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If an electron falls from n=3 to n=2, then how much energy is released?A. 10.2 eVB. 12.09 eVC. 1.9 eVD. 0.65 eV

Last updated date: 20th Jun 2024
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Hint: This question is related to Bohr’s postulates.
When an electron jumps from higher energy orbital to lower energy orbital, it releases a photon of the energy equal to the difference in energy of the orbitals. Energy of $n^{th}$ orbital in a hydrogen atom is inversely proportional to square of n.

Formula used:
Energy of $n^{th}$ orbital in hydrogen atom, ${{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}\,\text{eV}$

In 1913, Bohr proposed that electrons in an atom revolve around certain orbits known as stationary orbits. The energy of these orbits are constant and is given by
${{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}\,\text{eV}$
Where n is principal quantum number of the orbit
Energy of an electron decreases when it jumps from a higher orbital to lower orbital. So following law of conservation of energy, a photon of energy equivalent to change in energy of electron, is released.
When electron jumps from $3^{rd}$ orbit to the $2^{nd}$ orbit, change in its energy
$\Delta {{E}_{3 \text { to } 2}}=-13.6\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right)=-13.6\left( \dfrac{5}{36} \right)$
$\Delta {{E}_{3 \text { to } 2}}=-1.9eV$
When an electron jumps from higher orbital to lower orbital, a photon of energy equal to change in energy of electron is released. Therefore energy released by electron is 1.9 eV

So, the correct answer is “Option C”.

When electron jumps from higher orbit (${{n}_{i}}$ ) to lower orbit ( ${{n}_{f}}$ ) then, wavelength of emitted photon is given by
$\dfrac{1}{\lambda }=R\left( \dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}} \right)$
Where $R$ is Rydberg’s constant.