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JEE Advance - Physics (2009 - Paper 1 Offline - No. 18)

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 $$nt_0$$ 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; $${{{e^2}} \over {4\pi {\varepsilon _0}}} = 1.44 \times {10^9}$$ eVm.

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 $$nt_0$$ 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; $${{{e^2}} \over {4\pi {\varepsilon _0}}} = 1.44 \times {10^9}$$ eVm.

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 $$nt_0$$ 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; $${{{e^2}} \over {4\pi {\varepsilon _0}}} = 1.44 \times {10^9}$$ eVm.

Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion?
Deuteron density = $$2.0\times10^{12}~\mathrm{cm^{-3}}$$; Confinement time = $$5.0\times10^{-3}~\mathrm{s}$$.
Deuteron density = $$8.0\times10^{14}~\mathrm{cm^{-3}}$$; Confinement time = $$9.0\times10^{-1}~\mathrm{s}$$.
Deuteron density = $$4.0\times10^{23}~\mathrm{cm^{-3}}$$; Confinement time = $$1.0\times10^{-11}~\mathrm{s}$$.
Deuteron density = $$1.0\times10^{24}~\mathrm{cm^{-3}}$$; Confinement time = $$4.0\times10^{-12}~\mathrm{s}$$.

Explanation

To determine the most promising design based on the Lawson criterion, we need to calculate the Lawson number ($$nt_0$$) for each design and check if it exceeds $$5 \times 10^{14}$$ s/cm$$^3$$. We will do this for each of the given options:

Option A:

Deuteron density, $$n = 2.0 \times 10^{12}~\mathrm{cm^{-3}}$$

Confinement time, $$t_0 = 5.0 \times 10^{-3}~\mathrm{s}$$

Lawson number, $$nt_0 = n \cdot t_0 = (2.0 \times 10^{12}) \cdot (5.0 \times 10^{-3}) = 1.0 \times 10^{10}~\mathrm{s/cm^3}$$

Option B:

Deuteron density, $$n = 8.0 \times 10^{14}~\mathrm{cm^{-3}}$$

Confinement time, $$t_0 = 9.0 \times 10^{-1}~\mathrm{s}$$

Lawson number, $$nt_0 = n \cdot t_0 = (8.0 \times 10^{14}) \cdot (9.0 \times 10^{-1}) = 7.2 \times 10^{14}~\mathrm{s/cm^3}$$

Option C:

Deuteron density, $$n = 4.0 \times 10^{23}~\mathrm{cm^{-3}}$$

Confinement time, $$t_0 = 1.0 \times 10^{-11}~\mathrm{s}$$

Lawson number, $$nt_0 = n \cdot t_0 = (4.0 \times 10^{23}) \cdot (1.0 \times 10^{-11}) = 4.0 \times 10^{12}~\mathrm{s/cm^3}$$

Option D:

Deuteron density, $$n = 1.0 \times 10^{24}~\mathrm{cm^{-3}}$$

Confinement time, $$t_0 = 4.0 \times 10^{-12}~\mathrm{s}$$

Lawson number, $$nt_0 = n \cdot t_0 = (1.0 \times 10^{24}) \cdot (4.0 \times 10^{-12}) = 4.0 \times 10^{12}~\mathrm{s/cm^3}$$

Based on the calculations, the Lawson numbers for each option are:

  • Option A: $$1.0 \times 10^{10}~\mathrm{s/cm^3}$$
  • Option B: $$7.2 \times 10^{14}~\mathrm{s/cm^3}$$
  • Option C: $$4.0 \times 10^{12}~\mathrm{s/cm^3}$$
  • Option D: $$4.0 \times 10^{12}~\mathrm{s/cm^3}$$

The most promising design based on the Lawson criterion, which requires the Lawson number to be greater than $$5 \times 10^{14}~\mathrm{s/cm^3}$$, is Option B because its Lawson number exceeds the threshold.

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