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Energy E of a hydrogen atom with principal quantum number n is given by E = - 13 .6/n2eV. The energy of photon ejected when the electron jumps from n = 3 state to n = 2 state hydrogen is approximately
(a) 1.9 eV
(b) 1.5 eV
(c) 0.85 eV
(d) 3.4 eV
(a)
ΔE = E3 - E2
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