Question 14 Marks
Calculate the height of the potential barrier for a head-on collision of two deuterons.
(Hint: The height of the potetnial barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
(Hint: The height of the potetnial barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius 2.0 fm.)
Answer
View full question & answer→For a head-on collision, distance between the centers of two deuterons = r =2×radius
= 2×(2fm)
= 4fm
$=4 \times 10^{-5} m$
Charge (e) of each deuteron
$=1.6 \times 10^{-19} C$
When two deuterons are kept in mutual contact, the effective distance between them
$R =2 r=4 \times 10^{-15} m$
Hence the value of potential energy
$\begin{array}{l}=\frac{e^2}{4 \pi \in_0 r} \\ =\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{4 \times 10^{-15}} \text { Joule } \\ =360 KeV \end{array}$
Potential energy (P.E.)= 2 × kinetic energy of each deuteron
$=360 KeV$
Thus, height of Coulomb barrier of deuteron $=$ potential energy between two deuteron atom.
$=360 KeV$
= 2×(2fm)
= 4fm
$=4 \times 10^{-5} m$
Charge (e) of each deuteron
$=1.6 \times 10^{-19} C$
When two deuterons are kept in mutual contact, the effective distance between them
$R =2 r=4 \times 10^{-15} m$
Hence the value of potential energy
$\begin{array}{l}=\frac{e^2}{4 \pi \in_0 r} \\ =\frac{9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2}{4 \times 10^{-15}} \text { Joule } \\ =360 KeV \end{array}$
Potential energy (P.E.)= 2 × kinetic energy of each deuteron
$=360 KeV$
Thus, height of Coulomb barrier of deuteron $=$ potential energy between two deuteron atom.
$=360 KeV$