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determine the energy in ev of the photon emitted when an electron jumps down from the n = 3 orbit to the n = 2 orbit of a hydrogen atom.

Expert Solution

Step 1

In case of hydrogen atom, the energy of the emitted photon can be calculated by the formula

$E_{n} = \frac{- 13.6 e v}{n^{2}}$ where n= number of states

Here electron jump from n=3 orbit to n=2 orbit

so, initial orbit of electron n_i =3

final orbit n_f= 2

In our case the energy of photon become, E=E₃-E₂

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