First slide

Electrostatic force

Question

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

Moderate

Solution

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

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