Question 487: Atoms & Nuclei

Question

The shortest wavelength in the Balmer series of hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen like atom of atomic number $$\mathrm{z}$$. The value of $$\mathrm{z}$$ is

Accepted Answer

Using Rydberg's formula,
$$\frac{1}{\lambda}=R_H Z^2\left[\frac{1}{n^2}-\frac{1}{m^2}ight]$$
where $$R_H$$ is the Rydberg's constant
For calculating the shortest wavelength in the Balmer series of hydrogen atom, $$\mathrm{n}=2, \mathrm{~m}=\infty$$ and $$\mathrm{Z}=1$$
$$	herefore \quad \lambda_1=\frac{4}{\mathrm{R}_{\mathrm{H}}}$$
For hydrogen like atom, the shortest wavelength is given by,
In Brackett series
$$\begin{aligned}
& \mathrm{n}=4, \mathrm{~m}=\infty \
	herefore \quad & \frac{1}{\lambda_2}=\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2\left(\frac{1}{16}ight) \
	herefore \quad \lambda_2 & =\frac{16}{\mathrm{R}_{\mathrm{H}} \cdot \mathrm{Z}^2}
\end{aligned}$$
Given: $$\lambda_1=\lambda_2$$,
$$\begin{aligned}
& 	herefore \quad \frac{4}{R_H}=\frac{16}{R_H \cdot Z^2} \Rightarrow Z^2=4 \
& 	herefore \quad Z=2
\end{aligned}$$

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Atoms & Nuclei

The shortest wavelength in the Balmer series of hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen like atom of atomic number $\mathrm{z}. The value of \mathrm{z}$ is

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