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

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Question

(A) The $H_{\alpha}$ line of the Balmer series corresponds to the transition from $n_{2} = 3$ to $n_{1} = 2$.Using the Rydberg formula: $\frac{1}{\lam

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The $H_{\alpha}$ line of the Balmer series corresponds to the transition from $n_{2} = 3$ to $n_{1} = 2$.Using the Rydberg formula: $\frac{1}{\lambda_{\alpha}} = R \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = \frac{5R}{36}$.Thus,$\lambda_{\alpha} = \frac{36}{5R} = X$.The $H_{\beta}$ line of the Balmer series corresponds to the transition from $n_{2} = 4$ to $n_{1} = 2$.Using the Rydberg formula: $\frac{1}{\lambda_{\beta}} = R \left( \frac{1}{2^{2}} - \frac{1}{4^{2}} ight) = R \left( \frac{1}{4} - \frac{1}{16} ight) = \frac{3R}{16}$.Thus,$\lambda_{\beta} = \frac{16}{3R}$.Dividing $\lambda_{\beta}$ by $\lambda_{\alpha}$: $\frac{\lambda_{\beta}}{X} = \frac{16}{3R} 	imes \frac{5R}{36} = \frac{80}{108}$.Therefore,$\lambda_{\beta} = X \frac{80}{108} \ \mathring{A}$.

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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