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Question

Physics Question on Atoms

The shortest wavelength in Lyman series is 91.2nm91.2\,nm. The longest wavelength of the series is

A

121.6nm121.6\,nm

B

182.4nm182.4\,nm

C

243.4nm243.4\,nm

D

364.8nm364.8\,nm

Answer

121.6nm121.6\,nm

Explanation

Solution

The wavelength (λ)(\lambda) of lines is given by 1λ=R(1121n2)\frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right) For Lyman series, the shortest wavelength is for n=n=\infty and longest is for n=2n=2 1λS=R(112)\therefore \frac{1}{\lambda_{S}}=R\left(\frac{1}{1^{2}}\right) \ldots (i) 1λL=R(11122)=34R\frac{1}{\lambda_{L}}=R\left(\frac{1}{1}-\frac{1}{2^{2}}\right)=\frac{3}{4} R \ldots (ii) Dividing E (ii) by E (i), we get λLλS=43\frac{\lambda_{L}}{\lambda_{S}}=\frac{4}{3} Given, λS=91.2nm\lambda_{S}=91.2 \,n m λL=91.2×43\Rightarrow \lambda_{L}=91.2 \times \frac{4}{3} =121.6nm=121.6\, nm