Question

A monochromatic source of light is placed at a large distance d from a metal surface. Photoelectrons are ejected at rate $n$, the kinetic energy being $E$. If the source is brought nearer to distance $d$/2, the rate and kinetic energy per photoelectron become nearly:
A. $2n$ and $2E$
B. $4n$ and $4E$
C. $4n$ and $E$
D. $n$ and $4E$

Answer 1

Hint:Establish the dependence relationship between different parameters to solve this question. Number of photoelectrons emitted depends on the intensity of light, however the kinetic energy for the photoelectron depends only on the frequency and the work function of the metal.

Formula used:
Intensity I \[=\dfrac{P}{4\pi {{d}^{2}}}\], $KE=E-\phi $ and $E=h\upsilon $.

Complete step by step answer:
We know that intensity is defined as power(P) per unit area$\left( 4\pi {{r}^{2}} \right)$Intensity I\[=\dfrac{P}{4\pi {{d}^{2}}}\]
This shows that intensity(I) is inversely proportional to the distance of separation (d) between the source and the metal surface, $I\alpha \dfrac{1}{{{d}^{2}}}$. The rate of ejection of photoelectrons (r) is directly proportional to the intensity of the light (I), which implies that rate of ejection of photoelectron (R) is inversely proportional to the square of the distance (d).
$r\alpha I$
$\because I\alpha \dfrac{1}{{{d}^{2}}}$
$\Rightarrow r\alpha \dfrac{1}{{{d}^{2}}}$
Therefore, if the distance is halved that is it becomes $\dfrac{d}{2}$ then,
$r\alpha \dfrac{1}{{{\left( \dfrac{d}{2} \right)}^{2}}}$
$\therefore r=\dfrac{4}{{{d}^{2}}}$
So, the intensity, and so the rate of ejection becomes four times the initial rate $r$. But the kinetic energy of the photoelectron is given by the formula: $KE=E-\phi $ where $\phi $ is the work function of the metal and $E=h\upsilon $ is the energy of the incident radiation and $\upsilon $ is the frequency of the incident radiation and is independent of the distance of separation.

Therefore when the distance is halved neither the energy of the incident ray knows the work function $\phi $ of the metal changes so the kinetic energy of the photoelectron remains unchanged $KE=$ initial kinetic energy $=E$.

Note: It is important for the student to remember the dependence factor to solve this question. The rate of ejection of a photoelectron is inversely proportional to the square of distance whereas the kinetic energy of the photoelectron is independent of the distance. Also, intensity has no effect on the energy of the photon emitted electrons. For photoelectric effect to take place the frequency of the incident radiation must be greater than the threshold frequency of the metal.