Question

The shortest wavelength of the Bracket series of a hydrogen-like atom (atomic number = $Z$) is the same as the shortest wavelength of the Balmer series of hydrogen atom. The value of $Z$ is

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is A $2$

$\frac{1}{\lambda }=Z^{2}R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Shortest wavelength = Series limit
$n_2=\infty$
$\frac{1}{\lambda }=Z^{2}R\left(\frac{1}{n_1^2}\right)$
Brackett series , $n_2=4$
Balmer series, $n_2=2$

Given, for atomic no. Z,
$Z^{2}R\left(\frac{1}{4^2}\right)=R\left(\frac{1}{2^2}\right)$
$\Rightarrow Z=2$