Question

For a hypothetical hydrogen like atom, the potential energy of the system is given by $\text{U} \left(\text{r}\right) = \frac{- \left(\text{Ke}\right)^{2}}{\left(\text{r}\right)^{3}}$ , 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 :

• A. $\text{v} = \frac{\text{n}^{2} \text{h}^{3}}{\text{Ke}^{2} 8 \pi ^{3} \text{m}^{2}}$
• B. $\text{v} = \frac{\text{n}^{3} \text{h}^{3}}{8 \text{Ke}^{2} \pi ^{3} \text{m}^{2}}$
• C. $\text{v} = \frac{\text{n}^{3} \text{h}^{3}}{2 4 \text{Ke}^{2} \pi ^{3} \text{m}^{2}}$
• D. $\text{v} = \frac{\text{n}^{2} \text{h}^{3}}{2 4 \text{Ke}^{2} \pi ^{3} \text{m}^{2}}$

Answer 1

Answer: (C)

$\frac{\text{d} \left[\text{U} \left(\text{r}\right)\right]}{\text{dr}} = \frac{3 \left(\text{Ke}\right)^{2}}{\left(\text{r}\right)^{4}} ⇒ \text{Magnitude of the centripetal force (= Electrostatic force)}$
$∴ \, \, \frac{3 \text{Ke}^{2}}{\text{r}^{4}} = \frac{\text{mv}^{2}}{\text{r}}$
and we know $\text{mvr} = \frac{\text{nh}}{2 \pi }$ or $\text{r} = \frac{\text{nh}}{2 \pi \text{m} · \text{v}}$
$3 \text{Ke}^{2} \times \frac{8 \pi ^{3} \text{m}^{3} \text{v}^{3}}{\text{n}^{3} \text{h}^{3}} = \text{mv}^{2} , v = \frac{\text{n}^{3} \text{h}^{3}}{2 4 \text{Ke}^{2} \pi ^{3} \text{m}^{2}}$