Question

In a hydrogen atom, an electron of mass $9.1 \times 10^{-31}$ kg revolves about a proton in circular orbit of radius $0.53\, \mathring A$. The radial acceleration and angular velocity of electron are respectively

• A. $9 \times 10^{22} \; ms^{-2}, 4.1 \times 10^{16} \; s^{-1}$
• B. $ 4.1 \times 10^{16} \; ms^{-2}, 9 \times 10^{22} \; s^{-1}$
• C. $9 \times 10^{16} \; ms^{-2}, 4.1 \times 10^{22} \; s^{-1}$
• D. $4.1 \times 10^{22} \; ms^{-2}, 9 \times 10^{16} \; s^{-1}$

Answer 1

Answer: (A)

Given, mass of electron, $m_{e}=9.1 \times 10^{-31} kg$
and radius of circular orbit, $r=0.53 \times 10^{-10} m$
$\because$ An electron revolves about a proton in circular orbit.
So, centripetel force $=$ electrostatic force
$\frac{m_{e} v^{2}}{r} =\frac{k q_{1} q^{2}}{r^{2}} $
$m_{e} v^{2} r =k q_{1} q_{2} $
$\because v =r \omega$
$\therefore m_{e} r(r \omega)^{2}=k q_{1} q_{2}$
$\Rightarrow \omega^{2}=\frac{k q^{2}}{m_{e} r^{3}}$
$ \Rightarrow \omega=\sqrt{\frac{k q^{2}}{m_{e} r^{3}}}$
$\therefore $ angular velocity of electron,
$\omega=\sqrt{\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{9.1 \times 10^{-31} \times\left(0.53 \times 10^{-10}\right)^{3}}} $
$\omega=4.1 \times 10^{16} s ^{-1}$
Hence, the radial acceleration of electron $=m r \omega^{2}$
$=0.53 \times 10^{-10} \times\left(4.1 \times 10^{16}\right)^{2}$
$=9 \times 10^{22}\, ms ^{-2}$