Maximum velocity of the photoelectrons emitted by metal surface is $1.2 \times 10^{6} m s^{- 1}$ . assuming the specific charge of the electron to be $1.8 \times 10^{11} C k g^{- 1}$ , the value of the stopping potential in volt will be

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Solution

The correct option is C 4
We know that,
$\frac{1}{2} {M V}_{m a x}^{2} = e V_{s}$
$\Rightarrow V_{s} = \frac{{M V}_{m a x}^{2}}{2 e}$
$V_{s} = \frac{{(1.2 \times 10^{6})}^{2}}{(2 \times 1.8 \times 10^{11})}$
$V_{s} = 4 V$

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Maximum velocity of the photoelectrons emitted by metal surface is 1.2×106ms−1. assuming the specific charge of the electron to be 1.8×1011Ckg−1, the value of the stopping potential in volt will be

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Maximum velocity of the photoelectrons emitted by metal surface is $1.2 \times 10^{6} m s^{- 1}$ . assuming the specific charge of the electron to be $1.8 \times 10^{11} C k g^{- 1}$ , the value of the stopping potential in volt will be

The correct option is C 4
We know that,
$\frac{1}{2} {M V}_{m a x}^{2} = e V_{s}$
$\Rightarrow V_{s} = \frac{{M V}_{m a x}^{2}}{2 e}$
$V_{s} = \frac{{(1.2 \times 10^{6})}^{2}}{(2 \times 1.8 \times 10^{11})}$
$V_{s} = 4 V$