In the Balmer series,the wavelength of the fi | vedclass

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(A) The wavelength of a line in the Balmer series is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$ for $n = 3, 4, 5, \

Vedclass · Accepted Answer

$0.162$

Vedclass · Answer

(A) The wavelength of a line in the Balmer series is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} ight)$ for $n = 3, 4, 5, \dots$For the first line,$n = 3$:$\frac{1}{\lambda_1} = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36} \Rightarrow \lambda_1 = \frac{36}{5R}$The wavelength of a line in the Brackett series is given by $\frac{1}{\lambda} = R \left( \frac{1}{4^2} - \frac{1}{n^2} ight)$ for $n = 5, 6, 7, \dots$For the first line,$n = 5$:$\frac{1}{\lambda_2} = R \left( \frac{1}{16} - \frac{1}{25} ight) = R \left( \frac{25-16}{400} ight) = \frac{9R}{400} \Rightarrow \lambda_2 = \frac{400}{9R}$Dividing $\lambda_1$ by $\lambda_2$:$\frac{\lambda_1}{\lambda_2} = \frac{36}{5R} 	imes \frac{9R}{400} = \frac{36 	imes 9}{5 	imes 400} = \frac{324}{2000} = 0.162$

In the Balmer series,the wavelength of the first line is $\lambda_1$ and in the Brackett series,the wavelength of the first line is $\lambda_2$. Then,the ratio $\frac{\lambda_1}{\lambda_2}$ is:

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