For photoelectric affect in a metal, the graph of stopping potential V₀ (in volt) versus frequency v (in Hz) of the incident radiation is shown in figure. From the graph find,

(a) threshold frequency

(b) Planck's constant and

(c) work function of the metal.

OR

A beam of light consisting of two wavelengths l₁= 496 nm and l₂ = 620 nm falls on a metal surface. The maximum speeds of photoelectrons corresponding to l₁ and l₂ are v₁ and v₂ respectively. If v₁: v₂ = 2 : 1 and hc = 1240 eV nm, what is the work function of the metal?

eV₀ = h(v - v₀)

$\Rightarrow V_0=\frac{h}{e}\left(v-v_0\right)$

(a) It follows from Eq. (1) that $v=v_0$ if $V_0=0$. Hence ${\mathrm{v}}_0=4\times {10}^{15}\mathrm{Hz}$

(b) The slope of  $V_0$ versus v graph $=\frac{h}{e}$ , The slope of the given graph is $$\begin{gathered} \therefore =\frac{16.5-0}{(8-4)\times {10}^{15}}=4.125\times {10}^{-15} \\ \mathrm{h}=\mathrm{e}\times \text{ slope } \\ =\left(1.6\times {10}^{-19}\right)\times \left(4.125\times {10}^{-15}\right) \\ =6.6\times {10}^{-34}\mathrm{Js} \end{gathered}$$

(c) $W_0=h{v}_0=\left(6.6\times {10}^{-34}\right)\times \left(4\times {10}^{15}\right)$

$=26.4\times {10}^{-19}\mathrm{J}$ 

$\begin{array}{rr} & =16.5\mathrm{eV}\end{array}$

OR

$\begin{array}{rr} & {\mathrm{h}}_{\mathrm{v}}={\mathrm{K}}_{\max }+{\mathrm{W}}_0\\ & {\mathrm{K}}_{\max }=\mathrm{hv}-{\mathrm{W}}_0\\ & \frac{1}{2}{\mathrm{mv}}_{\max }^2=\frac{\mathrm{hc}}{\mathrm{\lambda }}-{\mathrm{W}}_0\\ & \mathrm{ }\therefore \mathrm{ }\frac{1}{2}{\mathrm{mv}}_1^2=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}-{\mathrm{W}}_0\\ & \text{ and }\mathrm{ }\frac{1}{2}{\mathrm{mv}}_2^2=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_2}-{\mathrm{W}}_0\end{array}$

Dividing (1) and (2), we get

$\frac{v_1^2}{v_2^2}=\frac{\frac{hc}{{\lambda }_1}-W_0}{\frac{hc}{{\lambda }_2}-W_0}$

Given $\frac{v_1}{v_2}=\sqrt{2}$ and  $\mathrm{hc}=1240\mathrm{eVnm}$. Substituting these values in (3), We have

$\begin{array}{rr} & 2=\frac{\frac{1240\mathrm{eVnm}}{496\mathrm{nm}}-W_0}{\frac{1240\mathrm{eVnm}}{620\mathrm{nm}}-W_0}\\ & 2=\frac{2.5\mathrm{eV}-W_0}{2\mathrm{eV}-W_0}\end{array}$

${\mathrm{W}}_0=1.5\mathrm{eV}$