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

Vedclass · Accepted Answer

$n = 4$ to $n = 2$

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(B) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$.For a hydrogen atom $(Z=1)$ transitioning from $n_i=2$ to $n_f=1$,we have $\frac{1}{\lambda_0} = R(1)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R \left[ 1 - \frac{1}{4} \right] = \frac{3R}{4}$.For a $He^+$ ion $(Z=2)$,we want the wavelength to be $\lambda_0$,so $\frac{1}{\lambda_0} = R(2)^2 \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$.Equating the two expressions: $\frac{3R}{4} = 4R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right] \Rightarrow \frac{3}{16} = \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$.Testing the options,for $n_i=4$ and $n_f=2$: $\frac{1}{2^2} - \frac{1}{4^2} = \frac{1}{4} - \frac{1}{16} = \frac{4-1}{16} = \frac{3}{16}$.Thus,the transition is from $n=4$ to $n=2$.

For a hydrogen atom,when an electron jumps from $n = 2$ to $n = 1$,the wavelength of the radiation emitted is found to be $\lambda_0$. For which transition of an electron in a $He^+$ ion will the wavelength of the radiation emitted be equal to $\lambda_0$?

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