For a hydrogen atom,lambda_1 and lambda_2 are | vedclass

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Question

(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = RZ^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right

Vedclass · Accepted Answer

$27$

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(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = RZ^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For a hydrogen atom,$Z = 1$.Transition $1$ is from $n = 3$ to $n = 1$. Thus,$\frac{1}{\lambda_1} = R(1)^2 \left[ \frac{1}{1^2} - \frac{1}{3^2} ight] = R \left[ 1 - \frac{1}{9} ight] = \frac{8R}{9}$.Transition $2$ is from $n = 2$ to $n = 1$. Thus,$\frac{1}{\lambda_2} = R(1)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = R \left[ 1 - \frac{1}{4} ight] = \frac{3R}{4}$.Now,find the ratio $\frac{\lambda_1}{\lambda_2}$:$\frac{\lambda_1}{\lambda_2} = \frac{\lambda_1}{1} 	imes \frac{1}{\lambda_2} = \left( \frac{9}{8R} ight) 	imes \left( \frac{3R}{4} ight) = \frac{27}{32}$.Given that $\frac{\lambda_1}{\lambda_2} = \frac{x}{32}$,we have $\frac{x}{32} = \frac{27}{32}$.Therefore,$x = 27$.

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 $\frac{x}{32}$. The value of $x$ is $..........$

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