Question

Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at the one having 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Answer 1

From Bohr’s theory the radius of the nth orbit of a hydrogen atom like species is as follows.
${{r}_{n}}=\dfrac{52.9{{n}^{2}}}{Z}$
Where ${{r}_{n}}$ = radius of the nth orbit
n = nth orbit in hydrogen atom
Z = Atomic number

Complete step by step answer:
- In the question it is given that the transition starts from the orbit having a radius of 1.3225 nm and ends at 211.6 pm.
- We have to find the name of the series the transition belongs to and the region of the spectrum.
- Here ${{r}_{1}}$ = 1.3225 nm = 1322.5 pm.
- Substitute the radius value in the below equation

& {{r}_{n}}=\dfrac{52.9{{n}^{2}}}{Z} \\\ & {{r}_{1}}=1322.5pm=\dfrac{52.9n_{1}^{2}}{Z}pm\to (1) \\\ \end{aligned}$$ \- Here ${{r}_{2}}$ = 211.6 pm $${{r}_{2}}=211.6pm=\dfrac{52.9n_{2}^{2}}{Z}pm\to (2)$$ \- Now divide the equation 1 with 2. $$\begin{aligned} & \dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{1322.5pm}{211.6pm}=\dfrac{\dfrac{52.9n_{1}^{2}}{Z}}{\dfrac{52.9n_{2}^{2}}{Z}} \\\ & \dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{n_{1}^{2}}{n_{2}^{2}}=6.25 \\\ & \dfrac{{{n}_{1}}}{{{n}_{2}}}=\dfrac{5}{2} \\\ \end{aligned}$$ \- Therefore the transition is from 5th orbit to 2nd orbit and it belongs to the Balmer series. \- As per Balmer formula $$\overset{-}{\mathop{\text{V}}}\,=\dfrac{1}{\lambda }={{R}_{H}}\left[ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right]$$ Where ${{R}_{H}}$ = Rydberg constant ${{n}_{1}}$ = lowest energy level orbital in the atom ${{n}_{2}}$ = highest energy level orbital in the atom. $\lambda $ = wavelength \- Substitute all the known values in the above equation to get the wavenumber of the region. $$\begin{aligned} & \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}{{{5}^{2}}} \right] \\\ & =(1.097\times {{10}^{7}})\left[ \dfrac{1}{4}-\dfrac{1}{25} \right] \\\ & =(1.097\times {{10}^{7}})\left[ \dfrac{25-4}{100} \right] \\\ & =(1.097\times {{10}^{7}})(0.21)/m \\\ & =23.037\times {{10}^{5}}/m \\\ \end{aligned}$$ \- We calculated wavenumber, from the wavenumber we have to calculate the wavelength of the region. $$\begin{aligned} & \lambda =\dfrac{1}{Wave\text{ }number} \\\ & =\dfrac{1}{23.037\times {{10}^{5}}} \\\ & =434\times {{10}^{-9}}m \\\ & =434nm \\\ \end{aligned}$$ \- 434 nm means the transition lines are in a visible region. **Note:** The visible region starts from 400 nm and ends at 800 nm in the Electromagnetic Spectrum (EMR). The light which is having the longest wavelength in the visible region is red and the light which is having the shortest wavelength is violet in the visible region.