A free hydrogen atom after absorbing a photon | vedclass

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(A) The energy of the absorbed photon is $\frac{hc}{\lambda_{a}} = 13.6 	ext{ eV} 	imes \left[\frac{1}{1^2} - \frac{1}{4^2}ight] = 13.6 	imes \fr

Vedclass · Accepted Answer

$2, 3$

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(A) The energy of the absorbed photon is $\frac{hc}{\lambda_{a}} = 13.6 	ext{ eV} 	imes \left[\frac{1}{1^2} - \frac{1}{4^2}ight] = 13.6 	imes \frac{15}{16} 	ext{ eV} \quad (i)$The energy of the emitted photon is $\frac{hc}{\lambda_{e}} = 13.6 	ext{ eV} 	imes \left[\frac{1}{m^2} - \frac{1}{4^2}ight] \quad (ii)$Given $\frac{\lambda_{a}}{\lambda_{e}} = \frac{1}{5}$,we divide $(ii)$ by $(i)$:$\frac{\lambda_{a}}{\lambda_{e}} = \frac{13.6 	imes [1 - 1/16]}{13.6 	imes [1/m^2 - 1/16]} = 5 \implies \frac{15/16}{1/m^2 - 1/16} = 5$$\frac{15}{16} = 5 	imes \left(\frac{1}{m^2} - \frac{1}{16}ight) \implies \frac{3}{16} = \frac{1}{m^2} - \frac{1}{16}$$\frac{1}{m^2} = \frac{4}{16} = \frac{1}{4} \implies m = 2$. Thus,option $(3)$ is correct.For kinetic energy,$K_n \propto \frac{1}{n^2}$. So,$\frac{K_{m=2}}{K_{n=1}} = \frac{1/2^2}{1/1^2} = \frac{1}{4}$. Thus,option $(2)$ is correct.For momentum,$\Delta p = \frac{h}{\lambda}$. So,$\frac{\Delta p_{a}}{\Delta p_{e}} = \frac{\lambda_{e}}{\lambda_{a}} = 5$. Thus,option $(4)$ is incorrect.

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