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Maximum velocity of the photoelectron emitted by a metal is 1.8 xx 10^(6) ms^(-1). Take the value of specific charge of the electron is 1.8 xx 10^(11) C kg^(-1). Thenthe stopping potential in volt is

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Solution :GIVEN `v=1.8xx10^(6)m//s, (E)/(m)=1.8xx10^(11) C//kg`
We have `eV_(0)=(1)/(2) mv^(2)`
`V_(0)(e)/(m)=(v^(2))/(2)`
`RARR V_(0)xx1.8xx10^(11)=(1.8xx1.8xx(10^(6))^(2))/(1)`
`V_(0)=9V`.


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