Question

The threshold frequency for a metal surface is $ \nu_0 $. A photoelectric current $ I $ is produced when it is exposed to a light of frequency $ \frac{11}{6} \nu_0 $ and intensity $ I_0 $. If both the frequency and intensity are halved, the new photoelectric current $ I_1 $ will become:

• A. \( I_1 = \frac{1}{4}I \)
• B. \( I_1 = 2I \)
• C. \( I_1 = 0 \)
• D. \( I_1 = \frac{1}{2}I \)

Hint:

For photoelectric emission to occur, the frequency of the light must be greater than or equal to the threshold frequency. If the frequency is lower than the threshold frequency, no emission takes place, and the current becomes zero.

Answer 1

The Correct Option is C

The photoelectric current is dependent on two factors: 
1. The intensity of the light (\( I \)), which is proportional to the number of photons striking the surface. 
2. The frequency of the light (\( \nu \)), which must be above the threshold frequency \( \nu_0 \) to cause emission of photoelectrons. The energy of a photon is given by: \[ E_{\text{photon}} = h \nu, \] where \( h \) is Planck's constant and \( \nu \) is the frequency of the light. For photoelectric emission to occur, the frequency \( \nu \) must be greater than or equal to the threshold frequency \( \nu_0 \). In this case, the frequency of the light is \( \frac{11}{6} \nu_0 \), which is above the threshold frequency, so photoelectric emission occurs. 
Step 1: Halving the frequency
When the frequency is halved, the new frequency becomes \( \frac{1}{2} \times \frac{11}{6} \nu_0 = \frac{11}{12} \nu_0 \). This new frequency is below the threshold frequency \( \nu_0 \), which means that the light will no longer have enough energy to emit photoelectrons. 
Step 2: Effect on the photoelectric current
Since the frequency is now below the threshold frequency, no photoelectrons will be emitted, and therefore the photoelectric current \( I_1 \) will be zero. 
Thus, the new photoelectric current is \( I_1 = 0 \).