Ncert Exercises & Solutions

13.28 Consider the D–T reaction (deuterium-tritium fusion)

\({ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+\mathrm{n}\)

(a) Calculate the energy released in MeV in this reaction from the data:

m(\({ }_{1}^{2} \mathrm{H}\))=2.014102 u

m(\({ }_{1}^{3} \mathrm{H}\)) =3.016049 u

(b) Consider the radius of both deuterium and tritium to be approximately 2.0 fm. What is the kinetic energy needed to overcome the coulomb repulsion between the two nuclei? To what temperature must the gas be heated to initiate the reaction?

(Hint: Kinetic energy required for one fusion event =average thermal kinetic energy available with the interacting particles = 2(3kT/2); k = Boltzman’s constant, T = absolute temperature.)

(a)
Hint: The energy released is given by the mass defect.
Step 1: Find the Q-value of the fission reaction.
Take the D-T nuclear reaction:

{}_{1}^{2}H + {}_{1}^{3}H \to {}_{2}^{4}H e + n

It is given that:

M a s s o f {}_{1}^{2}H, m_{1} = 2.014102 u

M a s s o f {}_{1}^{3}H, m_{2} = 3.016049 u

M a s s o f {}_{2}^{4}H, m_{3} = 4.002603 u

M a s s o f {}_{0}^{1}n, m_{4} = 1.008665 u

Q-value of the given D-T reaction is:

Q = [m_{1} + m_{2} - m_{3}] c^{2} = [2.014102 + 3.016049 - 4.002603 - 1.008665] c^{2} = [0.018883 c^{2}] u

B u t 1 u = 931.5 M e V / c^{2}

∴ Q = 0.018883 \times 931.5 = 17.59 M e V

(b)
Hint: The required kinetic energy should overcome the maximum potential energy of the nuclei.
Step 1: Find the maximum potential energy of the nuclei.
Radius of deuterium and tritium,

r \approx 2.0 f m = 2 \times 10^{- 15} m

Distance between the two nuclei at the moment when they touch each other,

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

Charge on the deuterium nucleus=e

Charge on the tritium nucleus =e

Hence, the repulsive potential energy between the two nuclei is given as:

V = \frac{e^{2}}{4 π \in_{0} (d)}

where,

\in_{0}

=Permittivity of free space

Of kinetic energy [KE] is needed to overcome the Coulomb repulsion between the two nuclei.

However, it is given that:

\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})}^{2}}{4 \times 10^{- 15}} = 5.76 \times 10^{- 14} J

= \frac{5.76 \times 10^{- 14}}{1.6 \times 10^{- 19}} = 3.6 \times 10^{5} e V = 360 K e V

H e n c e, 5.76 \times 10^{- 14} J o r 360 K e V o f k i n e t i c e n e r g y (K E) i s n e e d e d t o o v e r c o m e t h e C o u l o m b

r e p u l s i o n b e t w e e n t h e t w o n u c l e i .

Step 2: Find the temperature required using \(K E=2 \times \frac{3}{2} k T\).
It is given that:

\(K E=2 \times \frac{3}{2} k T\)
where k = Boltzmann constant = \(1.38 \times 10^{-23} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}{ }^{-2} K^{-1}\)
So, \(T=\frac{K E}{3k}\)
\(T=\frac{5.76 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}}=1.39 \times 10^{9} \mathrm{~K}\)

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