The shortest wavelength in the Balmer series | vedclass

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(A) Using Rydberg's formula,$\frac{1}{\lambda} = R_H Z^2 \left[ \frac{1}{n^2} - \frac{1}{m^2} \right]$,where $R_H$ is the Rydberg constant.For the sho

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$2$

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(A) Using Rydberg's formula,$\frac{1}{\lambda} = R_H Z^2 \left[ \frac{1}{n^2} - \frac{1}{m^2} \right]$,where $R_H$ is the Rydberg constant.For the shortest wavelength in the Balmer series of a hydrogen atom $(Z=1)$: $n=2, m=\infty$. Thus,$\frac{1}{\lambda_1} = R_H (1)^2 \left[ \frac{1}{2^2} - 0 \right] = \frac{R_H}{4}$,which gives $\lambda_1 = \frac{4}{R_H}$.For the shortest wavelength in the Brackett series of a hydrogen-like atom: $n=4, m=\infty$. Thus,$\frac{1}{\lambda_2} = R_H Z^2 \left[ \frac{1}{4^2} - 0 \right] = \frac{R_H Z^2}{16}$,which gives $\lambda_2 = \frac{16}{R_H Z^2}$.Given $\lambda_1 = \lambda_2$,we have $\frac{4}{R_H} = \frac{16}{R_H Z^2}$.Simplifying,$Z^2 = \frac{16}{4} = 4$,so $Z = 2$.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

The shortest wavelength in the Balmer series of a hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is:

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