Physics - Modern Physics Question with Solution | TestHub

Given, ${\lambda }_1=4972\mathring{A}$ and ${\lambda }_2=6216\mathring{A}$ and $\mathrm{I}=3.6\times 10^{-3}{\mathrm{Wm}}^{-2}$ Intensity associated with each wavelength $$=\frac{3.6\times 10^{-3}}{2}=1.8\times 10^{-3}{\mathrm{Wm}}^{-2}$$ Work function $\varphi =\mathrm{hv}=\frac{\mathrm{hc}}{\lambda }$ $$\begin{array}{rl}& =\frac{\left(6.62\times 10^{-34}\right)\left(3\times 10^8\right)}{\lambda }\\ & =\frac{12.4\times 10^3}{\lambda }\mathrm{ev}\end{array}$$ for different wavelengths $$\begin{array}{r}{\varphi }_1=\frac{12.4\times 10^3}{{\lambda }_1}=\frac{12.4\times 10^3}{4972}=2.493\mathrm{eV}\\ =3.984\times 10^{-19}\mathrm{J}\\ {\varphi }_2=\frac{12.4\times 10^3}{{\lambda }_2}=\frac{12.4\times 10^3}{6216}=1.994\mathrm{eV}\\ =3.184\times 10^{-19}\mathrm{J}\end{array}$$ Work function for metallic surface $\varphi =2.3$ $\mathrm{eV}$ (given) ${\varphi }_2<\varphi$ Therefore, ${\varphi }_2$ will not contribute in this process. Now, no. of electrons per ${\mathrm{m}}^2-\mathrm{s}=$ no. of photons per ${\mathrm{m}}^2-\mathrm{s}$ no. of electrons per ${\mathrm{m}}^2-\mathrm{s}$ $$\begin{array}{rl}& =\frac{1.8\times 10^{-3}}{3.984\times 10^{-19}}\times 10^{-4}\\ & \left(\because 1{\mathrm{cm}}^2=10^{-4}{\mathrm{m}}^2\right)=0.45\times 10^{12}\end{array}$$ So, the number of photo electrons liberated in 2 sec. $=0.45\times 10^{12}\times 2$ $=9\times 10^{11}$