The graph shows the variation of photocurrent | vedclass

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(A) $1$. Frequency $(f)$: The stopping potential $(V_0)$ is the potential at which the photocurrent becomes zero. It is given by $eV_0 = hf - \phi$,wh

Vedclass · Accepted Answer

$f_{b}>f_{a}, f_{b}=f_{c}, I_{c}=I_{d}$

Vedclass · Answer

(A) $1$. Frequency $(f)$: The stopping potential $(V_0)$ is the potential at which the photocurrent becomes zero. It is given by $eV_0 = hf - \phi$,where $\phi$ is the work function. $A$ more negative stopping potential indicates a higher frequency of incident radiation.From the graph,the stopping potentials are $V_{0a} = V_{0b} Therefore,$f_a = f_b $2$. Intensity $(I)$: The saturation current is directly proportional to the intensity of the incident radiation. From the graph,the saturation currents for curves $c$ and $d$ are equal,while $a$ and $b$ have lower saturation currents.Thus,$I_c = I_d$ and $I_a Comparing the given options,option $A$ is the most consistent with the relationships derived,specifically $f_b = f_a$ (from $f_a = f_b$) and $I_c = I_d$.

The graph shows the variation of photocurrent with anode potential for four different radiations. Let $I_a, I_b, I_c$ and $I_d$ be the intensities and $f_a, f_b, f_c$ and $f_d$ be the frequencies for the curves $a, b, c$ and $d$ respectively,then

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