When photons of energy h fall on an aluminium plate (of work function E_0 ), photoelectrons of maximum kinetic energy K are ejected. If the frequency - Physics MCQ Physics | ExamTest.live

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When photons of energy h ν 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 :

A $2 K + E_{0}$
B $2 K$
C $K$
D $K + h ν$

Solution & Step-by-step Explanation

According to Einstein's photoelectric equation: K = h ν - E_{0} ⟹ h ν = K + E_{0} .If the frequency is doubled, the new incident energy is 2 h ν .The new maximum kinetic energy K^{'} is: K^{'} = 2 h ν - E_{0} .Substituting h ν : K^{'} = 2 (K + E_{0}) - E_{0} = 2 K + 2 E_{0} - E_{0} = 2 K + E_{0} .(Note: Assuming Option A represents 2 K + E_{0} as commonly corrected from legacy typos.)

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When photons of energy h ν 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 :

2 K + E_{0}

2 K

K

K + h ν

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When photons of energy $h ν$ 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 :

Solution & Step-by-step Explanation

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When photons of energy hν fall on an aluminium plate (of work function E0​), 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 :

Solution & Step-by-step Explanation

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Discussion

When photons of energy $h ν$ 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 :