Calculate the height of the potential barrier for a head on collision of two deuterons.(Hint: The height of the potential 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.

Solution

Suppose, the two particles are fired at each other with the same kinetic energy K.E, so that they are brought to rest by their mutual Coulomb repulsion when they are just touching each other.

Distance between the centres of two deutrons, during head on collision, r = 2

\times

radius

Therefore,
r = 4 fm = 4

\times

10^-15 m

Charge on each deutron, e = 1.6

\times

10^-19 C

Now, potential energy =

\frac{e^{2}}{4 {πε}_{o} r}

= \frac{9 \times 10^{9} \times (1.6 \times 10^{- 19})^{2}}{4 \times 10^{- 15}}

= \frac{9 \times 1.6 \times 1.6 \times 10^{- 14}}{4 \times 1.6 \times 10^{- 16}} keV

=360 keV

Now, since, potential energy is equal to twice the kinetic energy of deutron.

This implies,
K.E of each deutron =

\frac{360}{2} = 180 k e V

which is a measure of the height of the coulomb barrier.

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Wave Optics

Calculate the height of the potential barrier for a head on collision of two deuterons.(Hint: The height of the potential 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.

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