Question

How many spectral lines in the UV region would be present during the de-excitation of an electron from the 6th orbit to the ground state in hydrogen emission spectra?

Answer 1

Hint: The number of spectral lines of the ultra-violet region can be calculated by the formula $\dfrac{({{n}_{1}}-{{n}_{2}})({{n}_{1}}-{{n}_{2}}+1)}{2}$ , where ${{n}_{1}}$ will be 6 and ${{n}_{2}}$ will be 1 because the question says from 6th orbit to the ground state.


Complete step by step answer:
 First, let us know about the hydrogen emission spectra:
The light is emitted on passing electric discharge at low pressure and hydrogen gas is taken in the discharge tube then the spectrum of hydrogen or line spectrum of hydrogen can be obtained which is examined with a spectroscope.
It is observed to have a large number of lines that are grouped into different series. These different series are named as Lyman series, Balmer series, Paschen series, Brackett series, and Pfund series. The emission spectrum is similar but that in place of dark lines, there are colored lines with dark space in between.
The Lyman series belongs to the ultraviolet region. The Balmer series belongs to the visible region. The Paschen series, Brackett series, and Pfund series belong to the infrared region.

Although a large number of lines are present in the hydrogen spectrum, Rydberg in 1890 gave a very simple theoretical equation for the calculation of the wavelength of these lines. The equation gives the calculation of the wavenumber ($\bar{v}$) of the lines by the formula:
 $\bar{v}=R\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$
Where R is a constant called Rydberg constant, and its value is equal to $109677\text{ c}{{\text{m}}^{-1}}$, ${{n}_{1}}\text{ and }{{n}_{2}}$ are whole numbers and for a particular series, ${{n}_{1}}$d is constant and ${{n}_{2}}$ varies.

So, the number of spectral lines of the ultraviolet region can be calculated by the formula $\dfrac{({{n}_{1}}-{{n}_{2}})({{n}_{1}}-{{n}_{2}}+1)}{2}$, where ${{n}_{1}}$ will be 6 and ${{n}_{2}}$ will be 1 because the question says from 6th orbit to the ground state.
So, putting the values, we get:
$\dfrac{({{n}_{1}}-{{n}_{2}})({{n}_{1}}-{{n}_{2}}+1)}{2}$
$\dfrac{(6-1)(6-1+1)}{2}$
$\dfrac{(5)\text{ x }(6)}{2}$
$\dfrac{30}{2}=15$
So, the number of spectral lines will be 15.

Note: In this question the 6th orbit was mentioned so we have used ${{n}_{1}}$ as 6, but if series is mentioned then the value of ${{n}_{1}}$ must be taken as follows:
For the Lyman series, ${{n}_{1}}=1$
For the Balmer series, ${{n}_{1}}=2$
For the Paschen series, ${{n}_{1}}=3$
For the Brackett series, ${{n}_{1}}=4$
For the Pfund series, ${{n}_{1}}=5$