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The kinetic energies of the photoelectro...

The kinetic energies of the photoelectrons ejected from a metal surface by light of wavelength `2000 Å` range from zero to `3.2 xx 10^(-19)` Joule. Then find the stopping potential (in V).

Text Solution

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`K_(max)=4.0xx10^(-19)Jxx(1eV)/(1.6xx10^(-19))=2.5 eV`.
Then, from `eV_(s)=K_(max),V_(s)=2.5V`
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