1.

if the wavelength is chaged to 4000A, then stopping potential will become

Answer»

1.36 V
3.40 V
1.60 V
1.97 V

Solution :LET energy CORRESPONDING to WAVELENGTH of `4000A` be E.
Then,
`(E)/(E')=(lamda')/(lamda)` or `(E)/(1.23)=(10000)/(4000)`
`E=1.23xx2.5=3.075eV`
But `hupsilon-hupsilon_0=eV_S`
`V_S=1.975V`


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