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If an electron falls from n=3 to n=2, then how much energy is released?
A. 10.2 eV
B. 12.09 eV
C. 1.9 eV
D. 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, ${{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}\,\text{eV}$

Complete step by step answer:
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
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”.

Additional Information:
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.

Note: Energy of each orbit is a constant. Energy of an electron in any orbital is negative. That is why it is bound to an atom.
When electrons gain energy, they move from a lower energy level to a higher energy level of an atom.
The energy of an orbital is usually denoted in electron Volts.