Question

A beam of electromagnetic radiation of intensity$6.4\times {10}^{-5}\mathrm{W}/{\mathrm{cm}}^2$ is comprised of wavelength,$\mathrm{\lambda }=310\mathrm{nm}$. It falls normally on a metal (work function $\mathrm{\phi }=2\mathrm{eV}$) of surface area of $1{\mathrm{cm}}^2$. If one in ${10}^3$ photons ejects an electron, total number of electrons ejected in $1\mathrm{s}$ is ${10}^{\mathrm{x}}$. ($\mathrm{hc}=1240\mathrm{eVnm}$, $1\mathrm{eV}=1.6\times {10}^{-19}\mathrm{J}$), then$\mathrm{x}$ is ___

Answer 1

Step 1: Given data

Intensity of radiation, $\mathrm{I}=6.4\times {10}^{-5}\mathrm{W}/{\mathrm{cm}}^2$

Wavelength, $\mathrm{\lambda }=310\mathrm{nm}$

Work function of given metal, $\mathrm{\phi }=2\mathrm{eV}$

Surface area, $\mathrm{A}=1{\mathrm{cm}}^2$

Step 2: Calculating the value of $\mathrm{x}$

We know that energy of incident radiation is given by:

$E=\frac{hc}{\lambda }$

$E=\frac{1240}{310}$ [$hc=1240$ given in question]

$E=4ev>2ev$[work function] [So photoelectric effect will take place]

$$\begin{gathered} E=4\times 1.6\times {10}^{-19}J \\ E=6.4\times {10}^{-19}J \end{gathered}$$

Now, applying the formula of power, $P$

$$\begin{gathered} P=\mathrm{nE}=\mathrm{IA} \\ \mathrm{n}=\frac{\mathrm{IA}}{\mathrm{E}} \end{gathered}$$ [Here, number of photons incident per second is $\mathrm{n}$.]

Now using the above formula we get,

$$\begin{gathered} \mathrm{n}=\frac{6.4\times {10}^{-5}}{6.4\times {10}^{-19}} \\ \mathrm{n}={10}^{14}\mathrm{per}\mathrm{second} \end{gathered}$$

Since one out of thousand photons is successful in ejecting an electron, therefore the number of electrons ejected in one second is $\frac{{10}^{14}}{{10}^3}={10}^{11}$ [Where, ${10}^3$ is number of photons needed to emit an electron

${10}^{11}={10}^x$

Therefore, comparing the values we get, $\mathrm{x}=11$