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

Einstein’s photoelectric equation states that the maximum kinetic energy of emitted electrons is: \[ K_{\text{max}} = h u - \phi, \] where $h u$ represents the photon energy and $\phi$ is the work function of the metal. For beam A, the maximum kinetic energy is calculated as: \[ K_{\text{max,A}} = 3.3 - 2.3 = 1.0 \, \text{eV}. \] For beam B, the maximum kinetic energy is calculated as: \[ K_{\text{max,B}} = 11.3 - 2.3 = 9.0 \, \text{eV}. \] The relationship between maximum kinetic energy and maximum speed is: \[ K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2. \] This implies the maximum speed is: \[ v_{\text{max}} = \sqrt{\frac{2 K_{\text{max}}}{m}}. \] The ratio of maximum speeds is derived as: \[ \frac{v_{\text{max,A}}}{v_{\text{max,B}}} = \sqrt{\frac{K_{\text{max,A}}}{K_{\text{max,B}}}}. \] Substituting the calculated values yields: \[ \frac{v_{\text{max,A}}}{v_{\text{max,B}}} = \sqrt{\frac{1.0}{9.0}} = \frac{1}{3}. \] Therefore, the ratio of the maximum speeds is: \[ \boxed{\frac{1}{3}}. \]