For a hydrogen atom, lambda_1 and lambda_2 ar | vedclass

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(B) The wavelength $\lambda$ for a transition from energy level $n_i$ to $n_f$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}

Vedclass · Accepted Answer

$27$

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(B) The wavelength $\lambda$ for a transition from energy level $n_i$ to $n_f$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$.For transition $1$ ($n_i = 3$ to $n_f = 1$):$\frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = R \left( 1 - \frac{1}{9} ight) = R \left( \frac{8}{9} ight) \implies \lambda_1 = \frac{9}{8R}$.For transition $2$ ($n_i = 2$ to $n_f = 1$):$\frac{1}{\lambda_2} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = R \left( \frac{3}{4} ight) \implies \lambda_2 = \frac{4}{3R}$.The ratio $\frac{\lambda_1}{\lambda_2} = \frac{9/8R}{4/3R} = \frac{9}{8} 	imes \frac{3}{4} = \frac{27}{32}$.Given that the ratio is $x/32$, we have $x/32 = 27/32$, which implies $x = 27$.

For a hydrogen atom, $\lambda_1$ and $\lambda_2$ are the wavelengths corresponding to the transitions $1$ and $2$ respectively, as shown in the figure. The ratio of $\lambda_1$ and $\lambda_2$ is $x/32$. The value of $x$ is

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