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Maximum velocity of photoelectron emitte...

Maximum velocity of photoelectron emitted is `4.8 ms^(-1)`. If e/m ratio of electron is `1.76xx10^(11)Ckg^(-1)`, then stopping potential is given by

A

`5xx10^(-10)` J/C

B

`3xx10^(-7)` J/C

C

`7xx10^(-11)` J/C

D

`2.5xx10^(2)` J/C

Text Solution

Verified by Experts

The correct Answer is:
C
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