The wavelength of the second line of the Balm | vedclass

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(A) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.For the second line of the Balmer ser

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$\frac{\lambda}{16}$

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(A) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.For the second line of the Balmer series of the $H$-atom $(Z=1, n_1=2, n_2=4)$:$\frac{1}{\lambda} = R(1)^2 \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{4} - \frac{1}{16} \right] = \frac{3R}{16}$ $\Rightarrow R = \frac{16}{3\lambda}$.For the first line of the Lyman series of the $He^{+}$ ion $(Z=2, n_1=1, n_2=2)$:$\frac{1}{\lambda_2} = R(2)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = 4R \left[ 1 - \frac{1}{4} \right] = 4R \left( \frac{3}{4} \right) = 3R$.Substituting $R = \frac{16}{3\lambda}$ into the expression for $\lambda_2$:$\frac{1}{\lambda_2} = 3 \left( \frac{16}{3\lambda} \right) = \frac{16}{\lambda}$ $\Rightarrow \lambda_2 = \frac{\lambda}{16}$.

The wavelength of the second line of the Balmer series of a hydrogen atom is $\lambda \ nm$. What is the wavelength of the first line of the Lyman series of the $He^{+}$ ion (in $nm$)?

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