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.

Asked by Abhisek | 1 year ago |  86

##### Solution :-

The radius of the n th orbit of hydrogen-like particles is given by,

$$r=\dfrac{0.529n^{2}}{Z}Å$$

$$r=\dfrac{5.29n^{2}}{Z}pm$$

For radius (r1) = 1.3225 nm

$$=1.32225\times 10^{-9}m\\ \\ =1322.25\times 10^{-12}m\\ \\ =1322.25\, pmn_{1}^{2}$$

$$=\dfrac{r_{1}Z}{52.9}\\ \\ n_{1}^{2}=\dfrac{1322.25Z}{52.9}$$

Similarly,

$$n_{2}^{2}=\dfrac{211.6Z}{52.9}\\ \\ \dfrac{n_{1}^{2}}{n_{2}^{2}}=\dfrac{1322.5}{211.6}\\ \\ \dfrac{n_{1}^{2}}{n_{2}^{2}}=6.25\\ \\ \dfrac{n_{1}}{n_{2}}=2.5\\ \\ \dfrac{n_{1}}{n_{2}}=\dfrac{25}{10}=\dfrac{5}{2}$$

⇒ n1 = 5 and n2 = 2

Therefore, the electron transition is from the 5th orbit to the 2nd orbit and it, therefore, corresponds to the Balmer series.

The wave number $$\bar{\nu}$$ of the transition is:

$$1.097\times 10^{7}(\dfrac{1}{2^{2}}-\dfrac{1}{5^{2}})m^{-1}\\ \\ =1.097\times 10^{7}m^{-1}(\dfrac{21}{100})\\ \\ =2.303\times 10^{6}m^{-1}$$

Wavelength (λ) of the emitted radiation is:

$$\lambda =\dfrac{1}{\nu }\\ \\ =\dfrac{1}{2.303\times 10^{6}m^{-1}}\\ \\ =0.434\times 10^{-6}m\, \lambda \\ \\ =434nm$$

Answered by Pragya Singh | 1 year ago

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