Assuming that about 20 MeV of energy is released per fusion reaction 1H2-1H2- 2He4-E- other particles- then the mass of 1H2 consumed per day in a fusion reactor of power 1 megawatt will approximately be

The correct option is B 1 g E = Δ m c^2 For one reactionMass Defect = Δ m #160; #160; #160; #160; #160; #160; #160; #160; # ...

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Standard X

Physics

Nuclear Fusion

Assuming that...

Question

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

0.001 g

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1 g

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10.0 g

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1000 g

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Solution

The correct option is B 1 g
$E = Δ m c^{2}$
For one reaction
Mass Defect = $Δ m$
$= 2 (m_{H}) - m_{H e} - m_{n}$
$= 2 (2.015) - 3.017 - 1.009$ amu
$= 0.004 a m u$
$1 a m u = 931.5 M e V / c^{2}$
Hence
$E = 0.004 \times 931.5 M e V = 3.724 M e V$
$E = 3.726 \times 1.6 \times 10^{- 13} J = 5.96 \times 10^{- 13} J$
.
Total Requirement = $1 M W = 10^{6} J / s$
Total fusions required = $\frac{10^{6}}{5.96 \times 10^{- 13}} = 1.67 \times 10^{18} s^{- 1}$
Mass rate = $\frac{1.67 \times 10^{18}}{N_{A}} \times 2 \times 2 = 1.1 \times 10^{- 5} g / s$
Total consumption in a day = $M \times (24 \times 3600) = 0.95 g m s$

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Similar questions

Q. Assuming that about

20 MeV

of energy is released per fusion reaction,

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

The mass of

_{1} H^{2}

consumed per day in a future fusion reactor of power

1 MW

would be approximately-

Q. The amount of energy released in the fusion of

_{1} H^{2}

to form a

_{2} H e^{4}

nucleus will be :
[Binding energy per nucleon of

_{1} H^{2} = 1.1 M e V

and

_{2} H e^{4} = 7 M e V

]

Q. The binding energy of deutron

_{1} H^{2}

1.112 M e V

per nucleon and an alpha particle

_{2} H e^{4}

has a binding energy of

7.047 M e V

per nucleon. Then, in the fusion rection

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

, the energy

Q

released is
(take mass of

_{1} H^{2} = 2.01478 u

;
mass of

_{2} H e^{4} = 4.00338 u

)

Q. The binding energy of deutron

_{1} H^{2}

1.112 M e V

per nucleon and an alpha particle

_{2} H e^{4}

has a binding energy of

7.047 M e V

per nucleon. Then, in the fusion rection

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

, the energy

Q

released is
(take mass of

_{1} H^{2} = 2.01478 u

;
mass of

_{2} H e^{4} = 4.00338 u

)

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

rating. If

25 %

of energy from the above reaction is used in the reactor, how many grams of deuterium will be needed per day?

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

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Assuming that about 20 MeV of energy is released per fusion reaction 1H2+1H2→2He4+E+ other particles, then the mass of 1H2 consumed per day in a fusion reactor of power 1 megawatt will approximately be

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