Question

The wavenumber for the longest wavelength transition in the Blamer series of atomic hydrogen is $1.523\times {{10}^{6}}{{m}^{-1}}$ .
Enter 1 if the statement is True or 0 if False.

Answer 1

As per Balmer formula
$\overset{-}{\mathop{\text{V}}}\,=\dfrac{1}{\lambda }={{R}_{H}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$
Where ${{R}_{H}}$ = Rydberg constant
${{n}_{1}}$ = lowest energy level orbital in the atom
${{n}_{2}}$ = highest energy level orbital in the atom.
$\lambda$ = wavelength
- In Balmer series ${{n}_{1}}$ is 2 and for longest wavelength transition for hydrogen wavenumber has to be small means ${{n}_{2}}$ should be minimum then ${{n}_{2}}$ is 3.

Complete step by step answer:
- In the given question there is a statement related to the Balmer series. We have to check whether the given statement is correct or not.
- From the formula we can say that wavelength is inversely proportional to wavenumber.
- Therefore for the Balmer series ${{n}_{1}}$ is 2 and lowest wavelength is possible in the Balmer series when ${{n}_{2}}$ is 3.
- Substitute all the known values in the below formula to get the wavenumber

& \overset{-}{\mathop{\text{V}}}\,={{R}_{H}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right] \\\ & =(1.097\times {{10}^{7}})\left[ \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{3}^{2}}} \right] \\\ & =(1.097\times {{10}^{7}})\left[ \dfrac{1}{4}-\dfrac{1}{9} \right] \\\ & =(1.097\times {{10}^{7}})\left[ \dfrac{9-4}{36} \right] \\\ & =1.5236\times {{10}^{6}}/m \\\ \end{aligned}$$ \- Therefore the wavenumber for the longest wavelength transition in the Blamer series of atomic hydrogen is $1.523\times {{10}^{6}}{{m}^{-1}}$. \- The given statement is correct. **\- So, we have to enter 1 for the True statement.** **Note:** Wave number is inversely proportional to the wavelength in the transition. Therefore, for the longest wavelength transition in the Balmer series, wavenumber should be the smallest. Therefore we have to take ${{n}_{2}}$ as 3. In the Balmer series the spectral lines of the hydrogen atom are formed due to the electron transition from higher energy levels to the lower energy level having a quantum number of 2.