Question

A neutron of energy $1MeV$ and mass $1.6\times {{10}^{-27}}kg$ passes a proton at such a distance that the angular momentum of neutron relative to the proton approximately equal to ${{10}^{-35}}Js$. The distance of closest approach neglecting the interaction between the particles is
A. $0.44nm$
B. $0.44mm$
C. $0.44\overset{\text{o}}{\mathop{\text{A}}}\,$
D. $0.44fm$

Answer 1

Hint: The energy of a neutron is given, and we can calculate the linear momentum associated with the neutron from the energy given and the mass given in the problem. If d is the distance of closest approach, then the linear momentum times the distance of closest approach gives the relative angular momentum of the neutron with respect to the proton.

Formula Used:
The linear momentum associated with a body which has a kinetic energy E and mass m is given by the formula,
$p=\sqrt{2mE}$
Angular momentum of a body of mass m, moving with a linear velocity v at a perpendicular distance r from the origin is given by,
L=mvr

Complete step by step answer:
In the question, it is given that the energy or the kinetic energy of a neutron is $1MeV$ and also the mass of the neutron is given $1.6\times {{10}^{-27}}kg$. So, from these factors, we can calculate the linear momentum associated with the neutron using the formula,
$p=\sqrt{2{{m}_{p}}E}$
Where ${{m}_{p}}$ is the mass of the proton. We can substitute the values in the above equation to get,
$p=\sqrt{2\times 1.6\times {{10}^{-27}}kg\times 1\times {{10}^{6}}eV}$
$p=\sqrt{2\times 1.6\times {{10}^{-27}}kg\times \left( 1\times {{10}^{6}}\times 1.6\times {{10}^{-19}} \right)J}$
$\therefore p=2.26\times {{10}^{-20}}kgm{{s}^{-1}}$
So, we now have the linear momentum associated with the neutron. The relative angular momentum of the neutron with respect to the proton can be represented as, (we can take the position of the proton as the origin)
Relative angular momentum=mvd=pd
Where d is the distance of the closest approach, p is the linear momentum of the neutron. We are given that the relative angular momentum of the neutron with the proton is ${{10}^{-35}}Js$, So we can write,
${{10}^{-35}}Js=pd$
$\Rightarrow {{10}^{-35}}Js=\left( 2.26\times {{10}^{-20}}kgm{{s}^{-1}} \right)d$
$\Rightarrow d=\dfrac{{{10}^{-35}}Js}{2.26\times {{10}^{-20}}kgm{{s}^{-1}}}$
$\therefore d=0.44fm$
So, the distance of the closest approach is 0.44 fm.
So, the answer to the question is option (D).

Note: In physics, angular momentum is the analogue of linear momentum for a rotating body. It is a pseudovector which means it transforms like a vector but is not really a vector. In three dimension it can be represented by the formula, \[\overrightarrow{L}=\overrightarrow{r}\times \overrightarrow{p}\], where L is the angular momentum of the body, r is the position vector from the origin and p is the linear momentum of the body.
Conservation of Angular Momentum: When no external torque acts on a rotating body or system executing a uniform rotational motion, then the angular momentum of the body is said to be conserved. Which means that the angular momentum does not change with time. The theorem is analogous to Newton’s third law.
The angular momentum of an extended body can be expressed as the product of the moment of inertia of the body and its angular velocity.