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

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 $=mr{\omega }^2$
$=0.53\times 10^{-10}\times {\left(4.1\times 10^{16}\right)}^2$
$=9\times 10^{22}m{s}^{-2}$