A hydrogen atom,a deuteron atom,a He^+ ion,an | vedclass

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Question

(A) The energy of a photon emitted during a transition from $n_i$ to $n_k$ is given by $\Delta E = \frac{hc}{\lambda} = 13.6 \ Z^2 \left( \frac{1}{n_k

Vedclass · Accepted Answer

$\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$

Vedclass · Answer

(A) The energy of a photon emitted during a transition from $n_i$ to $n_k$ is given by $\Delta E = \frac{hc}{\lambda} = 13.6 \ Z^2 \left( \frac{1}{n_k^2} - \frac{1}{n_i^2} ight) 	ext{ eV}$.Since $n_i = 2$ and $n_k = 1$ are constant for all cases,we have $\frac{hc}{\lambda} \propto Z^2$,which implies $\lambda \propto \frac{1}{Z^2}$ or $\lambda Z^2 = 	ext{constant}$.For Hydrogen $(Z=1)$,Deuteron $(Z=1)$,$He^+$ $(Z=2)$,and $Li^{++}$ $(Z=3)$:$\lambda_1 (1)^2 = \lambda_2 (1)^2 = \lambda_3 (2)^2 = \lambda_4 (3)^2$.Therefore,$\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$.

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