What is the ratio of the wavelengths emitted | vedclass

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(C) The Rydberg formula for the wavelength $\lambda$ of emitted radiation is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_

Vedclass · Accepted Answer

$4:9:36$

Vedclass · Answer

(C) The Rydberg formula for the wavelength $\lambda$ of emitted radiation is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the transition $2 	o 1$,the term $\left( \frac{1}{1^2} - \frac{1}{2^2} ight) = \left( 1 - \frac{1}{4} ight) = \frac{3}{4}$ is constant for all three species.Therefore,$\lambda \propto \frac{1}{Z^2}$.The atomic numbers are $Z_{Li} = 3$,$Z_{He} = 2$,and $Z_{H} = 1$.Thus,the ratio of wavelengths is $\lambda_{Li} : \lambda_{He^+} : \lambda_{H} = \frac{1}{3^2} : \frac{1}{2^2} : \frac{1}{1^2}$.This simplifies to $\frac{1}{9} : \frac{1}{4} : 1$.Multiplying by $36$ to clear the denominators,we get $4 : 9 : 36$.

What is the ratio of the wavelengths emitted during the $2 \to 1$ transition in $Li^{++}$,$He^+$,and $H$?

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