1.

When metal surface with 2.1 eV work function is irridated with photon of 6.0 eV energy.Value of stopping potential will be….

Answer»

`+8.1`
`-8.1V`
`+3.9V`
`-3.9 V`

Solution :Maximum kinetic ENERGY of electron emitted `(1)/(2)mv_(max)^(2)=hv-phi_(0)`
`therefore eV_(0)hv-phi_(0)[because (1)/(2)mv_(max)^(2)=eV_(0)`]
`therefore eV_(0)=(6.0-2.1)eV`
`therefore eV_(0)=3.9 eV`
Stopping POTENTIAL is NEGATIVE VOLTAGE.
`therefore V_(0)=-3.9V`


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