The figure shows the variation of photocurren | vedclass

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(A) According to Einstein's photoelectric equation,$hv - \phi_0 = eV_0$,where $V_0$ is the stopping potential.$1$. From the graph,curves $a$ and $b$ i

Vedclass · Accepted Answer

$f_a = f_b$ and $I_a 
eq I_b$

Vedclass · Answer

(A) According to Einstein's photoelectric equation,$hv - \phi_0 = eV_0$,where $V_0$ is the stopping potential.$1$. From the graph,curves $a$ and $b$ intersect the anode potential axis at the same point,which means they have the same stopping potential $(V_0)$. Since the stopping potential depends only on the frequency of the incident radiation,we conclude that $f_a = f_b$.$2$. The saturation current is directly proportional to the intensity of the incident radiation. From the graph,the saturation current for curve $b$ is higher than that for curve $a$. Therefore,the intensities are different,i.e.,$I_a 
eq I_b$.

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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