Ncert Exercises & Solutions

13.20 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.)

Hint: Their potential energy is maximum at the collision.
Step 1: Find the minimum distance at the collision between the deuteron nuclii.
When two deuterons collide head-on, the distance between their centers,
d = Radius of

1^{s t}

deutrons+Radius of

2^{n d}

deuteron

Radius of a deuteron nuleus=2 fm=

2 \times 10^{- 15} m

∴ d = 2 \times 10^{- 15} + 2 \times 10^{- 15} = 4 \times 10^{15} m

Step 2: Find the potential energy of the system.
Charge on a deuteron nucleus=Charge on an electron=

e = 1.6 \times 10^{- 19}

C
Potential energy of the two-deuteron system:

V = \frac{e^{2}}{4 π \in_{0} d}

where,

\in_{0}

=permittiity of free space

\frac{1}{4 π \in_{0}} = 9 \times 10^{9} N m^{2} C^{- 2}

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

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

= 30 k e V

Hence, the height of the potential barrier of the two-deuteron system is 360 keV.

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