For intensity I of a light of wavelength 5000A0 the photoelectron saturation current is 0.40 μA and stopping potential is 1.36 V, the work function of the metal is:
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a
2.47 eV
b
1.36 eV
c
1.10 eV
d
0.43 eV
answer is C.
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Detailed Solution
By using E=W0+KmaxE=123755000=2.475eV and Kmax=eV0=1.36eVSo2.475=W0+1.36⇒W0=1.1 eV.