A) \[2.28\times {{10}^{6}}\]
B) \[22.8\times {{10}^{6}}\]
C) \[8.23\times {{10}^{6}}\]
D) \[2.82\times {{10}^{6}}\]
Correct Answer: C
Solution :
| velocity of electron, \[{{V}_{n}}=2.19\times {{10}^{6}}\frac{Z}{n}m{{s}^{-1}}\] (Z=1 for H) |
| The distance, travelled by the electron in second Bohr?s orbit in \[{{10}^{-8}}s\]\[=\frac{2.19\times {{10}^{6}}\times 1.0\times {{10}^{-8}}}{2}m=1.095\times {{10}^{-2}}m\] |
| The circumference of second orbit \[=2\pi {{r}_{2}}=2\pi \times (0.529\times {{10}^{-10}}\times {{2}^{2}})\] |
| \[=13.30\times {{10}^{-10}}m\] |
| Number of revolutions \[=\frac{1.95\times {{10}^{-2}}}{13.3\times {{10}^{-10}}}=8.23\times {{10}^{6}}\] |
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