Question

Radiation of wavelength \( \lambda \) is incident on a photocell. The fastest emitted electron has speed \( v \). If the wavelength is changed to \( \frac{3\lambda}{4} \), then the speed of the fastest emitted electron will be

• A. \( \text{greater than } v\sqrt{\frac{4}{3}} \)
• B. \( \text{less than } v\sqrt{\frac{4}{3}} \)
• C. \( \text{equal to } v\sqrt{\frac{4}{3}} \)
• D. \( \text{equal to } v\sqrt{\frac{3}{4}} \)

Hint:

When scaling wavelengths in photoelectric effect problems, remember that because of the subtracting work function term (\( \phi \)), the kinetic energy (and thus the squared speed) always scales more than the inverse of the wavelength. Thus, when the wavelength is scaled by \( \frac{3}{4} \) (energy scaled by \( \frac{4}{3} \)), the speed must scale by more than \( \sqrt{\frac{4}{3}} \).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The maximum kinetic energy and hence the speed \( v \) of photoelectrons depends on the wavelength of incident radiation and the work function \( \phi \) of the photocell. We need to find the new maximum speed \( v' \) when the wavelength is decreased to \( \frac{3\lambda}{4} \).

Step 2: Key Formula or Approach:
According to Einstein's photoelectric equation:
\[ K_{\text{max}} = \frac{1}{2} m v^2 = \frac{hc}{\lambda} - \phi \]
where:
- \( h \) is Planck's constant.
- \( c \) is the speed of light.
- \( \phi \) is the work function of the metal.
- \( m \) is the mass of an electron.

Step 3: Detailed Explanation:
1. For the first case with wavelength \( \lambda \):
\[ \frac{1}{2} m v^2 = \frac{hc}{\lambda} - \phi \implies v^2 = \frac{2}{m} \left( \frac{hc}{\lambda} - \phi \right) \]
2. For the second case with wavelength \( \lambda' = \frac{3\lambda}{4} \):
\[ \frac{1}{2} m (v')^2 = \frac{hc}{\left(\frac{3\lambda}{4}\right)} - \phi = \frac{4hc}{3\lambda} - \phi \]
\[ (v')^2 = \frac{2}{m} \left( \frac{4}{3} \frac{hc}{\lambda} - \phi \right) \]
3. Let's rewrite the term inside the parenthesis to relate it to \( v^2 \):
\[ \frac{4}{3} \frac{hc}{\lambda} - \phi = \frac{4}{3} \left( \frac{hc}{\lambda} - \phi \right) + \frac{4}{3}\phi - \phi = \frac{4}{3} \left( \frac{hc}{\lambda} - \phi \right) + \frac{1}{3}\phi \]
Substitute this back:
\[ (v')^2 = \frac{4}{3} \left[ \frac{2}{m} \left( \frac{hc}{\lambda} - \phi \right) \right] + \frac{2\phi}{3m} \]
\[ (v')^2 = \frac{4}{3} v^2 + \frac{2\phi}{3m} \]
Since the work function \( \phi \) and the electron mass \( m \) are strictly positive, \( \frac{2\phi}{3m} > 0 \):
\[ (v')^2 > \frac{4}{3} v^2 \]
Taking the square root on both sides:
\[ v' > v \sqrt{\frac{4}{3}} \]

Step 4: Final Answer:
The speed of the fastest emitted electron will be greater than \( v\sqrt{\frac{4}{3}} \), which corresponds to option (A).