Question

A photo-cell is illuminated by a source of light, which is placed at a distance $d$ from the cell. If the distance becomes $d/2$, then the number of electrons emitted per second will be:
A) Remains same
B) Four times
C) Two times
D) One-fourth

Answer 1

The main concept we need to know for solving the problem is that intensity is directly proportional to the number of photoelectrons emitted. We need to first use the intensity formula to find the relation between intensity and the two distances and from that, we can say that how does it affect the number of electrons to arrive at the answer.

Complete answer:
Let us divide the question into two cases. In the first case let the intensity be ${I_1}$. And in the second case after the distance is reduced to $d/2$ the intensity is changed into ${I_2}$. Now the formula for the intensity is given as,
$I = \dfrac{E}{{At}}$
Here, $I$ is the intensity
$E$ is said to be the total energy of the source
$A$ is said to be the area of the illuminated surface
$t$ is the time taken.
We can substitute the formula for the area as , $A = 4\pi {r^2}$. Therefore the above equation becomes,
$I = \dfrac{E}{{4\pi {r^2}t}}$ ….. (1)
The ratio of the radius for both the cases is ${r_1}:{r_2} = 1:2$
Now taking the ratio between both the intensities,
$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{E}{{4\pi {r_1}^2t}} \times \dfrac{{4\pi {r_2}^2t}}{E}$
$\Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{r_2}^2}}{{{r_1}^2}}$
Substituting for the radii, we get,
$\Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{4}{1}$
$\Rightarrow \dfrac{{{I_2}}}{{{I_1}}} = \dfrac{1}{4}$
As we already said that intensity is directly proportional to the number of photoelectrons emitted, we can say that the number of electrons emitted initially is increased by four times.

Therefore, the correct option is (D).

Note: The effect of intensity on photoelectric current is for a fixed frequency, the photoelectric current increases linearly with the increase of light. In the photoelectric effect, one photon can only release one electron and so more intense light means more photons hence more electrons will be ejected from the metal surface and hence increasing the photocurrent.