When a point source of monochromatic light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are 0.6 volt and 18 mA respectively. If the same source is placed 0.6 m away from the photoelectric cell, then

Question

When a point source of monochromatic light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are 0.6 volt and 18 mA respectively. If the same source is placed 0.6 m away from the photoelectric cell, then (A) The stopping potential will be 0.2 V (B) The stopping potential will be 0.6 V (C) The saturation current will be 6 mA (D) The saturation current will be 18 mA

Tardigrade Admin · Accepted Answer

Cut-off voltage is independent of intensity and hence, remains the same.
Since distance becomes

3

times, intensity

(I)

becomes

\frac{I}{9}

.
Hence, photocurrent also decreases by this factor
i.e., becomes

\frac{18}{9} = 2 m A

.

Q. When a point source of monochromatic light is at a distance of 0.2m from a photoelectric cell, the cut-off voltage and the saturation current are 0.6 volt and 18mA respectively. If the same source is placed 0.6m away from the photoelectric cell, then

The stopping potential will be 0.2 V

The stopping potential will be 0.6 V

The saturation current will be 6 mA

The saturation current will be 18 mA

Cut-off voltage is independent of intensity and hence, remains the same. Since distance becomes 3 times, intensity (I) becomes 9I​. Hence, photocurrent also decreases by this factor i.e., becomes 918​=2mA.

Q. When a point source of monochromatic light is at a distance of $0.2 m$ from a photoelectric cell, the cut-off voltage and the saturation current are $0.6$ volt and $18 m A$ respectively. If the same source is placed $0.6 m$ away from the photoelectric cell, then

Cut-off voltage is independent of intensity and hence, remains the same.
Since distance becomes $3$ times, intensity $(I)$ becomes $\frac{I}{9}$ .
Hence, photocurrent also decreases by this factor
i.e., becomes $\frac{18}{9} = 2 m A$ .