Question

If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength $\lambda$. When it jumps from the 4th orbit to the 3rd orbit, what will be the corresponding wavelength of the photon?
A). $\dfrac{20}{13}\lambda $
B). $\dfrac{16}{25}\lambda $
C). $\dfrac{9}{16}\lambda $
D). $\dfrac{20}{7}\lambda $

Answer 1

Hint: 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. The energy of the nth orbital in the hydrogen atom is inversely proportional to the square of n.

Formula Used:
Energy of nth orbital in hydrogen atom, ${{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}\,\text{eV}$, energy of a photon, $E=h\nu =\dfrac{hc}{\lambda }$

Complete step-by-step solution:
In 1913, Bohr proposed that electrons in an atom revolve around certain orbits known as stationary orbits. The energy of these orbits is constant and is given by
${{E}_{n}}=\dfrac{-13.6}{{{n}^{2}}}\,\text{eV}$
Where n is the principal quantum number of the orbit
The energy of an electron decreases when it jumps from a higher orbital to a lower orbital. So following the law of conservation of energy, a photon of energy equivalent to the change in energy of the electron is released. The energy of a photon is related to its wavelength $\lambda$ by
$E=\dfrac{hc}{\lambda }$
Where $h$ is Planck’s constant and its value is $6.63\times {{10}^{-34}}{{m}^{2}}kg/s$
When an electron jumps from 3rd orbit to the 2nd orbit, change in its energy
$\Delta {{E}_{3to2}}=-13.6\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right)=-13.6\left( \dfrac{5}{36} \right)$
A photon is released having this change in energy, therefore
$E={{E}_{3to2}}$
Given that wavelength of photon released is $\lambda$, therefore
$\dfrac{hc}{\lambda }=-13.6\left( \dfrac{5}{36} \right)$ ……………...….(1)
Similarly when electron jumps from 4th orbit to 3rd, wavelength of photon released assuming it as $\lambda '$ is
$\dfrac{hc}{\lambda '}=-13.6\left( \dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{4}^{2}}} \right)=-13.6\left( \dfrac{7}{144} \right)$ ………………...….(2)
We divide (1) with (2) and simplify to get
$\dfrac{\lambda '}{\lambda }=\dfrac{20}{7}$
$\Rightarrow \lambda '=\dfrac{20}{7}\lambda $
Hence, option D is correct.

Note: 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. This relation can also be used to solve this type of problem.