Atoms & Nuclei

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Question 115 of 239

If Rydberg's constant is RR, the longest wavelength of radiation in Paschen series will be α7R\dfrac{\alpha }{7R}, where α=\alpha =______.

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Answer: 144

In the Paschen radiation series, the longest wavelength corresponds to the transition between n=3n=3 and n=4n=4.
The formula for the wavelength in the given transition can be written as,
1λ=RZ2(1n12−1n22)...(1)\dfrac{1}{\lambda }=R{Z}^{2}\left(\dfrac{1}{{n}_{1}^{2}}-\dfrac{1}{{n}_{2}^{2}}\right)...\left(1\right)
From equation (1), it follows that
1λ=RZ2(132−142)=RZ2(19−116)=7RZ29×16⇒λ=1447R\dfrac{1}{\lambda }=R{Z}^{2}\left(\dfrac{1}{{3}^{2}}-\dfrac{1}{{4}^{2}}\right) =R{Z}^{2}\left(\dfrac{1}{9}-\dfrac{1}{16}\right) =\dfrac{7R{Z}^{2}}{9\times 16} \Rightarrow \lambda =\dfrac{144}{7R}
Hence, α=144\alpha =144.

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