-10}}{6.64\times {10}^{-34}\times 3\times {10}^8}\approx 12\times {10}^{19}$

$\therefore$The number of electrons emitted $=1\%$ of incident Photons.

$\frac{N_e}{At}=\frac{1}{100}\times \left(\frac{n}{At}\right)=\frac{1}{100}\times 12\times {10}^{19}$

$\therefore \frac{N_e}{At}=12\times {10}^{17}$

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