Question

A light of wavelength${\text{12818}}\,{{\text{A}}^{\text{o}}}$is emitted when the electron of a hydrogen atom drop from fifth to third quantum level. Find the wavelength of the photon emitted when an electron falls from third to ground level?
A. ${\text{1025}}\,\,{\buildrel _{\circ} \over {\mathrm{A}}}$
B. ${\text{2050}}\,{\buildrel _{\circ} \over {\mathrm{A}}}$
C. ${\text{1320}}\,{\buildrel _{\circ} \over {\mathrm{A}}}$
D. ${\text{2460}}\,{\buildrel _{\circ} \over {\mathrm{A}}}$

Answer 1

We will determine the value of Rydberg constant by using the wavelength formula. Then by using the same formula and value of the calculated Rydberg constant we will determine the wavelength of the photon for the transition from third to ground level.
Formula used: $\dfrac{{\text{1}}}{{{\lambda }}} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_{{\text{Lower}}}^{\text{2}}}} - \dfrac{{\text{1}}}{{{\text{n}}_{{\text{higher}}}^2}}} \right]$

Complete answer
${\buildrel _{\circ} \over {\mathrm{A}}}$
The formula to determine the wavelength of transitions in hydrogen spectrum is as follows:
$\dfrac{{\text{1}}}{{{\lambda }}} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{\text{1}}}{{{\text{n}}_{{\text{Lower}}}^{\text{2}}}} - \dfrac{{\text{1}}}{{{\text{n}}_{{\text{higher}}}^2}}} \right]$
where,
${{\lambda }}$ is the wavelength.
${{\text{R}}_{\text{H}}}$ is the Rydberg constant.
The value of the Rydberg constant is .
${{\text{n}}_{{\text{lower}}}}$ is the principle quantum of the energy level in which the electron comes.
${{\text{n}}_{{\text{higher}}}}$ is the principle quantum of the energy level from which the electron comes.

Determine the Rydberg constant. Value as follows:
For the transition ${{\text{n}}_{\text{5}}}\, \to {{\text{n}}_{\text{3}}}$,
Substitute ${\text{12818}}{{\buildrel _{\circ} \over {\mathrm{A}}}}$ for wavelength, $3$ for ${{\text{n}}_{{\text{lower}}}}$ and $5$ for ${{\text{n}}_{{\text{higher}}}}$.
$\dfrac{{\text{1}}}{{12818\,{\buildrel _{\circ} \over {\mathrm{A}}}}} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{\text{1}}}{{{3^2}}} - \dfrac{{\text{1}}}{{{5^2}}}} \right]$
$\Rightarrow \dfrac{{\text{1}}}{{12818}} = \,{{\text{R}}_{\text{H}}}\left[ {\dfrac{{25 - 9}}{{225}}} \right]$
$\Rightarrow {{\text{R}}_{\text{H}}}\,\, = \,\dfrac{{225}}{{12818\,{\buildrel _{\circ} \over {\mathrm{A}}}\, \times \,16}}$
$\Rightarrow {{\text{R}}_{\text{H}}}\,\, = \dfrac{1.1 \times {10^{ - 3}}\,}{{\buildrel _{\circ} \over {\mathrm{A}}}}$
So, the Rydberg constant is $\dfrac{1.1 \times {10^{ - 3}}}{{\buildrel _{\circ} \over {\mathrm{A}}}}$.
For the transition ${{\text{n}}_3}\, \to {{\text{n}}_1}$,
Substitute for Rydberg constant, $1$ for ${{\text{n}}_{{\text{lower}}}}$ and $3$ for ${{\text{n}}_{{\text{higher}}}}$.
$\Rightarrow \dfrac{{\text{1}}}{{{\lambda }}} = \dfrac{1.1 \times {10^{ - 3}}\,}{{\buildrel _{\circ} \over {\mathrm{A}}}}\times\left[ {\dfrac{{\text{1}}}{{{3^2}}} - \dfrac{{\text{1}}}{{{1^2}}}} \right]$
$\Rightarrow \dfrac{{\text{1}}}{{{\lambda }}} = \dfrac{1.1 \times {10^{ - 3}}\,}{{\buildrel _{\circ} \over {\mathrm{A}}}}\times \left[ {\dfrac{{9 - 1}}{9}} \right]$
$\Rightarrow \dfrac{{\text{1}}}{{{\lambda }}} = \dfrac{1.1 \times {10^{ - 3}}}{{\buildrel _{\circ} \over {\mathrm{A}}}}\times \dfrac{8}{9}\,$
$\Rightarrow \dfrac{{\text{1}}}{{{\lambda }}} = \,\,9.75 \times {10^{ - 4}}/{\buildrel _{\circ} \over {\mathrm{A}}}$
$\Rightarrow {{\lambda }} = 1025\,{\buildrel _{\circ} \over {\mathrm{A}}}$
So, the wavelength of the photon emitted when electron falls from third to ground level is ${\text{1025}}\,\,{\buildrel _{\circ} \over {\mathrm{A}}}$.

Therefore, option (A) ${\text{1025}}\,\,{\buildrel _{\circ} \over {\mathrm{A}}}$ is correct.

Note: The relation between energy and wavelength is as follows: ${\text{E = }}\dfrac{{{\text{hc}}}}{{{\lambda }}}$ . Wavelength and energy are inversely proportional. Wavelength of ${{\text{n}}_{\text{5}}}\, \to {{\text{n}}_{\text{3}}}$transition is ${\text{12818}}\,\,{\buildrel _{\circ} \over {\mathrm{A}}}$ whereas the wavelength of ${{\text{n}}_3}\, \to {{\text{n}}_1}$transition is ${\text{1025}}\,{\buildrel _{\circ} \over {\mathrm{A}}}$ means high energy is required for ${{\text{n}}_3}\, \to {{\text{n}}_1}$ transition because lower energy are more stable. In hydrogen, the lines arise when electron comes from third energy level to lower energy level of Principle quantum number one ${{\text{n}}_3}\, \to {{\text{n}}_1}$ is known as Lyman series. The lines arise when electron comes from fifth energy level to lower energy level of Principle quantum number three ${{\text{n}}_{\text{5}}}\, \to {{\text{n}}_{\text{3}}}$ is known as Paschen series.