Question

The longest wavelength of $H{e^ + }$ in Paschen series is ’m’ then shortest wavelength of $B{e^{3 + }}$ in Paschen series in terms of ‘m’ is:

Answer 1

Use Rydberg formula to calculate where the atomic number Z of helium is 2 and beryllium is 3. In the Paschen series electron transition takes place from higher energy states where n = 4,5,6,7,8…to lower energy levels where n = 3.

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 spectral series contains three different series arranged in different wavelengths. The series are Lyman series, Balmer series, Paschen series.
The Rydberg formula shows the relation of the energy difference between various levels of Bohr’s model and wavelength which is absorbed by the photon or emitted by the photon.
The Rydberg formula is given as shown below.
$\dfrac{1}{\lambda } = R{Z^2}\left( {\dfrac{1}{{n_l^2}} - \dfrac{1}{{n_h^2}}} \right)$
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 Paschen series was observed in the year 1908, by a German physicist Friedrich Paschen. The Paschen series takes place when the electron transition takes place from a higher energy state where n = 4,5,6,7,8…to a lower energy level where n = 3.
Given,
The longest wavelength of $H{e^ + }$in Paschen series is ’m’
For $H{e^ + }$, atomic number is 2.
Substitute the values in the equation.
$\Rightarrow \dfrac{1}{m} = R \times 4\left( {\dfrac{1}{9} - \dfrac{1}{{16}}} \right)$
$\Rightarrow \dfrac{1}{m} = \dfrac{{7R}}{{36}}$…….(I)
For $B{e^{3 + }}$
$\Rightarrow \dfrac{1}{{{\lambda _{B{e^ + }}}}} = R \times 16\left( {\dfrac{1}{9} - \dfrac{1}{\infty }} \right)$
$\Rightarrow \dfrac{1}{{{\lambda _{B{e^ + }}}}} = \dfrac{{16R}}{9}$…….(II)
Dividing equation (i) by (ii) as shown below.
$\Rightarrow \dfrac{{{\lambda _{B{e^{3 + }}}}}}{m} = \dfrac{{7 \times 9}}{{16 \times 36}}$
$\Rightarrow {\lambda _{B{e^{3 + }}}} = \dfrac{7}{{64}}m$

Note: The Rydberg formula is valid for only hydrogen and hydrogen like elements. It gives reasonable value only if the highest energy level is greater than the lowest energy level.