The ratio of the wavelengths for 2 to 1 trans | vedclass

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(C) The Rydberg formula for the wavelength $\lambda$ of a transition between energy levels $n_2$ and $n_1$ is given by $\frac{1}{\lambda} = R Z^2 \lef

Vedclass · Accepted Answer

$4:9:36$

Vedclass · Answer

(C) The Rydberg formula for the wavelength $\lambda$ of a transition between energy levels $n_2$ and $n_1$ is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Since the transition $2 	o 1$ is the same for all three,the term $\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$ is constant.Therefore,$\lambda \propto \frac{1}{Z^2}$.For $Li^{++}$,$Z = 3$,so $\lambda_{Li^{++}} \propto \frac{1}{3^2} = \frac{1}{9}$.For $He^{+}$,$Z = 2$,so $\lambda_{He^{+}} \propto \frac{1}{2^2} = \frac{1}{4}$.For $H$,$Z = 1$,so $\lambda_{H} \propto \frac{1}{1^2} = \frac{1}{1}$.Thus,$\lambda_{Li^{++}} : \lambda_{He^{+}} : \lambda_{H} = \frac{1}{9} : \frac{1}{4} : 1$.Multiplying by $36$,we get $4 : 9 : 36$.

The ratio of the wavelengths for $2 \to 1$ transition in $Li^{++}, He^{+}$ and $H$ is

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