Question

If the wavelength of light that is emitted from hydrogen atom when an electron falls from orbit $n=2$ to orbit $n=1$ is $122\text{ nm}$, then minimum wavelength of the series is:
A.\[9150\text{ }\overset{\circ }{\mathop{\text{A}}}\,\]
B.\[812\text{ }\overset{\circ }{\mathop{\text{A}}}\,\]
C.\[915\text{ }\overset{\circ }{\mathop{\text{A}}}\,\]
D.\[405\text{ }\overset{\circ }{\mathop{\text{A}}}\,\]

Answer 1

Hint: When the electron falls orbit $n=2$ to orbit $n=1$, we can use the formula $\dfrac{1}{\lambda }={{R}_{H}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$ to calculate the minimum wavelength of the series. Here ${{n}_{1}}$ is the orbit with lowest energy while ${{n}_{2}}$ is an orbit with highest energy level. This formula gives us a relationship between wavelength and the number of orbits.
Formula used:
$\dfrac{1}{\lambda }={{R}_{H}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$

Complete answer:
In the question it is given that when an electron falls from orbit $n=2$ to orbit $n=1$, its wavelength is $122\text{ nm}$, therefore on substituting these values known to us in the formula $\dfrac{1}{\lambda }={{R}_{H}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$ we get:
 $\begin{align}
  & \dfrac{1}{122}={{R}_{H}}\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{2}^{2}}} \right] \\
 & \Rightarrow \dfrac{1}{122}={{R}_{H}}\left[ \dfrac{1}{1}-\dfrac{1}{4} \right] \\
 & \Rightarrow \dfrac{1}{122}={{R}_{H}}\left[ \dfrac{4-1}{4} \right] \\
 & \Rightarrow \dfrac{1}{122}=\dfrac{3{{R}_{H}}}{4}-\text{equation }1 \\
\end{align}$
As asked in the question, we need to find the minimum wavelength of the series. The wavelength will be minimum when orbit ${{n}_{1}}=1$ to orbit ${{n}_{2}}=\infty $. Substituting the values in the equation we get:
$\begin{align}
  & \dfrac{1}{{{\lambda }_{\min }}}={{R}_{H}}\left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{\infty } \right] \\
 & \Rightarrow \dfrac{1}{{{\lambda }_{\min }}}={{R}_{H}}\left[ \dfrac{1}{1}-0 \right] \\
 & \Rightarrow \dfrac{1}{{{\lambda }_{\min }}}={{R}_{H}}-\text{equation }2 \\
\end{align}$
Now dividing equation $1$ by equation $2$, we get:
$\begin{align}
  & \dfrac{\dfrac{1}{122}}{\dfrac{1}{{{\lambda }_{\min }}}}=\dfrac{\dfrac{3R}{4}}{R} \\
 & \Rightarrow \dfrac{{{\lambda }_{\min }}}{122}=\dfrac{3}{4} \\
 & \Rightarrow {{\lambda }_{\min }}=\dfrac{3}{4}\times 122 \\
 & \Rightarrow {{\lambda }_{\min }}=91.5\text{ nm} \\
 & \therefore {{\lambda }_{\min }}=915\text{ }\overset{\circ }{\mathop{\text{A}}}\, \\
\end{align}$

Hence the correct option is $C$.

Additional information:
The formula used in the question i.e., the Rydberg’s formula is only applicable when hydrogen-like atoms are used to emit the light for various transitions which means that the atoms must have only one valence electron in their valence shell so that we may use Rydberg’s formula for it. The term $1/\lambda $ as stated in the formula is known as the wavenumber as a whole where $\lambda $ is the wavelength of the light.

Note:
The term ${{R}_{H}}$ is the Rydberg constant whose value is $109678$. The formula that we used to solve the question is known as the Rydberg’s formula which helps us determine the wavelength of the emitted light for various transitions between the orbits.