In the nuclear fusion reaction - 21H- 31H - 42He-n - given that the repulsive potential energy between the two nuclei is - 7-7 - 10-14J- the temperature to which the gases must be heated to initiate the reaction is nearly Boltzmann-s constant k-1-38 x 10 -23J

The correct option is B 10^9K The repulsive potential energy =32kT 7.7× 10^ 14=32× 1.38×× 10^ 23× T → T = 3.7× 10^9 K ∼ 10^9 K ...

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Standard X

Physics

Nuclear Fusion

In the nuclea...

Question

In the nuclear fusion reaction : $_{1}^{2} H +_{1}^{3} H \to$ $_{2}^{4} H e + n$ , given that the repulsive potential energy between the two nuclei is $\sim 7.7 \times 10^{- 14}$ J, the temperature to which the gases must be heated to initiate the reaction is nearly
(Boltzmann's constant $k =$ 1.38 x 10 $^{- 23}$ J)

10^{7} K

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10^{5} K

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10^{3} K

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10^{9} K

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Solution

The correct option is B $10^{9} K$
The repulsive potential energy $= \frac{3}{2} k T$

$7.7 \times 10^{- 14} = \frac{3}{2} \times 1.38 \times \times 10^{- 23} \times T$

$\to T = 3.7 \times 10^{9} K \sim 10^{9} K$

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Similar questions

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

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In the nuclear fusion reaction : 21H+31H→ 42He+n , given that the repulsive potential energy between the two nuclei is ∼7.7×10−14J, the temperature to which the gases must be heated to initiate the reaction is nearly (Boltzmann's constant k=1.38 x 10−23J)

The correct option is B 109KThe repulsive potential energy =32kT7.7×10−14=32×1.38××10−23×T→ T=3.7×109K∼109K

The correct option is B $10^{9} K$
The repulsive potential energy $= \frac{3}{2} k T$

$7.7 \times 10^{- 14} = \frac{3}{2} \times 1.38 \times \times 10^{- 23} \times T$

$\to T = 3.7 \times 10^{9} K \sim 10^{9} K$