Solveeit Logo
Login

Question

Question: The shortest wavelength of the line in hydrogen atomic spectrum of Lyman series when \[{{R}_{H}}\]...

The shortest wavelength of the line in hydrogen atomic spectrum of Lyman series
when RH{{R}_{H}}= 109678 cm−1109678\,c{{m}^{-1}} is:
(a) 1002.7 A∘ \overset{\circ }{\mathop{A}}\,
(b) 1215.6 A∘ \overset{\circ }{\mathop{A}}\,
(c) 1127.30 A∘ \overset{\circ }{\mathop{A}}\,
(d) 911.7 A∘ \overset{\circ }{\mathop{A}}\,
(e) 1234.7 A∘ \overset{\circ }{\mathop{A}}\,

Explanation

Solution

Hint : Johannes Robert Rydberg then later in 1889, found several series of spectra that would fit a more general relationship which was similar to Balmer’s empirical formula, came to be known as the Rydberg formula. It is given by:
1λ=R[1nf2−1ni2]   ni>nf\dfrac{1}{\lambda }=R\left[ \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right]\,\,\,{{n}_{i}}>{{n}_{f}}
where, n1{{n}_{1}} and n2{{n}_{2}} = 1, 2, 3, 4, 5, ….
RR (Rydberg constant) =1.097×107m−11.097\times {{10}^{7}}{{m}^{-1}}
For the hydrogen atom, n1{{n}_{1}} = 1 corresponds to the Lyman series.

Complete step by step solution : We have been given in the question RH{{R}_{H}}= 109678 cm−1109678\,c{{m}^{-1}}
According to Rydberg formula, for Hydrogen we can write:
1λ=RH[1nf2−1ni2]  \dfrac{1}{\lambda }={{R}_{H}}\left[ \dfrac{1}{{{n}_{f}}^{2}}-\dfrac{1}{{{n}_{i}}^{2}} \right]\,\,
where, λ\lambda = wavelength of the emitted photon
RH{{R}_{H}} = Rydberg constant
nf{{n}_{f}} = final energy level of the transition
ni{{n}_{i}} = initial energy level of the transition
In our case, you have the ni=∞→nf=1{{n}_{i}}=\infty \to {{n}_{f}}=1 transition, which is part of the Lyman series.

After substituting the values, we get:
1λ=109678×[112−1∞2] \dfrac{1}{\lambda }=109678\times \left[ \dfrac{1}{{{1}^{2}}}-\dfrac{1}{{{\infty }^{2}}} \right]\,
1λ=109678×[1−0] \dfrac{1}{\lambda }=109678\times \left[ 1-0 \right]\,
λ=1109678\lambda =\dfrac{1}{109678}

& \lambda =9.117\times {{10}^{-6}}\, \\\ & \,\,\,\,\,=\,911.7\overset{\circ }{\mathop{A}}\, \\\ \end{aligned}$$ Therefore, the correct option is (d). **Note** : There are other series in the hydrogen atom also. Balmer series with $${{n}_{i}}$$ = 2, present in the ultraviolet lines. Paschen series with $${{n}_{i}}$$ = 3, present in the infrared region of the spectrum. Brackett and Pfund series corresponding to $${{n}_{i}}$$ = 4 and $${{n}_{i}}$$ = 5 respectively, present in the infrared region.