Question

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is \[\lambda \]. If an electron jumps from N-shell to the L-shell, the wavelength of emitted radiation will be?
A. \[\dfrac{{27}}{{20}}\lambda \]
B. \[\dfrac{{16}}{{25}}\lambda \]
C. \[\dfrac{{20}}{{27}}\lambda \]
D. \[\dfrac{{25}}{{16}}\lambda \]

Answer 1

Hint: We will use the expression for wavelength of the emitted radiation when an electron jumps one shell to another shell which gives us the relation between wavelength, Rydberg constant, quantum number for initial shell and final shell.

Complete step by step answer:Let us write the expression for the wavelength emitted when an electron jumps from M-shell to L-shell.
\[\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)\]……(1)
Here \[{n_1}\] is the quantum number of the L-shell and \[{n_2}\] is the quantum number of M-shell and R is the Rydberg constant.

We know that the quantum number for M-shell and L-shell can be written as below:
\[\begin{array}{l}
{n_1} = 2\\
{n_2} = 3
\end{array}\]

We will substitute 2 for \[{n_1}\] and 3 for \[{n_2}\] in equation (1) to get the value of R in terms of wavelength \[\lambda \].
\[\begin{array}{c}
\dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{3^2}}}} \right)\\
\dfrac{1}{\lambda } = \dfrac{5}{{36}}R\\
R = \dfrac{{36}}{5}\lambda
\end{array}\]

We can write the expression for wavelength emitted when an electron jumps from N-shell to L-shell.
\[\dfrac{1}{{\lambda '}} = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_3^2}}} \right)\]
Here \[{n_3}\] is the quantum number for N-shell and \[\lambda '\] is the wavelength emitted when an electron jumps from N-shell to L-shell.

We know that the quantum number for shell N is given as:
\[{n_3} = 4\]

We will substitute 2 for \[{n_1}\] and 4 for \[{n_2}\] in the above expression to get the value of R in terms of wavelength \[\lambda \].
\[\begin{array}{c}
\dfrac{1}{{\lambda '}} = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{4^2}}}} \right)\\
\dfrac{1}{{\lambda '}} = \dfrac{3}{{16}}R
\end{array}\]

We will substitute \[\dfrac{{36}}{5}\lambda \] for R in the above expression to get the final expression for wavelength \[\lambda '\].
\[\begin{array}{c}
\dfrac{1}{{\lambda '}} = \dfrac{3}{{16}}\left( {\dfrac{{36}}{5}\lambda } \right)\\
\lambda ' = \dfrac{{20}}{{27}}\lambda
\end{array}\]

Therefore, the wavelength of the emitted radiation when an electron jumps from N-shell to L-shell is \[\dfrac{{20}}{{27}}\lambda \], and option (C) is correct.

Note:We can remember the quantum number of different shells when an electron jumps from one shell to another shell by releasing radiation energy to solve similar kinds of problems.