Question

Light rays of wavelengths $6000\mathring{A}$ and of photon intensity $39.6$ $watts/m$${}^2$ incidents on a metal surface. If only $1\%$ of photons incident on the surface emit photoelectrons, then the number of electrons emitted per second per unit area from the surface will be approximately :
[Planck constant $h=6.64\times {10}^{-34}Js;$ Velocity of light $=3\times {10}^{8}m{s}^{-1}]$

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

Answer 1

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

The intensity of photons $I=\frac{Nhc}{A\lambda }$

Here $N$ is the number of photons and $A$ is the exposed area of the metal.

$\therefore \frac{N}{A}$ i.e. the number of photons striking the surface of the metal per second per unit area $=\frac{I\lambda }{hc}$

Substituting the values given in the question, we get $\frac{I\lambda }{hc}=12\times {10}^{19}/m^2$

But only $1$ % of photons cause photoemission.

So the number of electrons ejected from the surface of the metal per second per unit area $=12\times {10}^{17}/m^2$

So, the answer is option (C).