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Question
B 4.74×1012Hz C 2.32×1014Hz
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Solution
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Given, Frequency of light, υ=7.21×1014Hz Vmax=6.0×105m/s m=9×10−31kg applying Cinstein's photoelectric equation, K.E.max=12mV2max=h(υ−υ0) 12mV2max=h(υ−υ0) υ0=υ−MV2max2h υ0=7.21×1014−(9.1)×10−31×(6×105)22×(6.63×10−34) υ0=4.74×1014Hz
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