Question

A photocell emits $4\times {10}^{12}$ electrons per second with maximum KE of E when the source of light is at a distance d from the photocell. If the distance is made equal to 2d, the number of electrons emitted per second and maximum K.E.will be

• A. $2\times {10}^{12},E$
• B. ${10}^{12},E$
• C. $4\times {10}^{12},E$
• D. $2\times {10}^{12},2E$

Answer 1

The correct option is B ${10}^{12},E$

Number of emitted electrons per second is increasingly proportional to square of distance between source of light and photocell.
so, $n\alpha \frac{1}{d^2}$
so, $\frac{n_1}{n_2}=\frac{(d_2{)}^2}{(d_1{)}^2}$
$\frac{4\times {10}^{12}}{n_2}={\frac{2d}{d}}^2$
$n_2=\frac{4\times {10}^{12}}{4}={10}^{12}$ electrons per second.
Now, maximum kinetic energy only depend on the energy of incident light energy and work function of the material, which does not changes in the problem, so, $K.E{.}_{max}=E.$