What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of Hespectrum?

Asked by Pragya Singh | 1 year ago |  54

Solution :-

The wave number associated with the Balmer transition for the He+ ion (n = 4 to n = 2 ) is given by:

$$\bar{\nu}=\dfrac{1}{\lambda }=RZ^{2}(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}})$$

Where, n1 = 2 n2 = 4

$$\bar{\nu}=\dfrac{1}{\lambda }=R(2)^{2}(\dfrac{1}{4}-\dfrac{1}{16})\\ \\ =4R(\dfrac{4-1}{16})\\ \\ \bar{\nu}=\dfrac{1}{\lambda }=\dfrac{3R}{4}\\ \\ \Rightarrow \lambda =\dfrac{4}{3R}$$

As per the question, the desired transition in the hydrogen spectrum must have the same wavelength as that of Hespectrum.

$$R(1)^{2}[\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}]=\dfrac{3R}{4}\\ \\ \Rightarrow [\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}]=\dfrac{3}{4}\: \: \: \: \: …….(1)$$

This equality is true only when the value of n1 = 1 and that of n2 = 2.

The transition for n2 = 2 to n = 1 in the hydrogen spectrum would, therefore, have the same wavelength as the Balmer transition from n = 4 to n = 2 of the He+ spectrum.

Answered by Abhisek | 1 year ago

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