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

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The radius of $n^{th}$ orbit of H-like particles is given by:

$=0.529\times\frac{n^2}{Z} \AA$ Or $=52.9\times\frac{n^2}{Z}\ pm$

Here, starting radius, $r_{1} = 1.3225\ nm = 1322.5\ pm = 52.9n_{1}^2$

Ending radius, $r_{2} = 211.6\ pm = 52.9\left (\frac{n_{2}^2}{Z} \right )$

Therefore, $\frac{r_{1}}{r_{2}} = \frac{1322.5\ pm}{211.6\ pm} = \frac{n_{1}^2}{n_{2}^2}$

$\Rightarrow \frac{n_{1}^2}{n_{2}^2} = 6.25$

$\Rightarrow \frac{n_{1}}{n_{2}} = 2.5$

If $n_{2} = 2$ and $n_{1} = 5$ , then the transition is from $5^{th}$ orbit to $2^{nd}$ orbit.

Therefore, it belongs to the Balmer Series.

Frequency $\nu$ is given by:

$= 1.097\times10^7 m^{-1}\left ( \frac{1}{2^2}-\frac{1}{5^2} \right )$

$= 1.097\times10^7\left (\frac{21}{100} \right )m^{-1}$

Wavelength :

$\lambda = \frac{1}{\nu} = \frac{100}{1.097\times21\times10^7}m = 434\times10^{-9}m = 434\ nm$ .

Therefore, it lies in the visible range.

Posted by

Divya Prakash Singh

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