1.

In a photoelectric effect experiment irradiation of a metal with light of frequency `5.2 xx 10^(14) s^(-1)` yields electrons with maximum kinetic energy `1.3 xx 10^(-19) J`. Calculate the threshold frequency `(v_(0))` for the metal.

Answer» We known that, `hv=hv^(0)+KE" "` or `" " v^(0)=v-(KE)/(h)`
`KE =1.3 xx 10^(-19)J , v=5.2 xx 10^(14)s^(1) , h=6.626 xx 10^(-34)Js`
`:." "` Threshold frequency `(v^(0))=(5.2 xx 10^(14)s^(-1))-((1.3 xx 10^(-19)J))/((6.626 xx 10 ^(-34)Js))`
`=(5.2 xx 10^(14)s^(-1))-(1.96 xx 10^(14)s^(-1))=3.24 xx 10^(14 s^(-1)`


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