The wavelength of the first line of the Lyman | vedclass

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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

$10$

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(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 line of the Lyman series,$n_1 = 1$ and $n_2 = 2$,so $\frac{1}{\lambda} = R Z^2 \left( \frac{3}{4} \right)$.
For hydrogen $(Z = 1)$,$\lambda_H = 1216 \ \mathring{A} = \frac{4}{3R}$.
For $10$ times ionized sodium $(Na^{10+})$,$Z = 11$. The wavelength $\lambda_{Na}$ is given by $\frac{1}{\lambda_{Na}} = R (11)^2 \left( \frac{3}{4} \right)$.
Taking the ratio: $\frac{\lambda_{Na}}{\lambda_H} = \frac{R(1)^2(3/4)}{R(11)^2(3/4)} = \frac{1}{121}$.
Therefore,$\lambda_{Na} = \frac{1216}{121} \approx 10 \ \mathring{A}$.

The wavelength of the first line of the Lyman series for hydrogen is $1216 \ \mathring{A}$. The wavelength of the first line of the same series for a $10$ times ionized sodium atom $(Z = 11)$ will be ......... $\mathring{A}$.

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