Two beams, A and B whose photon energies are 3.3 eV and 11.3 eV respectively, illuminate a metallic surface (work function 2.3 eV ) successively. The ratio of maximum speed of electrons emitted due to beam $A$ to that due to beam $B$ is :

Question

YESPYQ · Accepted Answer

Correct answer: c) $\frac{1}{3}$. \begin{aligned} & \text{Given: photon energies } h\nu_A=3.3\ \mathrm{eV},\; h\nu_B=11.3\ \mathrm{eV},\; \text{work function } \phi=2.3\ \mathrm{eV}. \[4pt] & \text{Maximum kinetic energy of emitted electrons: } K_{\max}=h\nu-\phi. \[4pt] & K_{A}=3.3-2.3=1.0\ \mathrm{eV}, \qquad K_{B}=11.3-2.3=9.0\ \mathrm{eV}. \[6pt] & \text{Since } K=\tfrac{1}{2}mv^2 \Rightarrow v\propto\sqrt{K}, \[4pt] & \therefore\; \dfrac{v_A}{v_B}=\sqrt{\dfrac{K_A}{K_B}}=\sqrt{\dfrac{1.0}{9.0}}=\dfrac{1}{3}. \[8pt] & \boxed{\dfrac{v_A}{v_B}=\dfrac{1}{3}} \end{aligned}

Two beams, A and B whose photon energies are 3.3 eV and 11.3 eV respectively, illuminate a metallic surface (work functi…

Practice more Board Physics PYQs