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Question: The shortest wavelength of \( He \) atom in Balmer series is \( x, \) the longest wavelength in the ...

The shortest wavelength of HeHe atom in Balmer series is x,x, the longest wavelength in the Paschen series of Li+2 L{{i}^{+2}}~ is:
(A) 36x2\dfrac{36x}{2}
(B) 16x7\dfrac{16x}{7}
(C) 9x5\dfrac{9x}{5}
(D) 5x9\dfrac{5x}{9}

Explanation

Solution

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 balmer series is from n1=3→n2=∞{{n}_{1}}=3\to {{n}_{2}}=\infty The longest wavelength in the Paschen series of helium ions is from n1=3→n2=4{{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.
The wavelength of H-like an atom is given by
1λ=RZ2(1nl2−1nn2)\dfrac{1}{\lambda }=R{{Z}^{2}}\left( \dfrac{1}{n_{l}^{2}}-\dfrac{1}{n_{n}^{2}} \right)
The shortest wavelength in hydrogen atom of Lyman series is from n1=3→n2=∞{{n}_{1}}=3\to {{n}_{2}}=\infty Given that the shortest wavelength of hydrogen atom in the Lyman series is A. Substitute the values in the formula as shown below. For shortest wavelength x of Paschen series n1=3→n2=∞{{n}_{1}}=3\to {{n}_{2}}=\infty
1λ=R(3)2(132−1∞2)\dfrac{1}{\lambda }=R{{\left( 3 \right)}^{2}}\left( \dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right)
⇒x=99R=1R\Rightarrow x=\dfrac{9}{9R}=\dfrac{1}{R}
For longest wavelength of Paschen series n1=3→n2=4{{n}_{1}}=3\to {{n}_{2}}=4
1λ=R(3)2(132−142)\dfrac{1}{\lambda }=R{{\left( 3 \right)}^{2}}\left( \dfrac{1}{{{3}^{2}}}-\dfrac{1}{{{4}^{2}}} \right)
1λ=R(3)2(16−916×9)\dfrac{1}{\lambda }=R{{\left( 3 \right)}^{2}}\left( \dfrac{16-9}{16\times 9} \right)
⇒λ=167R=16x7\Rightarrow \lambda =\dfrac{16}{7R}=\dfrac{16x}{7}
Therefore, the correct answer is option B.

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