Question

In Bohr's theory of the hydrogen atom, an electron moves in a circular orbit about a proton of radius $0.53\mathring{A}$. 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

Answer: 2

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