A) \[10m{{s}^{-1}}\]
B) \[2\times {{10}^{-2}}m{{s}^{-1}}\]
C) \[4m{{s}^{-1}}\]
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 \[{{P}_{H-atom}}={{P}_{photon}}=\frac{{{E}_{radiated}}}{c}=\frac{13.6\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]eV}{c}\] \[1.6\times {{10}^{-27}}\times v=\frac{13.6\left[ \frac{1}{{{1}^{2}}}-\frac{1}{{{5}^{2}}} \right]\times 1.6\times {{10}^{-19}}}{3\times {{10}^{6}}}\] \[\Rightarrow \] \[v=4.352m/s=\,\,\approx 4m/s\]
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