Hydrogen (H),Deuterium (D),Helium (He^+),and | vedclass

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Question

(A) The energy of a photon emitted during a transition between energy levels $n_2$ and $n_1$ in a hydrogen-like atom is given by $\Delta E = 13.6 Z^2

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 between energy levels $n_2$ and $n_1$ in a hydrogen-like atom is given by $\Delta E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight) 	ext{ eV}$.Since the transition is from $n = 2$ to $n = 1$ for all cases,the term $\left( \frac{1}{1^2} - \frac{1}{2^2} ight)$ is constant.Therefore,$\Delta E \propto Z^2$.Since $\Delta E = \frac{hc}{\lambda}$,we have $\frac{hc}{\lambda} \propto Z^2$,which implies $\lambda Z^2 = 	ext{constant}$.For $H$ $(Z=1)$,$D$ $(Z=1)$,$He^+$ $(Z=2)$,and $Li^{2+}$ $(Z=3)$:$\lambda_1 (1)^2 = \lambda_2 (1)^2 = \lambda_3 (2)^2 = \lambda_4 (3)^2$.Thus,$\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$.

Hydrogen $(H)$,Deuterium $(D)$,Helium $(He^+)$,and Lithium $(Li^{2+})$ emit radiation of wavelengths $\lambda_1, \lambda_2, \lambda_3,$ and $\lambda_4$ respectively during the transition from $n = 2$ to $n = 1$. Then:

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