Question

The electron in the hydrogen atom is moving with a speed of $2.3\times {10}^{6}m/s$ in an orbit of radius $0.53\mathring{A}$. Calculate the period of revolution of electron. ($\pi =3.142$)

Answer 1

$v=2.3\times {10}^{6}m/s$

$r=0.53A^o=53\times {10}^{-8}m$

$T=?$

time taken by the electron to complete one revolution$=T=\frac{circumference}{speed}$

$\therefore T=\frac{2\pi r}{v}=\frac{2\times 3.142\times 53\times {10}^{-8}}{2.3\times {10}^6}=144.8\times {10}^{-14}sec$