Atoms And Nuclei Question 302
Question: If potential energy between a proton and an electron is given by $ |U|=ke^{2}/2R^{3} $ , where K is the charge of electron and R is the radius of atom, then radius of Bohr’s orbit is given by (h = Planck’s constant, k=constant)
Options:
A) $ \frac{ke^{2}m}{h^{2}} $
B) $ \frac{6{{\pi }^{2}}}{n^{2}}\frac{ke^{2}m}{h^{2}} $
C) $ \frac{2\pi }{n}\frac{ke^{2}m}{h^{2}} $
D) $ \frac{4{{\pi }^{2}}ke^{2}m}{n^{2}h^{2}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ U=-\frac{ke^{2}}{2R^{2}},F=-\frac{dU}{dR}=\frac{3ke^{2}}{2R^{4}} $ But, $ F=\frac{mv^{2}}{R}\Rightarrow \frac{mv^{2}}{R}=\frac{3ke^{2}}{2R^{4}} $ Also, $ mvR=\frac{nh}{2\pi } $ Solve to get: $ R=\frac{6{{\pi }^{2}}ke^{2}m}{n^{2}h^{2}} $
BETA
