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The wavelength of the first Balmer line caused by a transition from the $n=3$ level to the $n=2$ level in hydrogen is $\lambda_{1}$. the wavelength of the line caused by an electronic transition from $n=5$ to $n=3$ is :
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The correct answer is:
$\frac{125}{64} \lambda_{1}$
$\begin{array}{l}
\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{1}{2^{2}}-\frac{1}{3^{2}}}{\frac{1}{3^{2}}-\frac{1}{5^{2}}}=\frac{\frac{1}{4}-\frac{1}{9}}{\frac{1}{9}-\frac{1}{25}}=\frac{125}{64} \\
\lambda_{2}=\frac{125}{64} \lambda_{1}
\end{array}$
\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{1}{2^{2}}-\frac{1}{3^{2}}}{\frac{1}{3^{2}}-\frac{1}{5^{2}}}=\frac{\frac{1}{4}-\frac{1}{9}}{\frac{1}{9}-\frac{1}{25}}=\frac{125}{64} \\
\lambda_{2}=\frac{125}{64} \lambda_{1}
\end{array}$
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