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A photoelectron is moving with a maximum...

A photoelectron is moving with a maximum velocity of `10^(6) m//s`. Given `e=1.6xx10^(-19) c`, and `m=9.1xx10^(-31) kg`, the stopping potential is

A

`2.5V`

B

`2.8V`

C

`2.0V`

D

`1.4V`

Text Solution

Verified by Experts

The correct Answer is:
B
`1/2mv^(2)=eV_(0)`
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