Question
Calculate the wavelength for the emission transition if it starts from the orbit having radius $1.3225nm$ and ends at $211.6pm$. Name the series to which this transition belongs and the region of the spectrum.
✓
Answer
The radius of the $n^{th}$ orbit of hydrogen-like particles is given by,
$\text{r}=\frac{0.529\text{n}^2}{\text{Z}}\mathring{\text{A}}$
$\text{r}=\frac{52.9\text{n}^2}{\text{Z}}\text{pm}$
For radius $(r_1) = 1.3225nm$
$= 1.32225 \times 10^{–9}m$
$= 1322.25 \times 10^{–12}m$
$= 1322.25pm$
$\text{n}^2_1=\frac{\text{r}_1\text{Z}}{52.9}$
$\text{n}^2_1=\frac{1322.25\text{Z}}{52.9}$
Similarly,
$\text{n}^2_2=\frac{211.6\text{Z}}{52.9}$
$\frac{\text{n}_2^1}{\text{n}_2^2}=\frac{1322.5}{211.6}$
$\frac{\text{n}^2_1}{\text{n}_2^2}=6.25$
$\frac{\text{n}_1}{\text{n}_2}=2.5$
$\frac{\text{n}_1}{\text{n}_2}=\frac{25}{10}=\frac{5}{2}$
$\Rightarrow\text{n}_1=5 \text{ and }\text{n}_2=2$
Thus, the transition is from the $5^{th}$ orbit to the $2^{nd}$ orbit. It belongs to the Balmer series.
Wave number $(\bar{\text{v}})$ for the transition is given by,
$1.097\times10^7\text{m}^{-1}\Big(\frac{1}{2^2}-\frac{1}{5^2}\Big)$
$=1.097\times10^7\text{m}^{-1}\Big(\frac{21}{100}\Big)$
$=2.303\times10^6\text{m}^{-1}$
$\therefore$ Wavelength $(\lambda)$ associated with the emission transition is given by,
$\lambda=\frac{1}{\bar{\text{v}}}$
$=\frac{1}{2.303\times10^6\text{m}^{-1}}$
$=0.434\times10^{-6}\text{m}$
$\lambda=434\text{nm}$
This transition belongs to Balmer series and comes in the visible region of the spectrum.
$\text{r}=\frac{0.529\text{n}^2}{\text{Z}}\mathring{\text{A}}$
$\text{r}=\frac{52.9\text{n}^2}{\text{Z}}\text{pm}$
For radius $(r_1) = 1.3225nm$
$= 1.32225 \times 10^{–9}m$
$= 1322.25 \times 10^{–12}m$
$= 1322.25pm$
$\text{n}^2_1=\frac{\text{r}_1\text{Z}}{52.9}$
$\text{n}^2_1=\frac{1322.25\text{Z}}{52.9}$
Similarly,
$\text{n}^2_2=\frac{211.6\text{Z}}{52.9}$
$\frac{\text{n}_2^1}{\text{n}_2^2}=\frac{1322.5}{211.6}$
$\frac{\text{n}^2_1}{\text{n}_2^2}=6.25$
$\frac{\text{n}_1}{\text{n}_2}=2.5$
$\frac{\text{n}_1}{\text{n}_2}=\frac{25}{10}=\frac{5}{2}$
$\Rightarrow\text{n}_1=5 \text{ and }\text{n}_2=2$
Thus, the transition is from the $5^{th}$ orbit to the $2^{nd}$ orbit. It belongs to the Balmer series.
Wave number $(\bar{\text{v}})$ for the transition is given by,
$1.097\times10^7\text{m}^{-1}\Big(\frac{1}{2^2}-\frac{1}{5^2}\Big)$
$=1.097\times10^7\text{m}^{-1}\Big(\frac{21}{100}\Big)$
$=2.303\times10^6\text{m}^{-1}$
$\therefore$ Wavelength $(\lambda)$ associated with the emission transition is given by,
$\lambda=\frac{1}{\bar{\text{v}}}$
$=\frac{1}{2.303\times10^6\text{m}^{-1}}$
$=0.434\times10^{-6}\text{m}$
$\lambda=434\text{nm}$
This transition belongs to Balmer series and comes in the visible region of the spectrum.
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