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When photons of energy hν fall on an aluminium plate (of work function E_0), photoelectrons of maximum kinetic energy K are ejected. If the frequency of the radiation is doubled, the maximum kinetic energy of the ejected photoelectrons will be:

1. 2K
2. K
3. K + hν
4. K + E_0

Correct answer: K + hν

Solution

The maximum kinetic energy of photoelectrons is given by K = hν - E_0. If the frequency is doubled, the new energy becomes K' = h(2ν) - E_0 = 2hν - E_0. Subtracting the original K from K', we get K' = K + hν.

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