It is proposed to use the nuclear fusion reaction 1H2 - 1H2 - 2He4 in a nuclear reactor of 200 MW rating- If 25-mass of 1H2 - 2-0141 amu-mass of 2He4- 4-0026 amu- 1amu c2-931-5 MeV-

Energy released in the nuclear fusion reaction is given by, nbsp; Q = (Δ m)c^2 Here, Δ m=2M_(_1H^2) M_(_2He^4) ⇒Δ m=(2 × 2.0141 u) 4.00264 u=0.0256 u=0.0 ...

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Standard X

Physics

Nuclear Fission

It is propose...

Question

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

120.77 \frac{g}{day}

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220.77 \frac{g}{day}

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320.83 \frac{g}{day}

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420.83 \frac{g}{day}

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Solution

The correct option is A $120.77 \frac{g}{day}$
Energy released in the nuclear fusion reaction is given by,

$Q = (Δ m) c^{2}$

Here, $Δ m = 2 M_{(_{1} H^{2})} - M_{(_{2} {He}^{4})}$

$\Rightarrow Δ m = (2 \times 2.0141 u) - 4.00264 u = 0.0256 u = 0.0256 u \times c^{2}$

$\Rightarrow Q = 0.0256 \times 931.5 MeV$

$\Rightarrow Q = 23.8464 MeV = 23.8464 \times 1.6 \times 10^{- 13} J$

$\Rightarrow Q = 38.15424 \times 10^{- 13} J$

Since, only $25 %$ of energy is used in the reactor,

The effective energy used,

$Q^{'} = \frac{25}{100} \times Q$

$\Rightarrow Q^{'} = \frac{25}{100} \times 38.15424 \times 10^{- 13}$

$\Rightarrow Q^{'} = 9.53856 \times 10^{- 13} J$

Two deuterium nuclei are involved in a fusion reaction,

$∴$ The energy released per deuterium

$E_{1} = \frac{9.53856 \times 10^{- 13}}{2} J = 4.76928 \times 10^{- 13} J$

For $200 MW$ power per day, the number of deuterium nuclear required,

$n = \frac{P \times t}{E_{1}} = \frac{200 \times 10^{6} \times 86400}{4.76928 \times 10^{- 13}} [∵ 1 day = 86400 sec]$

$\Rightarrow n = 3.62319 \times 10^{25}$

We know that, $6 \times 10^{23}$ atoms are present in $2 g$ of deuterium.

Then, the mass of deuterium that contains $3.623 \times 10^{25}$ atoms is given by,

$m = \frac{2 \times 3.62319 \times 10^{25}}{6 \times 10^{23}} g = 120.77 g$

Hence, $(A)$ is the correct answer.

Why this question?

This question gives an idea about the mass of the fuel required in nuclear fusion reaction to produce a given amount of energy.

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Similar questions

Q. It is proposed to use the nuclear fusion reaction

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in a nuclear reactor with an electrical power rating of 200 MW. 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

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

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Q. Consider the following reaction

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It is proposed to use the nuclear fusion reaction 1H2+ 1H2→ 2He4 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 1H2=2.0141 amu,mass of 2He4=4.0026 amu,1amu c2=931.5 MeV]