The wavelength of the first Balmer line cause | vedclass

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(B) Using the Rydberg formula for hydrogen,$\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.For the first Balmer line ($n=3$ t

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$\frac{125}{64} \lambda_1$

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(B) Using the Rydberg formula for hydrogen,$\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.For the first Balmer line ($n=3$ to $n=2$): $\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight)$.For the transition from $n=5$ to $n=3$: $\frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{5^2} ight) = R \left( \frac{1}{9} - \frac{1}{25} ight) = R \left( \frac{25-9}{225} ight) = R \left( \frac{16}{225} ight)$.Taking the ratio: $\frac{\lambda_2}{\lambda_1} = \frac{R(5/36)}{R(16/225)} = \frac{5}{36} 	imes \frac{225}{16} = \frac{5 	imes 25}{4 	imes 16} = \frac{125}{64}$.Therefore,$\lambda_2 = \frac{125}{64} \lambda_1$.

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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