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Question

(C) The Rydberg formula for the wavelength of a spectral line is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

Vedclass · Accepted Answer

$36:9:4$

Vedclass · Answer

(C) The Rydberg formula for the wavelength of a spectral line is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the first Lyman line,$n_1 = 1$ and $n_2 = 2$. Thus,$\frac{1}{\lambda} = R Z^2 \left( 1 - \frac{1}{4} \right) = R Z^2 \left( \frac{3}{4} \right)$.This implies $\lambda \propto \frac{1}{Z^2}$.For Hydrogen $(Z=1)$,$He^+$ $(Z=2)$,and $Li^{2+}$ $(Z=3)$:$\lambda_1 : \lambda_2 : \lambda_3 = \frac{1}{1^2} : \frac{1}{2^2} : \frac{1}{3^2} = 1 : \frac{1}{4} : \frac{1}{9}$.Multiplying by $36$ to simplify the ratio:$\lambda_1 : \lambda_2 : \lambda_3 = 36 : 9 : 4$.

If the wavelengths of the first Lyman line for Hydrogen,$He^+$,and $Li^{2+}$ ions are $\lambda_1$,$\lambda_2$,and $\lambda_3$ respectively,then the ratio of these wavelengths is:

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