Explanation

(a) Intensity of light is inversely proportional to square of distance,
$ie,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I \propto \frac{1}{r^{2}}$
or $\frac{I_{2}}{I_{1}} = \frac{\left( r_{1} \right)^{2}}{\left( r_{2} \right)^{2}}$
Given, $r_{1} = 0.5\ $m, $r_{2} = 1.0\ $m
Therefore, $\frac{I_{1}}{I_{2}} = \frac{(0.5)^{2}}{(1)^{2}} = \frac{1}{4}$
Now, since number of photoelectrons emitted per second is directly proportional to intensity, so number of electrons emitted would decrease by factor of 4.