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Question: An \[a\] - particle of energy \[5\,{\text{MeV}}\] is scattered through \[180^\circ \] by the gold nu...

An aa - particle of energy 5 MeV5\,{\text{MeV}} is scattered through 180∘180^\circ by the gold nucleus. The distance of closest approach is of the order of:
A. 10−10 cm{10^{ - 10}}\,{\text{cm}}
B. 10−12 cm{10^{ - 12}}\,{\text{cm}}
C. 10−14 cm{10^{ - 14}}\,{\text{cm}}
D. 10−16 cm{10^{ - 16}}\,{\text{cm}}

Explanation

Solution

First of all, we will obtain the expression for potential energy and equate it with kinetic energy. We will substitute the values and manipulate it accordingly.

Stepwise solution:

In the given problem,
Energy associated with the alpha particle is 5 MeV5\,{\text{MeV}} .
This energy is kinetic in nature.
The alpha particle experiences the repulsion force by the force exerted by the nucleus of the gold atom. Since, the two charges are positive in nature, it is inevitable that both will experience repulsion force. Hence, the alpha particle is deflected away.
Atomic number of gold is 7979 .
Kinetic energy of the alpha particle is 5 MeV5\,{\text{MeV}} .
The energy can be converted into joules.
We have,

1 eV=1.6×10−19 J1\,{\text{eV}} = 1.6 \times {10^{ - 19}}\,{\text{J}}
So,
5\,{\text{MeV}} = 5 \times {10^6} \times 1.6 \times {10^{ - 19}}\,{\text{J}} \\\
5\,{\text{MeV}} = 8 \times {10^{ - 13}}\,{\text{J}} \\\
In the question we are asked to find the distance of closest approach. It is important to note that, at the closest approach, the overall kinetic energy of the alpha particle will be converted into potential energy. We know that energy can neither be destroyed nor be created, but it can only be converted from one form to another.
So,

K.E=P.EK.E = P.E …… (1)

However, the potential energy at the closest approach is given by the formula:
P.E=k×(Ze)×2erP.E = \dfrac{{k \times \left( {Ze} \right) \times 2e}}{r} …… (2)
Where,
kk indicates proportionality constant.
ZZ indicates atomic number of the element gold.
ee indicates charge of an electron.
rr indicates the closest approach.

From equation (1) and (2), we can write:
K.E=k×(Ze)×2erK.E = \dfrac{{k \times \left( {Ze} \right) \times 2e}}{r} …… (3)
Now, substituting the required values in the equation (3), we get:
K.E = \dfrac{{2k{e^2} \times Z}}{r} \\\ 8 \times {10^{ - 13}} = \dfrac{{2 \times 9 \times {{10}^9} \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2} \times 79}}{r} \\\ r = \dfrac{{18 \times {{10}^9} \times 2.56 \times {{10}^{ - 38}} \times 79}}{{8 \times {{10}^{ - 13}}}} \\\ r = 4.55 \times {10^{ - 14}}\,{\text{m}} \\\
The closest approach is found out to be 4.55×10−14 m4.55 \times {10^{ - 14}}\,{\text{m}} .
It is of the order of 10−14 m{10^{ - 14}}\,{\text{m}} .

{10^{ - 14}}\,{\text{m}} \\\ = {10^{ - 14}} \times {10^2}\,{\text{cm}} \\\ = {\text{1}}{{\text{0}}^{ - 12}}\,{\text{cm}} \\\ \end{gathered} $$ The order of the closest approach is $${\text{1}}{{\text{0}}^{ - 12}}\,{\text{cm}}$$ . **Hence, the correct option is B.** **Note:** The given problem is based on the order of closest approach. It can also be termed as the contact distance. It is important to note that at the closest approach, all the kinetic energy is converted to its equivalent potential energy. To obtain the desired result, we need to convert electron volt into joules. Failure to do so, will affect the answer obtained.