Question

In Bohr's theory of the hydrogen atom, an electron moves in a circular orbit about a proton of radius $0.53 \AA$. Speed of electron in Bohr's orbit is about $\frac{1}{N \times 69} \times$ speed of light. The value of $N$ is____

Answer 1

Force on electron causing centripetal acceleration is electrostatic force.
$\therefore \frac{m v^{2}}{r}=\frac{K e^{2}}{r^{2}} $
$\Rightarrow v=\sqrt{\frac{K e^{2}}{m r}}$
$=\sqrt{\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{9.1 \times 10^{-31} \times 0.53 \times 10^{-10}}}=2.19 \times 10^{6} m / s$
Now, $\frac{c}{v}=\frac{2.99 \times 10^{8}}{2.19 \times 10^{6}}$
$\Rightarrow v=\frac{c}{138} \Rightarrow v=\frac{c}{2 \times 69} m / s$
Hence, $N=2$