Question

A metal surface is illuminated by light of two different wavelengths $\text{248}\text{nm}$ and $\text{310}\text{nm}$ . The maximum speeds of the photoelectrons corresponding to these wavelengths are $u_1$ and $u_2$ respectively. If the ratio $\frac{u_1}{u_2}=\frac{2}{1}$ and $\text{hc}\text{=}\text{1240}\text{eVnm}$ , the work function of the metal is nearly

• A. 3.7 eV
• B. 3.2 eV
• C. 2.8 eV
• D. 2.5 eV

Answer 1

Answer: (A)

As $K.E\propto v^2$

ratio of kinetic energy $=\frac{4}{1}$ , as speed are in the ratio $=\frac{2}{1}$

K.E = Energy of incident photon - Work function (W)

$4KE=\frac{1240}{248}-W$ .....(i)

and $KE=\frac{1240}{310}-W$ .....(ii)

By solving (i) and (ii)

$W=3.7eV$