The wavelength of the first line of the Lyman | vedclass

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(D) For the first line of the Lyman series of hydrogen $(n_1=1, n_2=2)$:$\frac{1}{\lambda_1} = R(1)^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R \

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$2$

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(D) For the first line of the Lyman series of hydrogen $(n_1=1, n_2=2)$:$\frac{1}{\lambda_1} = R(1)^2 \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R \left(1 - \frac{1}{4}\right) = \frac{3R}{4}$For the second line of the Balmer series of a hydrogen-like ion $(n_1=2, n_2=4)$:$\frac{1}{\lambda_2} = R Z^2 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R Z^2 \left(\frac{1}{4} - \frac{1}{16}\right) = R Z^2 \left(\frac{3}{16}\right)$Given that $\lambda_1 = \lambda_2$,we have $\frac{1}{\lambda_1} = \frac{1}{\lambda_2}$:$\frac{3R}{4} = \frac{3RZ^2}{16}$Dividing both sides by $\frac{3R}{4}$:$1 = \frac{Z^2}{4}$$Z^2 = 4$$Z = 2$

The wavelength of the first line of the Lyman series for a hydrogen atom is equal to that of the second line of the Balmer series for a hydrogen-like ion. The atomic number $Z$ of the hydrogen-like ion is

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