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Energy E of a hydrogen atom with principal quantum number n is given by E = - 13 .6/n 2 eV. 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



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


Grade:10

1 Answers

Aditi Chauhan
askIITians Faculty 396 Points
7 years ago

(a)

ΔE = E3 - E2

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