The shortest wavelength in the Balmer series | vedclass

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(A) For the shortest wavelength in the Balmer series, the transition occurs from $n_{i} = \infty$ to $n_{f} = 2$.Using the Rydberg formula: $\frac{1}{

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$3646$

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(A) For the shortest wavelength in the Balmer series, the transition occurs from $n_{i} = \infty$ to $n_{f} = 2$.Using the Rydberg formula: $\frac{1}{\lambda} = R_{H} \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} ight)$.Substituting the values: $\frac{1}{\lambda} = 1.097 	imes 10^7 \left( \frac{1}{2^2} - \frac{1}{\infty^2} ight)$.$\frac{1}{\lambda} = 1.097 	imes 10^7 	imes \frac{1}{4} = 2742500 \ 	ext{m}^{-1}$.$\lambda = \frac{1}{2742500} \ 	ext{m} \approx 3.646 	imes 10^{-7} \ 	ext{m}$.Converting to $	ext{Å}$: $\lambda = 3.646 	imes 10^{-7} 	imes 10^{10} \ 	ext{Å} = 3646 \ 	ext{Å}$.

The shortest wavelength in the Balmer series of the hydrogen atom spectrum is approximately equal to (use $R_{H} = 1.097 \times 10^7 \ \text{m}^{-1}$) (in $\text{Å}$)

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