An electron is revolving in a circular orbit of a hydrogen atom with a radius of $2.12 \times 10^{-10}$ m. The kinetic energy of the electron in this orbit is approximately:
[Take $\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9$ N m $^2$ C $^{-2}$ and charge of electron $e = 1.6 \times 10^{-19}$ C]

Question

An electron is revolving in a circular orbit of a hydrogen atom with a radius of

2.12 \times 10^{-10}

m. The kinetic energy of the electron in this orbit is approximately:
[Take

\dfrac{1}{4\pi\epsilon_0} = 9 \times 10^9

N m

^2

C

^{-2}

and charge of electron

e = 1.6 \times 10^{-19}

C]

Quantrex Academy · Accepted Answer

Correct answer: C. 3.4 eV

The electrostatic force between the nucleus and the electron provides the necessary centripetal force for the electron to revolve in its orbit.

\dfrac{mv^2}{r} = \dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{r^2}

The kinetic energy

K

of the electron is given by:

K = \dfrac{1}{2}mv^2 = \dfrac{1}{4\pi\epsilon_0}\dfrac{e^2}{2r}

Substituting the given values, we get the kinetic energy in Joules:

K = \dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{2 \times 2.12 \times 10^{-10}}

J

To convert the kinetic energy into electron-volts (eV), divide by

1.6 \times 10^{-19}

C:

K = \dfrac{9 \times 10^9 \times 1.6 \times 10^{-19}}{2 \times 2.12 \times 10^{-10}}

eV

K = \dfrac{14.4 \times 10^{-10}}{4.24 \times 10^{-10}}

eV

K \approx 3.4

eV

Answer:

3.4

eV

An electron is revolving in a circular orbit of a hydrogen atom...