If the energy of H - atom in the ground state is −E, the velocity of photo electron emitted when a photon having energy Ep striked a stationary Li2+ ion in ground state, is given by:
A
v=√2(Ep−E)m
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B
v=√2(Ep+9E)m
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C
v=√2(Ep−9E)m
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D
v=√2(Ep−3E)m
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Solution
The correct option is Cv=√2(Ep−9E)m Energy of electron in the ground state of H− atom =−E=−13.6×Z2n2=−13.6×1212=−13.6 eV
Energy of electron in the ground state of Li2+=−13.6×Z2n2=−13.6×3212=−9E(∵−13.6eV=−E)
Energy supplied by photon =IE+12mv2 IE=−(−9E)=9E ∴Ep=9E+12mv2 ⇒v=√2(Ep−9E)m