Maximum velocity of photo electrons emitted by a photo meter is 1.8 x 106 m/s. Taking em=1.8×1011 C/kg for electrons, the stopping potential of emitter is _____

Maximum velocity of photo electrons emitted by a photo meter is 1.8 x 10⁶ m/s. Taking $\frac{e}{m} = 1 .8 \times 10^{11}$ C/kg for electrons, the stopping potential of emitter is _____

Solution:

$\frac{1}{2} {mv}^{2} = e . V = \frac{v^{2}}{2 (\frac{e}{m})} = \frac{(1 .8)^{2} \times 10^{12}}{2 \times 1 .8 \times 10^{11}} = 9 volt$

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