Question 18: Dual Nature of Matter and Radiation

Question

The threshold frequency for a metal surface is '$$n_0$$'. A photo electric current '$$I$$' is produced when it is exposed to a light of frequency $$\left(\frac{11}{6}\right) \mathrm{n}_{\mathrm{o}}$$ and intensity $$\mathrm{I}_{\mathrm{n}}$$. If both the frequency and intensity are halved, the new photoelectric current '$$\mathrm{I}^1$$' will become:

Accepted Answer

To solve this problem, let's go step by step and apply the photoelectric effect equations. According to the photoelectric effect, the photoelectric current $$I$$ depends on the number of photons hitting the surface and is directly proportional to the intensity of the light.
1.
Threshold Frequency $$(n_0)$$:
The minimum frequency of light required to eject electrons from the metal surface.
2.
Given Frequency:
The light has a frequency of $$\frac{11}{6} n_0$$, which is higher than the threshold frequency $$n_0$$.
3.
Initial Intensity $$(I_n)$$:
The initial intensity of the light is $$\mathrm{I}_{\mathrm{n}}$$, producing a photoelectric current $$I$$.
4.
Reducing Frequency and Intensity:
Both frequency and intensity are halved. Therefore, the new frequency is $$\frac{1}{2} \left( \frac{11}{6} n_0 \right) = \frac{11}{12} n_0$$. The new intensity becomes $$\frac{I_n}{2}$$.
When the frequency is reduced to $$\frac{11}{12} n_0$$, it becomes lower than the threshold frequency $$n_0$$, which means no electrons will be ejected from the metal surface. Hence, no photoelectric current will be produced.
Therefore, the new photoelectric current will be zero:
Option C: $$\mathrm{I^1=0}$$

Answer

$$\mathrm{I}^1=\frac{1}{4} \mathrm{I}$$

Answer

$$\mathrm{I}^1=2 \mathrm{I}$$

Answer

$$\mathrm{I^1=0}$$

Answer

$$\mathrm{I}^1=\frac{1}{2} \mathrm{I}$$

Dual Nature of Matter and Radiation

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