Assume that two deuteron nuclei in the core of fusion reactor at temperature are moving towards each other, each with kinetic energy , when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from

Question

Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, ${}_{1}^{2}H$ , known as deuteron and denoted by $D$ , can be thought of as a candidate for fusion reactor. The D-D reaction is ${}_{1}^{2}H + {}_{1}^{2}H \to {}_{2}^{3}{He} + n + energy .$ In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ${}_{1}^{2}H$ nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time $t_{0}$ before the particles fly away from the core. If $n$ is the density (number/volume) of deuterons, the product $n t_{0}$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than $5 \times 10^{14} s / {cm}^{3}$ . It may be helpful to use the following: Boltzmann constant $k = 8.6 \times 10^{- 5} eV / K; \frac{e^{2}}{4 π ε_{0}} = 1.44 \times 10^{- 9} eV m$ [2009]

Q. Assume that two deuteron nuclei in the core of fusion reactor at temperature $T$ are moving towards each other, each with kinetic energy $1.5 k T$ , when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature $T$ required for them to reach a separation of $4 \times 10^{- 15} m$ is in the range

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(1)

From conservation of mechanical energy

Loss of kinetic energy of two deuteron nuclei = gain in their potential energy.

$2 \times 1.5 k T = \frac{1}{4 π ε_{0}} \frac{e \times e}{r} = \frac{e^{2}}{4 π ε_{0}} \times \frac{1}{r}$

$\Rightarrow 2 \times 1.5 \times (8.6 \times 10^{- 5} \frac{eV}{K}) \times T = \frac{1.44 \times 10^{- 9} eV m}{4 \times 10^{- 15} m}$

$∴$ $T = \frac{1.44 \times 10^{- 9}}{2 \times 1.5 \times 8.6 \times 10^{- 5} \times 4 \times 10^{- 15}} = 1.4 \times 10^{9} K$

Solution

1 $1.0 \times 10^{9} K < T < 2.0 \times 10^{9} K$

2 $2.0 \times 10^{9} K < T < 3.0 \times 10^{9} K$

3 $3.0 \times 10^{9} K < T < 4.0 \times 10^{9} K$

4 $4.0 \times 10^{9} K < T < 5.0 \times 10^{9} K$

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Solution

1 1.0×109K<T<2.0×109K

2 2.0×109K<T<3.0×109K

3 3.0×109K<T<4.0×109K

4 4.0×109K<T<5.0×109K

Ans. (1) From conservation of mechanical energy Loss of kinetic energy of two deuteron nuclei = gain in their potential energy. 2×1.5kT=14πε0e×er=e24πε0×1r ⇒ 2×1.5×(8.6×10-5eVK)×T=1.44×10-9eV m4×10-15 m ∴ T=1.44×10-92×1.5×8.6×10-5×4×10-15=1.4×109K

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1 $1.0 \times 10^{9} K < T < 2.0 \times 10^{9} K$

2 $2.0 \times 10^{9} K < T < 3.0 \times 10^{9} K$

3 $3.0 \times 10^{9} K < T < 4.0 \times 10^{9} K$

4 $4.0 \times 10^{9} K < T < 5.0 \times 10^{9} K$