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Given that a photon of light of wavelength 1000 Å has an energy equal to 1.23 eV. When light of wavelength 5000 Å and intensity I0 falls on photoelectric cell, the saturation current is 0 . 40 x 106 amp. and the stopping potential is 1.36 V. Then the work-function is

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a

0.43 eV

b

1.10 eV

c

1.36 eV

d

2.47 eV

answer is B.

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Detailed Solution

hv=W+eV0 where e Vo = stopping Potential W=hv−eV0 Incident wavelength =5000Å hv= energy of 5000Å =2×1⋅23eV=2⋅46eV eV0=1⋅36eV W=(2⋅46−1⋅36)=1⋅10eV

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