In the nuclear fusion reaction
$$_1^2H + _1^3H \to _2^4He + n$$
given that the repulsive potential energy between the two nuclei is $$ \sim 7.7 \times {10^{ - 14}}J,$$ the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s Constant $$k = 1.38 \times {10^{ - 23}}J/K$$ ]
A.
$${10^7}K$$
B.
$${10^5}K$$
C.
$${10^3}K$$
D.
$${10^9}K$$
Answer :
$${10^9}K$$
Solution :
Average kinetic energy per molecule $$ = \frac{3}{2}kT$$
This kinetic energy should be able to provide the repulsive potential energy
$$\eqalign{
& \therefore \frac{3}{2}kT = 7.7 \times {10^{ - 14}} \cr
& \Rightarrow T = \frac{{2 \times 7.7 \times {{10}^{ - 14}}}}{{3 \times 1.38 \times {{10}^{ - 23}}}} = 3.27 \times {10^9}K \cr} $$
Releted MCQ Question on Modern Physics >> Atoms or Nuclear Fission and Fusion
In the nuclear fusion reaction
$$_1^2H + _1^3H \to _2^4He + n$$
given that the repulsive potential energy between the two
nuclei is $$ \sim 7.7 \times {10^{ - 14}}J,$$ the temperature at which the gases must be heated to initiate the reaction is nearly
[Boltzmann’s Constant $$k = 1.38 \times {10^{ - 23}}J/K$$ ]
The binding energy per nucleon of deuteron $$\left( {_1^2H} \right)$$ and helium nucleus $$\left( {_2^4He} \right)$$ is $$1.1\,MeV$$ and $$7\,MeV$$ respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is