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Nuclear Fusion

The correct e...

Question

The correct equation of nuclear fusion reaction is

A

_{1} H^{1} +_{1} H^{1} \to H_{2}

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B

_{1} H^{2} +_{1} H^{1} \to_{2} H e^{4} +_{0} n^{1}

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C

_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4} + e n e r g y

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D

2 H_{2} + 3 H_{2} \to 5 H_{2}

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Solution

The correct option is C $_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4} + e n e r g y$
In nuclear fusion reactions, the heavy isotopes of Hydrogen take part like deuterium $_{1} H^{2}$ and tritium $_{1} H^{3}$ as they have extra neutron. Therefore, the reaction C is correct out of the given options.

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Q. Assuming that about 20 MeV of energy is released per fusion reaction,

_{1} H^{2} +

_{1} H^{3}

⟶

_{0} n^{1} +

_{2} H e^{4} .

then the mass of

1_{H}^{1}

consumed per day in a fusion reactor of power 1 MW will approximately be :

10^{40}

deuterons. It produces energy via the process

_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} +_{1} H^{1}

_{1} H^{2} +_{1} H^{3} \to_{2} H e^{4} +_{0} n^{1}

If the average power radiated by the star is

10^{16}

W, then the deuteron supply of the star is exhausted in a time of the order of :

Q. The energy released in the reaction

_{1} H^{2} +_{1} H^{3}

\to_{2} H e^{4} +_{0} n^{1}

- ------- -

_{0} n^{1} = 1.00867 a m u

_{1} H^{2} = 2.01410 a m u

_{1} H^{3} = 3.0160 a m u

_{2} H e^{4} = 4.0026 a m u

Assertion: The reaction;

_{1} H^{2} +_{0} n^{1} \to_{2} H e^{4} +_{0} n^{1}

will not happen.

Reason: The equation given is not balanced.

Q. The atomic masses of the hydrogen isotopes are

m_{1} H^{1} = 1.007825 a m u

m_{1} H^{2} = 2.014102 a m u

m_{1} H^{3} = 3.016049 a m u

The energy released in the reaction

_{1} H^{2} +_{1} H^{2} \to_{1} H^{3} +_{1} H^{1}

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