A) about 106
B) about 1014
C) about 1016
D) about 102
Correct Answer: A
Solution :
Frequency of revolution in a Bohr's orbit is |
\[fn=\frac{-{{E}_{1}}}{h}\left( \frac{2}{{{n}^{3}}} \right)\] |
Where \[{{E}_{1}}\]=energy in \[n=1\]state. |
So, \[{{f}_{2}}=\frac{-{{E}_{1}}}{h}\times \frac{2}{{{2}^{3}}}=0.0823\times {{10}^{15}}rps\]\[=8.23\times {{10}^{14}}\,rps\] |
Number of revolutions made by electron in \[1\times {{10}^{-8}}s\]is |
\[\begin{align} & N={{f}_{2}}\times \Delta t=8.23\times {{10}^{14}}\times 1\times {{10}^{-8}} \\ & =8.23\times {{10}^{6}} \\ \end{align}\] |
Revolutions |
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