Question

If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of $$H{e}^{+}$$ is:
A. $${\displaystyle \frac{36A}{7}}$$
B. $${\displaystyle \frac{5A}{9}}$$
C. $${\displaystyle \frac{36A}{5}}$$
D. $${\displaystyle \frac{9A}{5}}$$

Answer 1

Hint: Use the Rydberg formula to solve this question which gives the relation of the energy difference between various levels of Bohr’s model and wavelength absorbed by the photon or emitted. The shortest wavelength in hydrogen atoms of the Lyman series is from $$n_1=1\to n_2=\mathrm{\infty }$$. The longest wavelength in the Paschen series of helium ions is from $$n_1=3\to n_2=4$$.

Complete step by step answer:
The spectral series are defined as the set of wavelengths arranged in a sequential fashion. The spectral series characterizes light or electromagnetic radiation which is emitted by energized atoms.
The Rydberg formula gives the relation of the energy difference between various levels of Bohr’s model and wavelength absorbed by the photon or emitted.
The Rydberg formula is given as shown below.
$${\displaystyle \frac{1}{\lambda }}=R{Z}^2\left({\displaystyle \frac{1}{n_l^2}}-{\displaystyle \frac{1}{n_h^2}}\right)$$
Where,
$$\lambda$$ is the wavelength
R is the Rydberg constant
Z is the atomic number
$$n_l$$ is the lower energy level.
$$n_h$$ is a higher energy level.
The shortest wavelength in hydrogen atom of Lyman series is from $$n_1=1\to n_2=\mathrm{\infty }$$
Given that the shortest wavelength of hydrogen atom in the Lyman series is A.
Substitute the values in the formula as shown below.
$$\Rightarrow {\displaystyle \frac{1}{A}}=R\left({\displaystyle \frac{1}{1}}-{\displaystyle \frac{1}{\mathrm{\infty }}}\right)1^2$$
$$\Rightarrow {\displaystyle \frac{1}{A}}=R$$
The longest wavelength in the Paschen series of helium ions is from $$n_1=3\to n_2=4$$.
To calculate the longest wavelength in Paschen series, substitute the values in the formula as shown below.
$$\Rightarrow {\displaystyle \frac{1}{{\lambda }^{\prime }}}=R\left({\displaystyle \frac{1}{3^2}}-{\displaystyle \frac{1}{4^2}}\right)2^2$$
$$\Rightarrow {\displaystyle \frac{1}{{\lambda }^{\prime }}}={\displaystyle \frac{4}{A}}\left({\displaystyle \frac{1}{9}}-{\displaystyle \frac{1}{16}}\right)$$
$$\Rightarrow {\displaystyle \frac{1}{{\lambda }^{\prime }}}={\displaystyle \frac{4}{A}}\times {\displaystyle \frac{7}{144}}$$
$$\Rightarrow {\lambda }^{\prime }={\displaystyle \frac{144A}{28}}$$
$$\Rightarrow {\lambda }^{\prime }={\displaystyle \frac{36A}{7}}$$

So, the correct answer is Option A.

Note: All the wavelengths in the Lyman series come under the Ultraviolet band of the electromagnetic spectrum. All the wavelength in the Paschen series comes under the infrared region of the electromagnetic spectrum.