Question

When a point of monochromatic light is at a distance of $0.2$ m from a photoelectric cell, the cutoff voltage and the saturation current are $0.6$ volt and $18$ mA respectively. If the same source is $0.6$ metres away from the photoelectric cell, then find the new stopping potential and saturation current.

Answer 1

Here we will use the concept that the current in the photo electric cell is inversely proportional to the square of the distance. Then will place the given data in the framed equation and simplify.

Complete step by step solution:
Frame the equation for the current in the photoelectric cell which is inversely proportional to the square of the distance.
So, $I\alpha \dfrac{1}{{{d^2}}}$
When there are two current inputs the above expression can be re-written as-
$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{{d_2}^2}}{{{d_1}^2}}$
Place the given values in the above expression –
$\dfrac{{18mA}}{{{I_2}}} = \dfrac{{{{(0.6)}^2}}}{{{{(0.2)}^2}}}$
Cross multiply the above expression, where the denominator of one side is multiplied with the numerator of the opposite side and vice-versa.
${I_2} = 18 \times {\left( {\dfrac{{0.2}}{{0.6}}} \right)^2}$
Simplify the above equation,
${I_2} = 18 \times {\left( {\dfrac{1}{3}} \right)^2} \\\ {I_2} = 18 \times \left( {\dfrac{1}{9}} \right) \\\$
Common factors from the numerator and the denominator cancels each other.
${I_2} = 2mA$
Therefore, the maximum kinetic energy only depends on the energy of the incident light energy and the work function of the material which is not changing the problem and hence the kinetic energy in the photoelectric cell remains the same.
Hence, the stopping potential remains the same that is $0.6V$

Note:
Always remember to follow the concepts and read the given data twice and frame the equation accordingly. Find the correlation between the known and the unknown terms and find the required justification.