A) \[10\,\,m{{s}^{-\,1}}\]
B) \[2\times {{10}^{-2}}\,m{{s}^{-}}^{1}\]
C) \[4\,m{{s}^{-1}}\]
D) \[8\,\,\times \,\,{{10}^{2}}\,m{{s}^{-1}}\]
Correct Answer: C
Solution :
The Hydrogen atom before the transition was at rest. Therefore from conservation of momentum. \[{{P}_{H}}-atom={{P}_{photon}}\,=\,\,\frac{{{E}_{radicated}}}{c}\] \[=\,\,\frac{13.6\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)eV}{c}\] \[1.6\,\,\times \,\,{{10}^{-27}}\,\times \,\,\nu \,\,=\,\,\frac{13.6\left( \frac{1}{{{1}^{2}}}-\frac{1}{{{5}^{2}}} \right)\,\,1.6\times {{10}^{-19}}}{3\times {{10}^{8}}}\] \[\Rightarrow \,\,\,\nu =4.352\,m/s \approx 4 m/s.\]
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