The ratio of the wavelengths of radiation emi | vedclass

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(B) The wavelength of radiation emitted from a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}

Vedclass · Accepted Answer

$20: 27$

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(B) The wavelength of radiation emitted from a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the first case ($n_2 = 4$ to $n_1 = 2$):$\frac{1}{\lambda_1} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} ight] = R \left[ \frac{1}{4} - \frac{1}{16} ight] = R \left[ \frac{4-1}{16} ight] = \frac{3R}{16}$.For the second case ($n_2 = 3$ to $n_1 = 2$):$\frac{1}{\lambda_2} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left[ \frac{9-4}{36} ight] = \frac{5R}{36}$.Now,taking the ratio of the wavelengths:$\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{5R/36}{3R/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{5 	imes 4}{9 	imes 3} = \frac{20}{27}$.Thus,the ratio is $20:27$.

The ratio of the wavelengths of radiation emitted when an electron in the hydrogen atom jumps from the $4^{th}$ orbit to the $2^{nd}$ orbit and from the $3^{rd}$ orbit to the $2^{nd}$ orbit is:

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