Question

The distance of the closest approach of the particle to the nucleus will be
(A) $6.4*{10^{ - 13}}m$
(B) $4.3*{10^{ - 13}}m$
(C) $2.1*{10^{ - 13}}m$
(D) $3.4*{10^{ - 13}}m$

Answer 1

Hint : First of all we have to use the equation $q = 2e$ in order to have the total charge that is on the particle approaching the nucleus. Now after finding the total charge in terms of coulombs we just have to imply all the values that are in the equation for the energy, $E = qV$ . Now after getting the value we will get the answer by comparing it with total potential energy.

Complete Step By Step Answer:
Here, first we have to sort out the given quantities:
The charge on the alpha particle will be: $+ 2$
Now, we have to use the equations to find the total charge,
Hence, the charge would be:
$q = 2e \\\ q = 2*(1.6*{10^{ - 19}})C \\\ q = 3.2*{10^{ - 19}}C \\\$
The alpha particle will be accelerated by a potential of nearly: $2*{10^6}V$
So to find the energy of the particle we have to imply the given equation :
$E = qV \\\ E = 3.2*{10^{ - 19}}*2*{10^6} \\\ E = 6.4*{10^{ - 13}}J \\\$
Now, for getting the point of closest approach we have to compare the energy calculated with the formula for the total potential energy.
$PE = {\text{total energy}} \\\ 6.4*{10^{ - 13}} = \dfrac{1}{{4\pi { \in _o}}}*\dfrac{{47*2{e^2}}}{d} \\\ \\\$
On solving the above equation we will get:
$d = 3.384*{10^{ - 14}}m$

Note :
In the case of an alpha particle we know that the total charge would be calculated by comparing it with the electrons as we know that they are just a pair of the protons. And we know that the charge on the electrons and the protons are basically just the same and only the signs are changed.