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(A) From the graph,we observe that the stopping potential $V_{0}$ is the same for curves $1$ and $2$. However,for curve $3$,the stopping potential is

Vedclass · Accepted Answer

$\gamma_{1}=\gamma_{2}$ and $I_{1} 
eq I_{2}$

Vedclass · Answer

(A) From the graph,we observe that the stopping potential $V_{0}$ is the same for curves $1$ and $2$. However,for curve $3$,the stopping potential is greater in magnitude than that for curves $1$ and $2$.We know that the stopping potential is given by the relation: $e V_{0} = E_{\max} = h \gamma - \phi_{0}$,where $\phi_{0}$ is the work function of the metal.Since $V_{0}$ is the same for curves $1$ and $2$,it implies that the incident frequencies are equal,i.e.,$\gamma_{1} = \gamma_{2}$.For curve $3$,the stopping potential is higher,which means $\gamma_{3} > \gamma_{1} = \gamma_{2}$.Regarding the saturation current,it depends on the intensity of the incident light. Curves $2$ and $3$ reach the same saturation current,meaning $I_{2} = I_{3}$,while curve $1$ has a lower saturation current,so $I_{1} Therefore,comparing the options,$\gamma_{1} = \gamma_{2}$ and $I_{1} 
eq I_{2}$ is the correct statement.

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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