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Find the maximum velocity of photoelectr...

Find the maximum velocity of photoelectrons emitted by radiation of frequency `3 xx 10^(15)` Hz from a photoelectric surface having a work function of 4.0 eV. Given `h=6.6 xx 10^(-34)` Js and 1 eV = `1.6 xx 10^(-19)` J.

Text Solution

Verified by Experts

`1/2mv_"max"^2=hv-phi`
`=6.63xx10^(-34)xx3xx10^15 - 4xx1.6xx10^(-19)`
`v_"max"^2=(2[19.89xx10^(-19)-6.4xx10^(-19)])/(9.1xx10^(-31))`
`=(26.98xx10^(-19))/(9.1xx10^(-31))`
`=2.96xx10^12`
`v_"max"=1.72xx10^6` m/s
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