| A free hydrogen atom after absorbing a photon of wavelength \[{{\lambda }_{a}}\] gets excited from the state \[n=1\] to the state \[n=4.\] Immediately after that the electron jumps to \[n=m\] state by emitting a photon of wavelength \[{{\lambda }_{e}}.\] Let the change in momentum of atom due to the absorption and the emission are \[\vartriangle {{P}_{a}}\] and \[\vartriangle {{P}_{e}},\] respectively. If \[{{\lambda }_{a}}\text{/}{{\lambda }_{e}}\,=\,\frac{1}{5}\] which of the option(s) is/ are correct? |
| [Use \[hc=1242\text{ }eVnm;\]\[1nm=~{{10}^{-9}}m,h\] and c are Plank's constant and speed of light, respectively] |
A) \[\vartriangle {{p}_{a}}/\vartriangle {{p}_{e}}=\,\frac{1}{2}\]
B) The ratio of kinetic energy of the electron in the State \[n=m\] to the state \[n=1\] is \[\frac{1}{4}\]
C) \[{{\lambda }_{e}}=\,418\,\text{nm}\]
D) \[m=2\]
Correct Answer: B , D
Solution :
| \[{{\text{E}}_{a}}=\frac{hc}{\lambda }\] |
| \[\frac{{{\lambda }_{a}}}{{{\lambda }_{e}}}=\frac{{{E}_{4}}-{{E}_{1}}}{{{E}_{4}}-{{E}_{m}}}=\frac{1-\frac{1}{16}}{\frac{1}{{{n}^{2}}}-\frac{1}{16}}=\frac{1}{5}\] |
| \[\Rightarrow \] \[m=2\] |
| \[(4\to 2){{\text{E}}_{e}}=\frac{hc}{\lambda }\] |
| \[\lambda =\frac{h}{\sqrt{2m\text{KE}}}\] |
| \[\Rightarrow \] \[{{\lambda }_{e}}=\frac{12400}{{{\text{E}}_{e}}}=\frac{12400}{2.75}\] |
| [b] \[n=2\,\to \,-\,3.4\] |
| \[n=1\to \,-\,136.\] |
| \[\therefore \] \[\frac{{{E}_{2}}}{{{E}_{1}}}=\,\frac{3.4}{13.6}=\,\frac{1}{4},\] |
Solution :
Same as above
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