An electron in the hydrogen atom initially in the fourth excited state makes a transition to $n^{\text{th}}$ energy state by emitting a photon of energy 2.86 eV. The integer value of n will be \_\_\_\_.

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Modern Physics

An electron in the hydrogen atom initially in the fourth excited state makes a transition to

n^{th}

energy state by emitting a photon of energy 2.86 eV. The integer value of n will be ____.

Answer: 2

Solution

The fourth excited state of hydrogen is

n_{i} = 5

. The energy released in a downward transition is

E = 13.6 (\frac{1}{n _{f}^{2}} - \frac{1}{n _{i}^{2}}) eV

. Setting this equal to

2.86 eV

\frac{1}{n _{f}^{2}} = \frac{2.86}{13.6} + \frac{1}{25} = 0.21 + 0.04 = 0.25

, so

n_{f}^{2} = 4

. The electron lands in the state

n = 2

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Integer answer

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