The wavelength of the first line of the Lyman | vedclass

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(B) The wavelength $\lambda$ of the first line of the Lyman series for a hydrogen atom $(Z=1)$ is given by the Rydberg formula:$\frac{1}{\lambda} = R(

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$2$

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(B) The wavelength $\lambda$ of the first line of the Lyman series for a hydrogen atom $(Z=1)$ is given by the Rydberg formula:$\frac{1}{\lambda} = R(1)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R \left[ 1 - \frac{1}{4} \right] = \frac{3R}{4}$The wavelength $\lambda'$ of the second line of the Balmer series for a hydrogen-like ion with atomic number $Z$ is given by:$\frac{1}{\lambda'} = R Z^2 \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R Z^2 \left[ \frac{1}{4} - \frac{1}{16} \right] = R Z^2 \left[ \frac{4-1}{16} \right] = \frac{3 R Z^2}{16}$According to the problem,$\lambda = \lambda'$,which implies $\frac{1}{\lambda} = \frac{1}{\lambda'}$:$\frac{3R}{4} = \frac{3 R Z^2}{16}$Dividing both sides by $\frac{3R}{4}$:$1 = \frac{Z^2}{4}$$Z^2 = 4$$Z = 2$Thus,the atomic number of the hydrogen-like ion is $2$.

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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