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In a photoelectric effect experiment if the frequency of light is doubled, the stopping potential will
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become more than double
The energy equation in photoelectric effect is given as
$e V_0=h v-\phi_0 \Rightarrow V_0=\frac{h v}{e}-\frac{\phi_0}{e}$ $\ldots$ (i)
When the frequency is doubled,
$V_0^{\prime}=\frac{2 h v}{e}-\frac{\phi_0}{e} \Rightarrow V_0^{\prime}=2\left(\frac{h v}{e}-\frac{\phi_0}{e}\right)+\frac{\phi_0}{e}$
$V_0^{\prime}=2 V_0+\frac{\phi_0}{e} \quad$ [using Eq. (i)]
$\Rightarrow \quad V_0^{\prime}>2 V_0$
Thus, the stopping potential will become more than double.
$e V_0=h v-\phi_0 \Rightarrow V_0=\frac{h v}{e}-\frac{\phi_0}{e}$ $\ldots$ (i)
When the frequency is doubled,
$V_0^{\prime}=\frac{2 h v}{e}-\frac{\phi_0}{e} \Rightarrow V_0^{\prime}=2\left(\frac{h v}{e}-\frac{\phi_0}{e}\right)+\frac{\phi_0}{e}$
$V_0^{\prime}=2 V_0+\frac{\phi_0}{e} \quad$ [using Eq. (i)]
$\Rightarrow \quad V_0^{\prime}>2 V_0$
Thus, the stopping potential will become more than double.
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