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In hydrogen atom if the difference in the energy of the electron in n=2 and n=3 orbits is E, then ionization energy of hydrogen atom isA) 13.2EB)7.2EC)5.6ED)3.2E |
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Answer» we know the energy of electron in n th orbit of Hydrogen En = -K/n^2ev, k = constant So the difference in the energy of the electron in n=2 and n=3 orbits is E = E3 – E2 = -K[1/3^2-1/2^2] = K*5/36 =>K = 7.2E Now ionization energy of hydrogen atom is the energy required to promote electron from n=1 to n= ∞ So This ionization energy = -K{1/∞^2-1/1^2]=K =7,2E Hence option (B) is accepted |
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