A) \[4.22\,\times {{10}^{-3}}\,J-\sec \]
B) \[2.11\,\times {{10}^{-34}}\,J-\sec \]
C) \[3.16\,\times {{10}^{-34}}\,J-\sec \]
D) \[1.05\,\times {{10}^{-34}}\,J-\sec \]
Correct Answer: D
Solution :
Electron after absorbing 10.2 eV energy will go to its first excited state \[(n=2)\] from ground state \[(n=1)\] \[\therefore \] Increase in momentum \[=2\frac{h}{2\pi }-\frac{h}{2\pi }=\frac{h}{2\pi }\] \[=\frac{6.6\times {{10}^{-34}}}{2\times 3.14}=\frac{6.6\times {{10}^{-34}}}{6.28}\] \[=1.05\times {{10}^{-34}}\,J-s\]
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