The wave number of the ${{H}_{\alpha }}$ – line in balmer series of hydrogen spectrum is:
A. 5R/36
B. 3R
C. 21R/100
D. 3R/4
Answer
558.9k+ views
Hint:Balmer series is going to form in a hydrogen spectrum when an electron does transitions from third energy level to second energy level.
We can calculate the wave number of ${{H}_{\alpha }}$ – line in the balmer series of the hydrogen spectrum by using the following formula.
\[\overset{-}{\mathop{v}}\,=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]\]
$\overset{-}{\mathop{v}}\,$ = Wave number
R = Rydberg constant
${{n}_{1}}$ = lower energy level
${{n}_{2}}$ = higher energy level
Complete step-by-step answer:- In the question it is given that to find the wave number of the ${{H}_{\alpha }}$ – line in the balmer series of hydrogen spectrum.
- We know that ${{H}_{\alpha }}$ – line in the balmer series of the hydrogen spectrum is going to form when an electron does transitions from third energy level to second level.
- So, by using the following formula we can calculate the wavenumber corresponding to balmer series.
$\overset{-}{\mathop{v}}\,=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$
$\overset{-}{\mathop{v}}\,$ = Wave number
R = Rydberg constant
${{n}_{1}}$ = lower energy level in ${{H}_{\alpha }}$ – line in balmer series of hydrogen spectrum = 2
${{n}_{2}}$ = higher energy level in ${{H}_{\alpha }}$ – line in balmer series of hydrogen spectrum = 3
\[\begin{align}
& \overset{-}{\mathop{v}}\,=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right] \\
& \overset{-}{\mathop{v}}\,=R\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right] \\
& \overset{-}{\mathop{v}}\,=\dfrac{5R}{36} \\
\end{align}\]
- Therefore the wave number of the ${{H}_{\alpha }}$ – line in the balmer series of the hydrogen spectrum is 5R/36.
So, the correct option is A.
Note: Wavenumber is the reciprocal of the wavelength and wavelength is going to be denoted with a symbol $\lambda $ and wave number is going to be denoted with $\overset{-}{\mathop{v}}\,$. ${{H}_{\alpha }}$ – line in the balmer series of hydrogen spectrum appears in deep-red visible color in the hydrogen spectrum.
We can calculate the wave number of ${{H}_{\alpha }}$ – line in the balmer series of the hydrogen spectrum by using the following formula.
\[\overset{-}{\mathop{v}}\,=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]\]
$\overset{-}{\mathop{v}}\,$ = Wave number
R = Rydberg constant
${{n}_{1}}$ = lower energy level
${{n}_{2}}$ = higher energy level
Complete step-by-step answer:- In the question it is given that to find the wave number of the ${{H}_{\alpha }}$ – line in the balmer series of hydrogen spectrum.
- We know that ${{H}_{\alpha }}$ – line in the balmer series of the hydrogen spectrum is going to form when an electron does transitions from third energy level to second level.
- So, by using the following formula we can calculate the wavenumber corresponding to balmer series.
$\overset{-}{\mathop{v}}\,=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$
$\overset{-}{\mathop{v}}\,$ = Wave number
R = Rydberg constant
${{n}_{1}}$ = lower energy level in ${{H}_{\alpha }}$ – line in balmer series of hydrogen spectrum = 2
${{n}_{2}}$ = higher energy level in ${{H}_{\alpha }}$ – line in balmer series of hydrogen spectrum = 3
\[\begin{align}
& \overset{-}{\mathop{v}}\,=R\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right] \\
& \overset{-}{\mathop{v}}\,=R\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right] \\
& \overset{-}{\mathop{v}}\,=\dfrac{5R}{36} \\
\end{align}\]
- Therefore the wave number of the ${{H}_{\alpha }}$ – line in the balmer series of the hydrogen spectrum is 5R/36.
So, the correct option is A.
Note: Wavenumber is the reciprocal of the wavelength and wavelength is going to be denoted with a symbol $\lambda $ and wave number is going to be denoted with $\overset{-}{\mathop{v}}\,$. ${{H}_{\alpha }}$ – line in the balmer series of hydrogen spectrum appears in deep-red visible color in the hydrogen spectrum.
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