1.

Maximum speed of electron emitted from metal surface is 5xx10^(6 )ms^(-1) .If specifi charge of electron is 1.8xx10^(11) ckg^(-1),then value of stopping potential will be…..

Answer»

2V
3V
7V
4V

Solution :`(1)/(2)mv_(MAX)^(2)=eV_(0)`
`V_(0)=(mv_(max)^(2))/(2E)`
`(v_(max)^(2))/(2(e//m))`
`((5XX10^(6))^(2))/(2xx1.8xx10^(11))`
`=6.95V~~7.0V`


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