Question

In a simplified model of a hydrogen atom, the electron moves around the proton nucleus in a circular orbit of radius 0.53×10−10m0.53×10−10m. Mass of the electron is 9.11×10−31kg9.11×10−31kg.
b) Determine the radial acceleration of the electron.

c)Determine the speed of the electron.

d)Determine the period of the circular motion.

Answer 1

Answer

given,

radius of the circular orbit, r = 0.53 x 10⁻¹⁰ m

mass of electron, M = 9.11 x 10⁻³¹ Kg

charge of electron, q₁ = 1.6 x 10⁻¹⁹ C

                                q₂ = 1.6 x 10⁻¹⁹ C

we know, force between two charges

$F = \dfrac{kq_1q_2}{r^2}$

$F = \dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(0.53\times 10^{-10})^2}$

  F = 8.20 x 10⁻⁸ N

b) using newton's second law

F = m a

m a =  8.20 x 10⁻⁸

$a =\dfrac{8.20\times 10^{-8}}{9.11\times 10^{-31}}$

    a = 9 x 10²² m/s²

c) speed of the electron

 $a =\dfrac{v^2}{r}$

 $9\times 10^{22} =\dfrac{v^2}{0.53\times 10^{-10}}$

   v² = 4.77 x 10¹²

  v = 2.18 x 10⁶ m/s

d) the period of the circular motion.

    $T=\dfrac{2\pi}{\omega}$

    $T=\dfrac{2\pi r}{v}$

    $T=\dfrac{2\pi\times 0.53\times 10^{-10}}{2.18\times 10^6}$

          T = 1.53 x 10⁻¹⁶ s

Answer 2

Answer:

(a) $a = 9.00\times 10^{22}\ m/s^{2}$

(b) $v = 3.089\times 10^{6}\ m/s$

(c) $T = 1.078\times 10^{-16}\ s$

Solution:

As per the question:

Radius of the orbit, $R = 0.53\times 10^{- 10}\ m$

Mass of an electron, $m_{e} = 9.11\times 10^{- 31}\ kg$

Now,

To calculate the radial acceleration, force due to the acceleration of the electron is balanced by the electrostatic force:

$F = F_{e}$               (1)

where

$F = ma$

$F_{e} = k\frac{QQ'}{R^{2}}$

Since, the charge particles are electrons, thus:

$F_{e} = k\frac{e^{2}}{R^{2}}$

where

k = electrostatic constant = $9.0\times 10^{9}\ Nm^{2}C^{- 2}$

Use eqn (1):

$m_{e}a = k\frac{e^{2}}{R^{2}}$

$a = k\frac{e^{2}}{m_{e}R^{2}}$

where

a = radial acceleration

e = charge on an electron = $1.6\times 10^{- 19}\C$

Therefore,

$a = 9.0\times 10^{9}\times \frac{(1.6\times 10^{- 19})^{2}}{9.11\times 10^{- 31}\times (0.53\times 10^{- 10})^{2}}$

a = $9.00\times 10^{22}\ m/s^{2}$

(b) To calculate the speed, 'v' of an electron in the circular orbit:

Here, the force is given by the centripetal force:

$ma = m\frac{v^{2}}{R}$

$a = \frac{v^{2}}{R}$

$v = \sqrt{2aR} = \sqrt{2\times 9.00\times 10^{22}\times 0.53\times 10^{- 10}}$

$v = 3.089\times 10^{6}\ m/s$

(c) To calculate the period:

$T = \frac{2\pi}{\omega}$

where

$\omega = \frac{v}{R}$ = angular velocity

Thus

$T = \frac{2\pi}{\frac{v}{R}} = \frac{2\pi R}{v}$

$T = \frac{2\pi \times 0.53\times 10^{- 10}}{3.089\times 10^{6}}$

$T = 1.078\times 10^{-16}\ s$