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It is propose...

Question

It is proposed to use the nuclear fusion reaction
$_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4}$
in a nuclear reactor with an electrical power rating of $200 M W$ . If the energy from the above reaction is used with $25$ percent efficiency in the reactor, how many grams of deuterium fuel will be needed per day? (The masses of $_{1} H^{2}$ and $_{2} H e^{4}$ are $2.0141 a m u$ and $4.0026 a m u$ respectively.)

A

130 g m

.

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B

140 g m

.

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C

150 g m

.

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D

120 g m

.

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Solution

The correct option is D $120 g m$ .

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Similar questions

Q. It is proposed to use the nuclear fusion reaction

_{1} H^{2} +_{1} H^{2} \to_{2} {He}^{4}

in a nuclear reactor of

200 MW

25 %

of energy from the above reaction is used in the reactor. It is found that nearly

12 x

grams of deuterium will be needed per day. The value of

x

is (integer only)

[Take, mass of_{1} H^{2} = 2.0141 amu, mass of_{2} {He}^{4} = 4.0026 amu, 1 amu c^{2} = 931.5 MeV]

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

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

other particles, then the mass of

_{1} H^{2}

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

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

Q. Binding energy per nucleon of

_{1} H^{2}

_{2} H e^{4}

1.1 M e V

7.0 M e V

, respectively. Energy released in the process

_{1} H^{2} +_{1} H^{2} ⟶_{2} H e^{4}

Q. Binding energy per nucleon of

_{1} H^{2}

_{2} H e^{4}

1.1

7.0

MeV respectively. Energy released in the process

_{1} H^{2} +_{1} H^{2} ⟶_{2} H e^{4}

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