A) \[{{10}^{20}}\]
B) \[{{10}^{19}}\]
C) \[{{10}^{17}}\]
D) \[{{10}^{15}}\]
Correct Answer: D
Solution :
\[mvr=\frac{h}{2\pi }\](for first orbit) \[\Rightarrow m\omega {{r}^{2}}=\frac{h}{2\pi }\Rightarrow m\times 2\pi \nu \times {{r}^{2}}=\frac{h}{2\pi }\]\[\Rightarrow \nu =\frac{h}{4{{\pi }^{2}}m{{r}^{2}}}\] \[=\frac{6.6\times {{10}^{-34}}}{4{{(3.14)}^{2}}\times 9.1\times {{10}^{-31}}\times {{(0.53\times {{10}^{-10}})}^{2}}}\]\[=6.5\times {{10}^{15}}\frac{rev}{sec}\]
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