Atomic Structure Result Question 3-1
3. The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about $9$ . The spectral series are
(2019 Main, 10 April II)
(a) Lyman and Paschen
(b) Brackett and Pfund
(c) Paschen and Pfund
(d) Balmer and Brackett
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Answer:
Correct Answer: 3. ( a )
Solution:
- According to Rydberg’s equation,
$ \dfrac{1}{\lambda}=\dfrac{R_{\mathrm{H}}}{h c}\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right) \quad \text { or } \quad \dfrac{1}{\lambda} \propto\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right) $
For shortest wavelength, i.e. highest energy spectral line, $n _2$ will be $(\infty )$.
For the given spectral series, ratio of the shortest wavelength of two spectral series can be calculated as follows :
(a) $\dfrac{\lambda_{\mathrm{L}}}{\lambda_{\mathrm{P}}}=\dfrac{\dfrac{1}{3^2}-\dfrac{1}{\infty^2}}{\dfrac{1}{1^2}-\dfrac{1}{\infty^2}}=\dfrac{\dfrac{1}{9}-0}{1-0}=\dfrac{1}{9}$
(b) $\dfrac{\lambda_{\mathrm{Bk}}}{\lambda_{\mathrm{Pf}}}=\dfrac{\dfrac{1}{5^2}-\dfrac{1}{\infty^2}}{\dfrac{1}{4^2}-\dfrac{1}{\infty^2}}=\dfrac{1}{25} \times \dfrac{16}{1}=\dfrac{16}{25}$
(c) $\dfrac{\lambda_{\mathrm{P}}}{\lambda_{\mathrm{Pf}}}=\dfrac{\dfrac{1}{5^2}-\dfrac{1}{\infty^2}}{\dfrac{1}{3^2}-\dfrac{1}{\infty^2}}=\dfrac{1}{25} \times \dfrac{9}{1}=\dfrac{9}{25}$
(d) $\dfrac{\lambda_{\mathrm{B}}}{\lambda_{\mathrm{Bk}}}=\dfrac{\dfrac{1}{4^2}-\dfrac{1}{\infty^2}}{\dfrac{1}{2^2}-\dfrac{1}{\infty^2}}=\dfrac{1}{16} \times \dfrac{4}{1}=\dfrac{1}{4}$
Note Lyman $= \mathrm{L}\left(\begin{array}{ll}n _1 =1\end{array}\right)$, Balmer $= \mathrm{B}\left(\begin{array}{ll}n _1 = 2\end{array}\right)$
Paschen $P\left(\begin{array}{ll}n _1 = 3\end{array}\right)$, Brackett $= \operatorname{Bk}\left(n _1 = 4\right)$
Pfund $\operatorname{Pf}\left(n _1 = 5\right)$
BETA
