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. Find the value of Z.

Answer 1

Hint: For the balmer series we will find the value of shortest wavelength for the balmer series. We will also find the shortest wavelength value for the bracket series of hydrogen-like atoms. As said in question both the wavelengths are equal we will equate both the wavelengths and hence obtain the answer.

Formula used:
Formula to calculate wavelength is
$\dfrac{1}{\lambda }=R{{Z}^{2}}\left( \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right)$

Complete step by step answer:
First, we will calculate the value of the shortest wavelength for the balmer series in a hydrogen like atom.
For hydrogen atom to calculate wavelength for balmer series we will have

$\begin{align}
  & {{n}_{1}}=2\text{ and }{{n}_{2}}=\infty \\
 & \dfrac{1}{{{\lambda }_{1}}}=R\left( \dfrac{1}{{{2}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right) \\
 & =\dfrac{1}{{{\lambda }_{1}}}=\dfrac{R}{4} \\
\end{align}$

Now, for hydrogen like atom of atomic no z we will calculate shortest wavelength for brackett series

$\begin{align}
  & {{n}_{1}}=4\text{ and }{{n}_{2}}=\infty \\
 & \dfrac{1}{{{\lambda }_{2}}}=R{{Z}^{2}}\left( \dfrac{1}{{{4}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right) \\
 & =\dfrac{1}{{{\lambda }_{2}}}=\dfrac{R{{Z}^{2}}}{16} \\
\end{align}$

According to the question both the wavelengths are equal.
Hence we will equate both the wavelengths .
$\begin{align}
  & \dfrac{R}{4}=\dfrac{R{{Z}^{2}}}{16} \\
 & \Rightarrow {{Z}^{2}}=4 \\
 & \Rightarrow Z=2 \\
\end{align}$
Hence from the above we obtain the value of Z is 2.
Hence, the correct answer is option Z=2.

Additional Information
Hydrogen atom is the simplest atomic system found in nature, thus it produces the only of those series. When the beam of sunshine or any radiation is formed to enter the device through a slit, each individual component of the sunshine or radiation form images of the source. These images can be visualised when resolved under the spectroscope. The images will be in the form of parallel lines arranged next to each other with regular spacing. The lines will be apart in the higher wavelength side and they come closer gradually when moved from higher to lower wavelength side. The shortest wavelength will possess least spaced spectral lines and it's named as series limit.

Note:
The Balmer series is the name given to a series with a principal quantum no. 2 of spectral emission lines of the hydrogen atom. This in turn results from electron transitions from higher levels down to the energy level. When an electron jumps between the fourth and higher energy levels of the hydrogen atom a series of spectral lines known as bracket series are released.