Q.

If a photocell is illuminated with a radiation of 1240 , 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,5200

b

2eV,6200

c

3eV,7200

d

4eV,4200

answer is B.

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Detailed Solution

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W=hveVs

hv= energy of incident photon  Here hv=124001240eV=10eV W=108=2eV

 So,  λ0= Threshold wavelength 

 λ0=124002eV=6200

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