Question

The shortest wavelength of the Brackett series of a hydrogen-like atom (atomic number $= Z$) is the same as the shortest wavelength of the Balmer series of hydrogen atoms. The value of $Z$ is:
(A) $2$
(B) $3$
(C) $4$
(D) $6$

Answer 1

Hint
the value of the $Z$ can be determined by using the wave length formula of the atomic spectra. The equation of the wavelength of the brackett series of hydrogen is equated with the equation of the wavelength of the Balmer series, then the value of $Z$ can be determined.
The wavelength formula for the Balmer series is given by,
$\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Where, $\lambda$ is the wavelength, $R$ is the Rydberg constant and $n$ is the number of spectrums.
The wavelength formula for the Bracket’s series is given by,
$\Rightarrow \dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right)$
Where, $\lambda$ is the wavelength, $R$ is the Rydberg’s constant, $Z$ is the atomic number and $n$ is the number of spectrums.

Complete step by step answer
Given that, The shortest wavelength so ${n_2} = 0$ for both the series.
The value of ${n_1}$ for the Balmer series is $2$
The value of ${n_1}$ for the Bracket series is $4$
By substituting the values of ${n_1}$ and ${n_2}$ in the Balmer series wavelength equation, then
$\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{{{2^2}}} - \dfrac{1}{{{\infty ^2}}}} \right)$
On further simplification in the above equation, then
$\Rightarrow \dfrac{1}{\lambda } = R\left( {\dfrac{1}{4}} \right)\,....................\left( 1 \right)$
By substituting the values of ${n_1}$ and ${n_2}$ in the Bracket series wavelength equation, then
$\Rightarrow \dfrac{1}{\lambda } = {Z^2}R\left( {\dfrac{1}{{{4^2}}} - \dfrac{1}{{{\infty ^2}}}} \right)$
On further simplification in the above equation, then
$\Rightarrow \dfrac{1}{\lambda } = {Z^2}R\left( {\dfrac{1}{{16}}} \right)\,....................\left( 2 \right)$
By equating the equation (1) and equation (2), then
$\Rightarrow R\left( {\dfrac{1}{4}} \right) = {Z^2}R\left( {\dfrac{1}{{16}}} \right)$
By cancelling the same terms in the above equation, then
$\Rightarrow \dfrac{1}{4} = \dfrac{{{Z^2}}}{{16}}$
By rearranging the terms in the above equation, then
$\Rightarrow {Z^2} = \dfrac{{16}}{4}$
On dividing the above equation, then
$\Rightarrow {Z^2} = 4$
By taking square root on both sides, then
$\Rightarrow Z = 2$
Hence, the option (A) is the correct answer.

Note
In atomic spectra, there are a number of series available and each series have separate $n$ values. But the Rydberg’s constant value is the constant value which is common for all the atomic spectra series. The wavelength equation of the Balmer series and Bracket series is equated because in the question it is given that both are the same.