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An alkali metal was irradiated by using 2 different photons of frequency x and 2x. If the K.E of emitted photoelectron in these cases is y and 3y respectively. Then the threshold frequency of the metal is

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answer is 2.

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Detailed Solution

givenand KE1 = y and KE2 = 3yKE1=hν1-w=hν1-hν0 y = hx-hν0 y = hx-v0 KE2=hν2-hν0 KE2 =h2x-ν0 3y=h2x-v0 3=2x-v0x-v0 3x-3v0=2x-v0 x=2v0 v0=x2

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