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  • In the line of spectra of the hydrogen atom, the difference between the largest and the shortest wavelengths of the Lyman series is 304 . The corresponding diffe

In the line of spectra of the hydrogen atom, the difference between the largest and the shortest wavelengths of the Lyman series is 304 \dot{A}. The corresponding difference for the Paschan is series in\dot{A} is_________ (Give answer in closest integer)
Option: 1 10553
Option: 2 10567
Option: 3 9876
Option: 4 122234

Answers (1)

best_answer

\dpi{150} \lambda=\frac{c}{\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)}
for Lyman series
\\ \lambda_{1}=\frac{c}{\frac{1}{1^{2}}-\frac{1}{\infty^{2}}}=c(n=\infty$ to $n=1)$ \\ \\ $\lambda_{2}=\frac{c}{\frac{1}{1^{2}}-\frac{1}{2^{2}}}=\frac{4 c}{3}(n=2$ to $n=1)$\\ \\ $\Delta \lambda=\lambda_{2}-\lambda_{1}=\frac{\mathrm{c}}{3}=304 \ A^0 \\ \Rightarrow \mathrm{c}=912 \ A^0

for Paschen series
\begin{array}{l} \lambda_{1}=\frac{c}{\frac{1}{3^{2}}-\frac{1}{\infty^{2}}}=9 c \ \ (n=\infty \text { to } n=3) \\ \\ \lambda_{2}=\frac{c}{\frac{1}{3^{2}}-\frac{1}{4^{2}}}=\frac{144 c}{7} \ \ (n=4 \text { to } n=3) \\ \\ \Delta \lambda=\lambda_{2}-\lambda_{1}=\frac{144 c}{7}-9 c=\frac{81 c}{7} \\ \Delta \lambda=\frac{81 \times 912}{7} =10553.14 \ A^0\approx 10553 \ A^0 \end{array}

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