Question

When a certain metal surface is illuminated with light of frequency $ \nu $, the stopping potential for photoelectric current is $ V_0 $. When the same surface is illuminated by light of frequency $ \frac{\nu}{2} $, the stopping potential is $ \frac{V_0}{4} $. The threshold frequency for photoelectric emission is

• A. \( \frac{\nu}{2} \)
• B. \( \frac{\nu}{4} \)
• C. \( \frac{3\nu}{4} \)
• D. \( \nu \)

Hint:

The threshold frequency for photoelectric emission is the frequency at which the stopping potential becomes zero. Use the photoelectric equation to solve for the threshold frequency in terms of the given frequencies.

Answer 1

The Correct Option is B

The photoelectric equation is given by: \[ K.E = h \nu - \phi \] where:
- \( K.E \) is the kinetic energy of the emitted electrons,
- \( h \) is Planck's constant,
- \( \nu \) is the frequency of the incident light,
- \( \phi \) is the work function of the metal. The stopping potential \( V_0 \) is related to the kinetic energy of the emitted electrons by: \[ K.E = eV_0 \] When the frequency of light is \( \nu \), the stopping potential is \( V_0 \), and the equation becomes: \[ eV_0 = h \nu - \phi \] When the frequency of light is \( \frac{\nu}{2} \), the stopping potential becomes \( \frac{V_0}{4} \), and the equation becomes: \[ e\left(\frac{V_0}{4}\right) = h \left(\frac{\nu}{2}\right) - \phi \] Simplifying: \[ \frac{eV_0}{4} = \frac{h\nu}{2} - \phi \] Now, solving the two equations:
1. \( eV_0 = h\nu - \phi \)
2. \( \frac{eV_0}{4} = \frac{h\nu}{2} - \phi \) By comparing these equations, we can find that the threshold frequency \( \nu_0 \), which is the frequency at which the stopping potential is zero, is: \[ \nu_0 = \frac{\nu}{2} \]
Thus, the threshold frequency for photoelectric emission is: \[ \text{(B) } \frac{\nu}{2} \]