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The stopping potential of a photoelectri...

The stopping potential of a photoelectric diode is 9 volts. `e//m = 1.8 xx 10^(11) C kg^(-1)` , then what is its velocity ?

A

`32. 4 xx 10^(11)` m/s

B

`18 xx 10^5` m/s

C

`16.2 xx 10^5` m/s

D

`4.02 xx 10^5` m/s

Text Solution

Verified by Experts

The correct Answer is:
B

Let `v_(max)` be the maximum velocity of the photoelectrons.
`therefore 1/2 mv_("max")^(2) = eV_0` (where `V_0` is the stopping potential)
`therefore v_("max") = sqrt(2V_0 . (e)/(m))=sqrt(2 xx9 xx1.8xx 10^(11))`
`=18 xx 10^5 ms^(-1)`
`= 18 xx 10^5 ms^(-1)`
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