The wavelength of radiation emitted is lambda | vedclass

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(D) According to Rydberg's formula,$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the first case,the electron jumps from

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$27$

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(D) According to Rydberg's formula,$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the first case,the electron jumps from the $2^{	ext{nd}}$ excited state $(n_2 = 3)$ to the $1^{	ext{st}}$ excited state $(n_1 = 2)$:$\frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{5}{36} ight)$.For the second case,the electron jumps from the $3^{	ext{rd}}$ excited state $(n_2 = 4)$ to the $2^{	ext{nd}}$ orbit $(n_1 = 2)$:$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{4} - \frac{1}{16} ight) = R \left( \frac{3}{16} ight)$.Dividing the two equations:$\frac{\lambda_0}{\lambda} = \frac{R(5/36)}{R(3/16)} = \frac{5}{36} 	imes \frac{16}{3} = \frac{5 	imes 4}{9 	imes 3} = \frac{20}{27}$.Therefore,$\lambda = \frac{27}{20} \lambda_0$. Wait,re-evaluating the ratio: $\frac{1}{\lambda} = \frac{3}{16}R$ and $\frac{1}{\lambda_0} = \frac{5}{36}R$. $\frac{\lambda}{\lambda_0} = \frac{5/36}{3/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{20}{27}$.Thus,$\lambda = \frac{20}{27} \lambda_0$. Comparing with $\frac{20}{x} \lambda_0$,we get $x = 27$.

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