Nuclear Physics And Radioactivity Question 31
Question: The number of revolutions per second made by an electron in the first Bohr orbit of hydrogen atom is of the order of [AMU 1995]
Options:
A) $ 10^{20} $
B) $ 10^{19} $
C) $ 10^{17} $
D) $ 10^{15} $
Show Answer
Answer:
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}}mr^{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} $
BETA
