if e/m of electron is `1.76xx10^(11)C(kg)^(-1)` andn stopping potential is 0.71 V, then the maximum velocity of the photoelectron is

Question

if e/m of electron is `1.76xx10^(11)C(kg)^(-1)` andn stopping potential is 0.71 V, then the maximum velocity of the photoelectron is

if e/m of electron is `1.76xx10^(11)C(kg)^(-1)` andn stopping potential is 0.71 V, then the maximum velocity of the photoelectron is
A. `150kms^(-1)`
B. `200kms^(-1)`
C. `500kms^(-1)`
D. `250kms^(-1)`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - C
We can write, `(1)/(2)mv^(2)=eV`
`implies v=sqrt((2eV)/(m))=sqrt(2Vxx(e)/(m))`
or `v=sqrt(2xx0.71xx1.76xx10^(11))`
`=5xx10^(5)ms^(-1)=500kms^(-1)`

4 views Answered 2023-02-05 15:29