SATHEE: Atomic Structure Question 490

Atomic Structure Question 490

Question: For a hypothetical hydrogen like atom, the potential energy of the system is given by $U (r) = \frac{- k e^{2}}{r^{4}}$ 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) $\frac{n h}{16, k e π^{2}, m^{3 / 2}}$

B) $\frac{n^{2} h^{2}}{18, k^{2} e^{2} π^{2}, m^{3}}$

C) $\frac{n^{3} h^{3}}{18, k^{2} e^{2} π^{3}, m^{4}}$

D) $\frac{n^{2} h^{2}}{4 \sqrt{2}, k e π^{2} m^{3 / 2}}$

Show Answer

Answer:

Correct Answer: D

Solution:

$\frac{d [U (r)]}{d r} = \frac{4 k e^{2}}{r^{5}} = f o r c e$ $\frac{4 k e^{2}}{r^{5}} = \frac{m v^{2}}{r}$ and $m v r = \frac{n h}{2 π}$ or $r = \frac{n h}{2 π m v} \Rightarrow \frac{1}{r} = \frac{2 π m v}{n h}$ $4 k e^{2} \times \frac{1}{r^{5}} = \frac{m v^{2}}{r}$ $2 k e^{2} \times \frac{1}{r^{4}} = m v^{2}$ $2 k e^{2} \times \frac{16 π^{4} m^{4} v^{4}}{n^{4} h^{4}} = m v^{2}$ $v^{2} = n^{4} h^{4} / 32 k e^{2} π^{4} m^{3}$ $v = \frac{n^{2} h^{2}}{4 \sqrt{2} k e π^{2} m^{3 / 2}}$

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