The work functions of two metals are 2.75 eV and 2 eV respectively. If these are irradiated by photo

Question

The work functions of two metals are 2.75 eV and 2 eV respectively. If these are irradiated by photons of energy 3 eV , the ratio of maximum momenta of the photoelectrons emitted respectively by them is\

Professor PC Thomas Classes · Answer

The maximum energy of photoelectrons\ $\mathrm{K}_{	ext {max }_{1}}=\mathrm{h} v-\phi_{1}=3 \mathrm{eV}-2.75 \mathrm{eV}$\ $=0.25 \mathrm{eV}$\ and $\mathrm{K}_{	ext {max }_{2}}=\mathrm{h} \mathrm{v}-\phi_{2}=3 \mathrm{eV}-2 \mathrm{eV}$\ $=1 \mathrm{eV}$\ Momenta of photoelectron is given by\ $\mathrm{p}=\sqrt{2 \mathrm{mK}}$\ $	herefore \frac{\mathrm{p}_{	ext {max }_{1}}}{\mathrm{p}_{	ext {max }_{2}}}=\sqrt{\frac{\mathrm{K}_{	ext {max }_{1}}}{\mathrm{~K}_{	ext {max }_{2}}}}=\sqrt{\frac{0.25 \mathrm{eV}}{1 \mathrm{eV}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$\