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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 isA. `150kms^(-1)`B. `200kms^(-1)`C. `500kms^(-1)`D. `250kms^(-1)` |
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Answer» 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)` |
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