Question

An experimental study of the photoelectric effect involves a metal of work function $\phi_0$. What is the smallest wavelength of the incident photon to photoemit an electron of mass $m$ which has the same de Broglie wavelength as that of the incident photon? [Given $h$ is the Planck's constant, $c$ is the speed of light, and $\phi_0 \ll mc^2$]

• A. $\frac{h}{mc} \left( 1 + \sqrt{1 - \frac{2\phi_0}{mc^2}} \right)^{-1}$
• B. $\frac{h}{mc} \left( 1 - \sqrt{1 - \frac{2\phi_0}{mc^2}} \right)^{-1}$
• C. $\frac{h}{mc} \left( 1 - \sqrt{1 - \frac{\phi_0}{mc^2}} \right)^{-1}$
• D. $\frac{h}{mc} \left( 1 + \sqrt{1 - \frac{\phi_0}{mc^2}} \right)^{-1}$

Hint:

In algebraic physics problems with quadratic equations, always correlate "smallest value of variable" with the "largest value of its reciprocal" to determine whether to use the positive or negative root.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This problem combines the Einstein Photoelectric Equation with the de Broglie wavelength hypothesis.
Key Formula or Approach:
1. Photoelectric Eq: \( \frac{hc}{\lambda} = \phi_{0} + K \).
2. de Broglie Eq: \( \lambda_{e} = \frac{h}{p} \). Given \( \lambda_{e} = \lambda \), then \( p = \frac{h}{\lambda} \).
3. Kinetic Energy: \( K = \frac{p^{2}}{2m} = \frac{h^{2}}{2m\lambda^{2}} \).

Step 2: Detailed Explanation:
1. Substitute into equation:
\[ \frac{hc}{\lambda} = \phi_{0} + \frac{h^{2}}{2m\lambda^{2}} \]
2. Rearrange into a quadratic in \( (1/\lambda) \):
\[ \frac{h^{2}}{2m} \left( \frac{1}{\lambda} \right)^{2} - hc \left( \frac{1}{\lambda} \right) + \phi_{0} = 0 \]
3. Solve using Quadratic Formula for \( x = 1/\lambda \):
\[ x = \frac{hc \pm \sqrt{(hc)^{2} - 4(\frac{h^{2}}{2m})\phi_{0}}}{2(\frac{h^{2}}{2m})} = \frac{mc}{h} \left( 1 \pm \sqrt{1 - \frac{2\phi_{0}}{mc^{2}}} \right) \]
4. Choose smallest wavelength: Smallest \( \lambda \) corresponds to largest \( 1/\lambda \), so we take the positive root.
\[ \frac{1}{\lambda} = \frac{mc}{h} \left( 1 + \sqrt{1 - \frac{2\phi_{0}}{mc^{2}}} \right) \implies \lambda = \frac{h}{mc} \left( 1 + \sqrt{1 - \frac{2\phi_{0}}{mc^{2}}} \right)^{-1} \]

Step 3: Final Answer:
The wavelength is \( \frac{h}{mc} \left( 1 + \sqrt{1 - \frac{2\phi_{0}}{mc^{2}}} \right)^{-1} \).