In the experiment of P.E.E. (Photoelectric Ef | vedclass

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(D) $1$. Saturation current $(i_s)$ is directly proportional to the intensity of incident light. Since the intensity is doubled, the saturation curren

Vedclass · Accepted Answer

$i_s = 10\,mA$ and $V_s > 20\,V$

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(D) $1$. Saturation current $(i_s)$ is directly proportional to the intensity of incident light. Since the intensity is doubled, the saturation current becomes $2 	imes 5\,mA = 10\,mA$.$2$. Stopping potential $(V_s)$ is given by Einstein's photoelectric equation: $eV_s = h
u - \Phi$, where $\Phi$ is the work function.$3$. Initially, $eV_{s1} = h
u - \Phi = 10\,eV$.$4$. When frequency is doubled $(2
u)$, the new stopping potential $V_{s2}$ satisfies $eV_{s2} = h(2
u) - \Phi = 2h
u - \Phi$.$5$. Since $h
u = 10 + \Phi$, we have $eV_{s2} = 2(10 + \Phi) - \Phi = 20 + 2\Phi - \Phi = 20 + \Phi$.$6$. Since $\Phi > 0$, it follows that $V_{s2} > 20\,V$.$7$. Therefore, $i_s = 10\,mA$ and $V_s > 20\,V$.

In the experiment of $P.E.E.$ (Photoelectric Effect), the saturation current is $5\,mA$ and the stopping potential is $10\,V$. If the intensity and frequency of the incident light are both doubled, then what will be the new saturation current $(i_s)$ and stopping potential $(V_s)$?

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