An electron in the hydrogen atom initially in the fourth excited state makes a transition to n energy state by emitting a photon of energy 2.86 eV. The integer value of n will be ________. [2025]

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Solution

Q.
An electron in the hydrogen atom initially in the fourth excited state makes a transition to n energy state by emitting a photon of energy 2.86 eV. The integer value of n will be ________. [2025]

Ans.
(2)

$E = 13.6 (\frac{1}{n_{1}^{2}} - \frac{1}{n_{1}^{2}})$

$2.86 = 13.6 (\frac{1}{n^{2}} - \frac{1}{5^{2}})$

$\frac{1}{n^{2}} = 0.21 + \frac{1}{25}$

$n^{2} = 4 \Rightarrow n = 2$

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