1.

If a photocell is illuminated with a radiation of `1240 A`, then stopping potential is found to be 8 V. The work function of the emitter and the threshold wavelength areA. 1 eV,`5200A`B. 2 eV,`6200A`C. 3 eV,`7200A`D. 4 eV,`4200A`

Answer» Correct Answer - B
`W=hc-eV_S`
`hupsilon=` energy of incident photon
Here `hupsilon=(12400)/(1240)eV=10eV`
`W=10-8=2eV`
So, `lamda_0=` Threshold wavelength
`=(12400)/(2eV)A=6200A`


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