JEE Main 2023 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

In a hypothetical fusion reactor, two deuterium nuclei fuse to form a helium-4 nucleus. Take the binding energy per nucleon as 1.0 MeV for deuterium…

Numerical answer

In a hypothetical fusion reactor, two deuterium nuclei fuse to form a helium-4 nucleus. Take the binding energy per nucleon as

1.0\ \mathrm{MeV}

for deuterium and

7.0\ \mathrm{MeV}

for helium-4. The number of fusion reactions per second needed to generate

3.84\ \mathrm{W}

of power is

x\times10^{11}

. Taking

1\ \mathrm{MeV}=1.6\times10^{-13}\ \mathrm J

, the value of

x

Show answer and solution

Answer: 10

Energy released per reaction

=

(binding energy of

{}^4\mathrm{He}

)

-

(binding energy of two deuterons)

=4\times7.0-2\times(2\times1.0)=28-4=24\ \mathrm{MeV}=24\times1.6\times10^{-13}=3.84\times10^{-12}\ \mathrm J

. Reactions per second

=\dfrac{3.84}{3.84\times10^{-12}}=10^{12}=10\times10^{11}

. Hence

x=10

. Using

28\ \mathrm{MeV}

would give

x\approx8.6

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