The figure shows the variation of photocurren | vedclass

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(A) In the photoelectric effect,the stopping potential depends only on the frequency of the incident radiation,while the saturation current depends on

Vedclass · Accepted Answer

${f_a} = {f_b}$ and ${I_a} 
eq {I_b}$

Vedclass · Answer

(A) In the photoelectric effect,the stopping potential depends only on the frequency of the incident radiation,while the saturation current depends on the intensity of the incident radiation.$1$. From the graph,curves $a$ and $b$ intersect the anode potential axis at the same point,which means they have the same stopping potential. Therefore,the frequencies are equal: ${f_a} = {f_b}$.$2$. The saturation current (the maximum value of photocurrent) for curve $b$ is higher than that for curve $a$. Since saturation current is directly proportional to the intensity of incident radiation,we have ${I_a} $3$. Thus,${f_a} = {f_b}$ and ${I_a} 
eq {I_b}$.

The figure shows the variation of photocurrent with anode potential for a photo-sensitive surface for three different radiations. Let ${I_a}, {I_b}$ and ${I_c}$ be the intensities and ${f_a}, {f_b}$ and ${f_c}$ be the frequencies for the curves $a, b$ and $c$ respectively.

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The following graph represents the variation of photocurrent with anode potential for a metal surface. Here $I_{1}, I_{2}$ and $I_{3}$ represent intensities and $\gamma_{1}, \gamma_{2}, \gamma_{3}$ represent frequencies for curves $1, 2$ and $3$ respectively,then

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