Question

What is the amount of energy released by deuterium and tritium fusion?
A. \[{{60}}{{.6 eV}}\]
B. \[{{123}}{{.6 eV}}\]
C. \[{{17}}{{.6 eV}}\]
D. \[{{28}}{{.3 eV}}\]

Answer 1

Hint: Nuclear fusion nucleus combines to form a new one. Deuterium and tritium are the major ingredients in most fusion reactions, they combine together to form helium and a neutron releasing energy.

Complete step by step answer:
Nuclear fusion is the process when two nuclei of two lighter atoms combine to form a new nucleus producing energy. Energy is produced according to Einstein’s mass-energy equivalence. The combination of isotopes of Hydrogen Deuterium and Tritium, to give Helium, releasing a neutron and releasing 17 MeV of energy is an example of a fusion reaction. There are several advantages of nuclear fusion: some it is, it is much more cost effective and sustainable, the amount of greenhouse gases produced by it is minimal. It is a safer source of energy which can be used to produce electricity. The availability of fuel is very abundant.
Deuterium and tritium fusion can be written as,
\[_{\text{1}}^{\text{2}}{\text{H + }}_{\text{1}}^{\text{3}}{\text{H = }}_{\text{1}}^{\text{4}}{\text{He + }}_{\text{0}}^{\text{1}}{\text{N}}\]
Mass can be written as,
\[{\text{M}}\,{\text{of}}\;\,_{\text{1}}^{\text{2}}{\text{H}}\,{\text{ + }}\,_{\text{1}}^{\text{3}}{\text{H}}\,\,\,{\text{ = }}\,{\text{2}}{\text{.014102}}\,{\text{ + }}\,{\text{3}}{\text{.016050}}\,{\text{ = }}\,{\text{5}}{\text{.030152}}\]
\[{\text{M}}\,{\text{of}}\;_{\text{1}}^{\text{4}}{\text{He + }}_{\text{0}}^{\text{1}}{\text{N}}\,{\text{ = }}\,{\text{4}}{\text{.002603}}\,{\text{ + }}\,{\text{1}}{\text{.008665}}\,{\text{ = }}\,\,{\text{5}}{\text{.011268}}\]
The change in mass \[{{\Delta m}}\] is,
\[\,{{\Delta m}}\,{\text{ = }}\,{\text{5}}{\text{.030152}}\,{\text{ - }}\,\,{\text{5}}{\text{.011268}}\,\,{\text{ = 0}}{\text{.018884a}}{\text{.m}}{\text{.u}}\]
\[{\text{So, 0}}{\text{.018884 a}}{\text{.m}}{\text{.u are converted to energy for every nucleus of deuterium}}\] for that,
$
{{E = m }}{{{c}}^{{2}}}{{ = }}\left( {{{0}}{{.018884 a}}{{.m}}{{.u}}{{.}}} \right){{ }}\left( {\dfrac{{{{1}}{{.66056 \times 1}}{{{0}}^{{{ - 27}}}}{{ kg}}}}{{{{1 a}}{{.m}}{{.u}}{{.}}}}{{ }}} \right){{ }}\left( {{{3 \times 1}}{{{0}}^{{8}}}{{m}}{{{s}}^{{{ - 1}}}}{{ }}} \right){{2 }} \\
{{ = }}\left( {{{2}}{{.82 \times 1}}{{{0}}^{{{ - 12}}}}{{ J}}} \right){{ }}\left( {\dfrac{{{{6}}{{.242 \times 1}}{{{0}}^{{{12}}}}{{ MeV}}}}{{{{1 J }}}}{{ }}} \right){{ = 17}}{{.6 MeV/nucleus}} \\
$
Thus, the energy released is \[{{17}}{{.6 MeV}}\].

So, the correct answer is Option C.

Additional Information:
Nuclear binding energy is the amount of energy needed to split the nuclei into its component atoms. Iron has the largest binding energy. Any one of the subatomic particles (proton or a neutron) is defined as a Nucleon.

Note: Isotopes are elements that have the same atomic number and different mass number. There are three isotopes of hydrogen, hydrogen, deuterium, and tritium. They differ in the number of neutrons. Hydrogen has no neutron; deuterium has one neutron and tritium has two neutrons.