Calculate the Height of the Potential Barrier for a Head on Collision of Two Deuterons. - Physics


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.)



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" = "e"^2/(4pi in_0 "d")`


`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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Chapter 13: Nuclei - Exercise [Page 464]


NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 13.20 | Page 464
NCERT Physics Class 12
Chapter 13 Nuclei
Exercise | Q 20 | Page 464



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Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy release in MeV in the deuterium-tritium fusion reaction :


Using the data :

m(`""_1^2H`) = 2.014102 u

m(`""_1^3H`) = 3.016049 u

m(`""_2^4He`) = 4.002603 u

mn = 1.008665 u

1u = 931.5 MeV/c2

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where MZ,N = mass of an atom with Z protons and N neutrons in the nucleus and MB = mass of a hydrogen atom. This energy is known as proton-separation energy.

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(a) `""_1^2H + ""_1^2H → ""_1^3H + ""_1^1H`

(b) `""_1^2H + ""_1^2H → ""_2^3H + n`

(c) `""_1^2H + ""_1^3H → _2^4H + n`.

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