Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Calculate the height of the potential barrier for a head on collision of two deuterons.

seo-qna
Last updated date: 14th Apr 2024
Total views: 33.6k
Views today: 1.33k
MVSAT 2024
Answer
VerifiedVerified
33.6k+ views
Hint: The height of potential barrier here means the electrostatic potential energy between the two deuterons so, use the equation of electrostatic potential energy between two charged particles i.e.
$U = \dfrac{1}{{4\pi { \in _ \circ }}}\dfrac{{{q_1}{q_2}}}{r}$
Where U= potential energy or barrier, ${q_1},{q_2}$ are the two charges and r= distance between the two charges

Complete step by step solution:
Step1: The potential energy for the collision of two charged particles can be given by-
$U = \dfrac{1}{{4\pi { \in _ \circ }}}\dfrac{{{q_1}{q_2}}}{r}$ ………………(1)
Where U= potential barrier, ${q_1},{q_2}$ are the two charges and r= distance between the two charges
 Since the deuterons are the charged particles whose magnitude of charge is equal to that of an electron.
Therefore, ${q_1} = {q_2} = 1.6 \times {10^{ - 19}}C$
Also the distance between their centres on collision is given by-
r= radius of 1st deuteron radius of 2nd deuteron
radius of nucleus is given by-
$r = r_0 A^{1/3}$
Where ${r_ 0}$ = fermi constant = 1.6 fm, A= mass number
Since deuteron has a mass number of 2, therefore A=2
Hence radius of deuteron nucleus is given by,
$r$ = $1.6 \times 2^{1/3}$
So the radius of a deuteron nucleus = $2fm$ (Fermi meter)
Therefore \[r = 2 + 2 = 4fm\]
Since, $1fm = {10^{ - 15}}m$
\[ \Rightarrow r = 4 \times {10^{ - 15}}m\]
$\dfrac{1}{{4\pi { \in _ \circ }}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$

Step2: Substituting all the values in equation (1) and simplify
$\Rightarrow$ $U = 9 \times {10^9}\dfrac{{{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{4 \times {{10}^{ - 15}}}}J$.
This potential energy is in J. In order to convert it in eV divides it with charge of electron i.e. $1.6 \times {10^{ - 19}}C$.
Therefore,
$
\Rightarrow U = 9 \times {10^9}\dfrac{{{{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{4 \times {{10}^{ - 15}}\left( {1.6 \times {{10}^{ - 19}}} \right)}}eV \\
\Rightarrow U = 360 \times {10^3}eV \\
\Rightarrow U = 360keV \\ $

Hence the height of the potential barrier for a head on collision of two deuterons is 360keV.

Note: Always remember that there is a difference between the electrostatic potential and electrostatic potential energy. Electrostatic potential, at any point in an electric field, is defined as the amount of work done in moving a unit positive charge between infinity and that point, without any acceleration, against the electric force. Every point around a source charge is characterized with electric potential, which is given by-
$V = \dfrac{1}{{4\pi { \in _ \circ }}}\dfrac{q}{r}$
Where V= Electric potential
If instead of bringing a unit positive, we bring a charge 'q' from infinity to that point, work done W in doing so is given by-
\[W = q \times V = \dfrac{1}{{4\pi { \in _ \circ }}}\dfrac{{{q^2}}}{r}\]
This work is termed as 'electrostatic potential energy' of the charge q in the field of charge ‘q'. We can also say that the electric potential, at any point, in an electric field is the potential energy of a unit positive charge placed at that point.