If a photocell is illuminated with a radiation of 1240 Å, then stopping potential is found to be 8 V. The work function of the emitter and the threshold wavelength are
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a
1eV,5200Å
b
2eV,6200Å
c
3eV,7200Å
d
4eV,4200Å
answer is B.
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Detailed Solution
W=hv−eVshv= energy of incident photon Here hv=124001240eV=10eV∴ W=10−8=2eV So, λ0= Threshold wavelength λ0=124002eVÅ=6200Å