Question

The frequency of one of the lines in the paschen series of hydrogen atoms is \[2.33 \times {10^{14}}\] Hz. The quantum number n2 which causes this transition is:

Answer 1

Hint: the spectral lines are given by the emission spectrum. The hydrogen atom emission spectrum can be divided into various spectral lines depending upon the transfer of electrons from the one energy shell to the other, either lower or higher.

Complete answer:
There are different spectral lines of the hydrogen atom emission spectrum like – lyman series, Balmer series, paschen series, brackett Pfund series. All these series vary in terms of their electron movement between the atomic shells.
So, let’s understand the basics about the Paschen series first.
The Paschen series is displayed when electron transition takes place from higher energy states (\[{n_h}\] =\[4,5,6,7,8\] ) to \[{n_1}\] = \[3\] energy state.
Let’s have a look at the question now,
We are given the frequency of the paschen series of hydrogen atom as \[2.33 \times {10^{14}}\] Hz
Also, we know the formula that, \[\nu {\text{ = }}\dfrac{c}{\lambda }\]
We can also write it, \[\lambda {\text{ = }}\dfrac{c}{\nu }\]
Now, putting the values in the above equation we will get the wavelength of this frequency:
As, \[
  \lambda {\text{ = }}\dfrac{{3 \times {{10}^8}}}{{2.33 \times {{10}^{14}}}} \\
  \lambda = 1.288 \times {10^{ - 6}}m \\
 \]
Now as we know that for paschen series: \[{n_1}\] \[ = 3\]
\[{n_2} = ?\]
Also we know that the Rydberg’s formula, so by outing the values in the formula we will get:
\[
  \dfrac{1}{\lambda } = R{z^2}\left( {\dfrac{1}{{{n_1}^2}} - \dfrac{1}{{{n_2}^2}}} \right) \\
  \dfrac{1}{\lambda } = R{\left( 1 \right)^2}\left( {\dfrac{1}{{{3^2}}} - \dfrac{1}{{{n^2}}}} \right) \\
  776666.667 = 10967800\left( {\dfrac{{{n^2} - 9}}{{9{n^2}}}} \right) \\
  0.637 \times {n^2} = {n^2} - 9 \\
  9 = {n^2}\left( {0.363} \right) \\
  n = 5 \\
 \]
Thus by solving the equation we get the transition number as \[n = 5\]
Therefore we can say that. The quantum number n2 which causes this transition is \[5\].

Note:
As we know that hydrogen emission spectral lines are divided into many regions and each region depicts or lies in the difference in the radiation like , so, the paschen series lies in the ‘infrared radiation’ range.