Answer to Question #103796 in Atomic and Nuclear Physics for sridhar

Question

In a system a particle A of mass m and charge -2q is moving in the nearest orbit around a very heavy particle B having charge +q .Assuming Bohr's model of the atom to be applicable to this system the orbital angular velocity of the particle A is

Answer: (2πmq^4)/(€²h^3)

Answer 1

$F=ma$

$\frac{2q^2}{4\pi \epsilon_0r^2}=m\frac{v^2}{r}\to m^2v^2=\frac{2mq^2}{4\pi \epsilon_0r}$

According to Bohr's model of the atom

$mvr=n\frac{h}{2\pi}$. In our case $n=1$

$\frac{2mq^2}{4\pi \epsilon_0r}=\frac{h^2}{4\pi^2r^2}\to r=\frac{\epsilon_0h^2}{2\pi mq^2}$

$v=\frac{h}{2\pi mr}=\frac{2\pi hmq^2}{2\pi m\epsilon_0h^2}=\frac{q^2}{\epsilon_0h}$

So, we have

$v=\omega r\to \omega=\frac{v}{r}=\frac{q^2}{\epsilon_0h}\cdot \frac{2\pi mq^2}{\epsilon_0h^2}=\frac{2\pi mq^4}{\epsilon^2_0h^3}$ Answer.