Question

Calculate the wavelength of $1\mathrm{st}$ and $2\mathrm{nd}$ line in the Paschen series.

Answer 1

Step 1: Given

Step 1: Formula used

The wave number of the Paschen series is given by,
$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{3^2}-\frac{1}{{\mathrm{n}}^2}\right)$

$\mathrm{\lambda }=$wavelength

$\mathrm{R}=$Rydberg constant

$\mathrm{n}=$orbit number

The value of the rydberg constant is $\mathrm{R}=1.1\times {10}^7{\mathrm{m}}^{-1}$

Step 2: Determine the wavelength of the Paschen series

In the hydrogen spectrum, various lines can be seen in the spectrum due to the change in the wavelength of the photons emitted from the atom.

This is called the Paschen series when the electron jumps from a higher orbit to the third orbit and reaches a line in the spectrum.

The first line in the Paschen series is the line when the electron jumps from the $4$th to the $3$ rd.

The second line of the orbit and the Passhen series is the line found when the electron jumps from the $5$th orbit to $3$rd orbit.

Currently, the wavenumber of the Paschen series is given by the following equation,

$\frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{3^2}-\frac{1}{{\mathrm{n}}^2}\right)$

So, putting the value for first-line $\mathrm{n}=4$ we have,
$\frac{1}{{\mathrm{\lambda }}_1}=\mathrm{R}\left(\frac{1}{3^2}-\frac{1}{4^2}\right)$

$\frac{1}{{\mathrm{\lambda }}_1}=1.1\times {10}^7\left(\frac{1}{3^2}-\frac{1}{4^2}\right)$
Upon simplifying we have,
${\mathrm{\lambda }}_1=\frac{144}{7\times 1.1}\times {10}^{-7}$

${\lambda }_1=1870\mathrm{nm}$
Now, putting the value for the second line $\mathrm{n}=5$ we have,
$\frac{1}{{\mathrm{\lambda }}_2}=1.1\times {10}^7\left(\frac{1}{3^2}-\frac{1}{5^2}\right)$

${\lambda }_2=1258\mathrm{nm}$
Hence, the wavelength of the first lines of the Paschen series is $1870\mathrm{nm}$ and the wavelength for the second lines is $1258\mathrm{nm}$.