Question

The kinetic energy of $\alpha$-particles at a distance $5\times {10}^{-14}$m from the nucleus will be(in Joules)

• A. $6.4\times {10}^{-13}$
• B. $4.3\times {10}^{-13}$
• C. $2.1\times {10}^{-13}$
• D. $3.4\times {10}^{-14}$

Answer 1

The correct option is C $2.1\times {10}^{-13}$

The charge on an alpha particle is $+2$.

Hence $q=2e=3.2\times {10}^{-19}C$

This alpha particle is accelerated by a potential of $2\times {10}^{6}V$

Hence Energy of the particle $=qV=3.2\times {10}^{-19}\times 2\times {10}^6=6.4\times {10}^{-13}J$

This is the total energy.

Potential energy when it is at a distance of $5\times {10}^{-14}m$ from the nucleus

$PE=\frac{1}{4\pi {ϵ}_0}\frac{q_1{q}_2}{d}$

Nucleus contains 47 protons.

Hence, $q_2=47e=47\times 1.6\times {10}^{-19}C$

Substituting these values:

We get PE = $4.3\times {10}^{-13}J$

$KE+PE=TE$

$KE=TE-PE=(6.4-4.3)\times {10}^{-13}J=2.1\times {10}^{-13}J$