The wavelength of H_{alpha} line of Balmer se | vedclass

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(B) The Rydberg formula for the Balmer series $(n_1 = 2)$ is given by $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{n_2^2} \right)$.For th

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$X \ \frac{80}{108} \ \mathring{A}$

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(B) The Rydberg formula for the Balmer series $(n_1 = 2)$ is given by $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{n_2^2} \right)$.For the $H_{\alpha}$ line,$n_2 = 3$: $\frac{1}{X} = RZ^2 \left( \frac{1}{4} - \frac{1}{9} \right) = RZ^2 \left( \frac{5}{36} \right) \implies RZ^2 = \frac{36}{5X}$.For the $H_{\beta}$ line,$n_2 = 4$: $\frac{1}{\lambda_{\beta}} = RZ^2 \left( \frac{1}{4} - \frac{1}{16} \right) = RZ^2 \left( \frac{3}{16} \right)$.Substituting $RZ^2$: $\frac{1}{\lambda_{\beta}} = \left( \frac{36}{5X} \right) \left( \frac{3}{16} \right) = \frac{108}{80X}$.Therefore,$\lambda_{\beta} = X \left( \frac{80}{108} \right) \ \mathring{A}$.

The wavelength of $H_{\alpha}$ line of Balmer series is $X \ \mathring{A}$. What is the wavelength of $H_{\beta}$ line of Balmer series?

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