Atomic Structure Question 490
Question: For a hypothetical hydrogen like atom, the potential energy of the system is given by $ U(r)=\frac{-ke^{2}}{r^{4}} $ where r is the distance between the two particles. If Bohr’s model of quantization of angular momentum is applicable, then velocity of particle is given by
Options:
A) $ \frac{nh}{16,ke{{\pi }^{2}},{m^{3/2}}} $
B) $ \frac{n^{2}h^{2}}{18,k^{2}e^{2}{{\pi }^{2}},m^{3}} $
C) $ \frac{n^{3}h^{3}}{18,k^{2}e^{2}{{\pi }^{3}},m^{4}} $
D) $ \frac{n^{2}h^{2}}{4\sqrt{2},ke{{\pi }^{2}}{m^{3/2}}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- $ \frac{d[U(r)]}{dr}=\frac{4ke^{2}}{r^{5}}=force $ $ \frac{4ke^{2}}{r^{5}}=\frac{mv^{2}}{r} $ and $ mvr=\frac{nh}{2\pi } $ or $ r=\frac{nh}{2\pi mv}\Rightarrow \frac{1}{r}=\frac{2\pi mv}{nh} $ $ 4ke^{2}\times \frac{1}{r^{5}}=\frac{mv^{2}}{r} $ $ 2ke^{2}\times \frac{1}{r^{4}}=mv^{2} $ $ 2ke^{2}\times \frac{16{{\pi }^{4}}m^{4}v^{4}}{n^{4}h^{4}}=mv^{2} $ $ v^{2}=n^{4}h^{4}/32ke^{2}{{\pi }^{4}}m^{3} $ $ v=\frac{n^{2}h^{2}}{4\sqrt{2}ke{{\pi }^{2}}{m^{3/2}}} $
