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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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#### Solution

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

Radius of 1^{st} deuteron + Radius of 2^{nd} 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" = "e"^2/(4pi in_0 "d")`

Where,

`in_0` = Permittivity of free space

`1/(4piin_0) = 9 xx 10^9 "N" "m"^2"C"^(-2)`

`therefore "V" = (9 xx 10^9 xx (1.6 xx 10^(-19))^2)/(4 xx 10^(-15)) "J"`

`= (9 xx 10^9 xx (1.6 xx 10^(-19))^2)/(4 xx 10^(-15) xx (1.6 xx 10^(-19)) " eV"`

= 360 keV

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

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