Atoms & Nuclei

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Question 125 of 239

In a hydrogen spectrum λ\lambda be the wavelength of first transition line of Lyman series. The wavelength difference will be "aλa\lambda" between the wavelength of 3rd{3}^{rd} transition line of Paschen series and that of 2nd{2}^{nd} transition line of Balmer Series where a=a= _____.

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Answer: 5

Using Rydberg formula,
1λ=R(1nf2−1ni2)\dfrac{1}{\lambda }=R\left(\dfrac{1}{{n}_{f}^{2}}-\dfrac{1}{{n}_{i}^{2}}\right), where, RR is Rydberg constant and nn is quantum number.
For first line of Lyman series
1λ=R(1−14)=R(34)\dfrac{1}{\lambda }=R\left(1-\dfrac{1}{4}\right)=R\left(\dfrac{3}{4}\right)
⇒λ=43R...(1)\Rightarrow \lambda =\dfrac{4}{3R}...\left(1\right)For 3rd{3}^{rd} line of Paschen series
1λ3=R(132−162)=R9×34\dfrac{1}{{\lambda }_{3}}=R\left(\dfrac{1}{{3}^{2}}-\dfrac{1}{{6}^{2}}\right)=\dfrac{R}{9}\times \dfrac{3}{4}
For 2nd{2}^{nd} line of Balmer series
1λ2=R(122−142)=R4×34\dfrac{1}{{\lambda }_{2}}=R\left(\dfrac{1}{{2}^{2}}-\dfrac{1}{{4}^{2}}\right)=\dfrac{R}{4}\times \dfrac{3}{4}
Thus, aλ=λ3−λ2=12R−163R=203Ra\lambda ={\lambda }_{3}-{\lambda }_{2}=\dfrac{12}{R}-\dfrac{16}{3R}=\dfrac{20}{3R}
Putting equation (1)\left(1\right)
a(43R)=203R⇒a=5a\left(\dfrac{4}{3R}\right)=\dfrac{20}{3R}\Rightarrow a=5

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