Question

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

• A. 3
• B. 9
• C. $\frac{1}{3}$
• D. $\frac{1}{9}$

Hint:

For photoelectric problems, use $K_{\text{max}} = h \nu - \phi$ to calculate the kinetic energy of emitted electrons and relate it to their speed using $v_{\text{max}} = \sqrt{\frac{2 K_{\text{max}}}{m}}$.

Answer 1

The Correct Option is C

The maximum kinetic energy of emitted electrons is given by Einstein’s photoelectric equation: \[ K_{\text{max}} = h \nu - \phi, \] where: \begin{itemize} \item $h \nu$ is the photon energy, \item $\phi$ is the work function of the metal. \end{itemize} For beam A: \[ K_{\text{max,A}} = 3.3 - 2.3 = 1.0 \, \text{eV}. \] For beam B: \[ K_{\text{max,B}} = 11.3 - 2.3 = 9.0 \, \text{eV}. \] The maximum speed of the emitted electrons is related to the kinetic energy by: \[ K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2. \] Thus: \[ v_{\text{max}} = \sqrt{\frac{2 K_{\text{max}}}{m}}. \] The ratio of speeds is: \[ \frac{v_{\text{max,A}}}{v_{\text{max,B}}} = \sqrt{\frac{K_{\text{max,A}}}{K_{\text{max,B}}}}. \] Substitute the values: \[ \frac{v_{\text{max,A}}}{v_{\text{max,B}}} = \sqrt{\frac{1.0}{9.0}} = \frac{1}{3}. \] Thus, the ratio of maximum speeds is: \[ \boxed{\frac{1}{3}}. \]