For a hydrogen atom,lambda_1 and lambda_2 are | vedclass

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(C) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$.F

Vedclass · Accepted Answer

$27$

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(C) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} ight]$.For transition $1$ ($n_i = 3$ to $n_f = 1$): $\frac{1}{\lambda_1} = R \left[ \frac{1}{1^2} - \frac{1}{3^2} ight] = R \left[ 1 - \frac{1}{9} ight] = \frac{8}{9} R$.For transition $2$ ($n_i = 2$ to $n_f = 1$): $\frac{1}{\lambda_2} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = R \left[ 1 - \frac{1}{4} ight] = \frac{3}{4} R$.Taking the ratio of the two wavelengths:$\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{\frac{3}{4} R}{\frac{8}{9} R} = \frac{3}{4} 	imes \frac{9}{8} = \frac{27}{32}$.Given that $\frac{\lambda_1}{\lambda_2} = \frac{x}{32}$,we have $\frac{x}{32} = \frac{27}{32}$,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 $\frac{x}{32}$. The value of $x$ is:

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