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Q.

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
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