Atoms & Nuclei

JEE Main level, numerical answer, medium

135 of 239

Question 135 of 239

In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is 305A∘305A^{^{\circ}}. The corresponding difference for the Paschan series in A∘A^{^{\circ}} is:________

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Answer: 10553.14

1λmin⁡=R(1)=R;n=∞\dfrac{1}{{\lambda }_{\min }}=R\left(1\right)=R ; n=\infty to 11
1λmax⁡=R{1−14}=3R4\dfrac{1}{{\lambda }_{\max }}=R\left\{1-\dfrac{1}{4}\right\}=\dfrac{3R}{4} ; n=2n=2 to 11
⇒\Rightarrow λmax⁡−λmin⁡=43R−1R{\lambda }_{\max }-{\lambda }_{\min }=\dfrac{4}{3R}-\dfrac{1}{R}
304=13R304=\dfrac{1}{3R} ......(a)\left(a\right)
Paschen : Yλ′min⁡=R(19){Y}_{\lambda {'}_{\min }}=R\left(\dfrac{1}{9}\right)
and Yλ′max⁡=R(19−116)=7R16×9{Y}_{\lambda {'}_{\max }}=R\left(\dfrac{1}{9}-\dfrac{1}{16}\right)=\dfrac{7R}{16\times 9}
λ′max⁡−λ′min⁡=16×97R−9R=817R{{\lambda }^{'}}_{\max }-{{\lambda }^{'}}_{\min }=\dfrac{16\times 9}{7R}-\dfrac{9}{R}=\dfrac{81}{7R} ...(b)\left(b\right)
(b)(a)=x3×304=817\dfrac{\left(b\right)}{\left(a\right)}=\dfrac{x}{3\times 304}=\dfrac{81}{7}
⇒\Rightarrow x=10553.14x=10553.14

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