Atoms & Nuclei

JEE Main level, single correct, medium

97 of 239

Question 97 of 239

In hydrogen atom spectrum, (R→(R \rightarrow Rydberg's constant )) A. the maximum wavelength of the radiation of Lyman series is 43R\frac{4}{3 R} B. the Balmer series lies in the visible region of the spectrum C. the minimum wavelength of the radiation of Paschen series is 9R\frac{9}{R} D. the minimum wavelength of Lyman series is 54R\frac{5}{4 R} Choose the correct answer from the options given below :

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Answer: (B) A, B and C Only

A. Maximum wavelength of Lyman series (n=1n=1 to n=2n=2):
1λ=R(1−14)=3R4\frac{1}{\lambda} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4}
λ=43R\lambda = \frac{4}{3R} TRUE

B. The Balmer series lies in the visible region of the spectrum. TRUE

C. Minimum wavelength of Paschen series (n=3n=3 to n=∞n=\infty):
1λ′=R(19)\frac{1}{\lambda'} = R\left(\frac{1}{9}\right)
λ′=9R\lambda' = \frac{9}{R} TRUE

D. Minimum wavelength of Lyman series (n=1n=1 to n=∞n=\infty):
1λ=R(1)\frac{1}{\lambda} = R\left(1\right)
λ=1R\lambda = \frac{1}{R}, NOT 54R\frac{5}{4R} FALSE

So A, B and C are correct.

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