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If e/m of electron is 1.76 xx 10^(11) "C...

If e/m of electron is `1.76 xx 10^(11) "C kg"^(-1)` and the stopping potential is 0.71 V, then the maximum velocity of the photo-electron is

A

150 `kms^(-1)`

B

200 `kms^(-1)`

C

500 `kms^(-1)`

D

250 `kms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
C
From condition of stoping potential
`(1)/(2)mv^(2) = ev`
`v = sqrt((2eV)/(m)) = sqrt(2 xx 1.76 xx 10^(11) xx 0.71)`
`= 5 xx 10^(5) ms^(-1) = 500 kms^(-1)`
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