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

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:

When two deuterons collide head-on, the distance between their centres, d is given as:

Radius of 1st deuteron + Radius of 2nd deuteron

Radius of a deuteron nucleus = 2 fm = 2 × 10−15 m

d = 2 × 10−15 + 2 × 10−15 = 4 × 10−15 m

Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10−19 C

Potential energy of the two-deuteron system:

$V=\frac{e^{2}}{4 \pi \in_{0} d}$

Where,

$\epsilon_{0}=$ Permittivity of free space

$\frac{1}{4 \pi \in_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2}$

$\therefore V=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{4 \times 10^{-15}} \mathrm{~J}$

$=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{4 \times 10^{-15} \times\left(1.6 \times 10^{-19}\right)} \mathrm{eV}$

$=360 \mathrm{keV}$

Hence, the height of the potential barrier of the two-deuteron system is

 

360 keV.

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