
If \[1gm\] of hydrogen is converted into \[0.993gm\] of helium in a thermonuclear reaction, the energy released in the reaction is
\[(A)63 \times {10^7}J\]
\[(B)63 \times {10^{10}}J\]
\[(C)63 \times {10^{14}}J\]
\[(D)\]\[63 \times {10^{20}}J\]
Answer
516.9k+ views
Hint:Each helium atom is made up of four hydrogen atoms fused together. Some of the mass is turned into energy during the process.
The proton-proton cycle and the carbon-nitrogen-oxygen cycle are two chains of processes that are used to convert hydrogen to helium.
Two hydrogen nuclei (protons) are fused in the proton-proton cycle, and one of these protons is transformed to a neutron by beta decay (see radioactivity) to form a deuterium nucleus (one proton and one neutron). The light isotope of helium, helium-3, is created by adding a third proton to deuterium. When two helium-3 nuclei collide, a nucleus of conventional helium, helium-4 (two protons and one neutron), is formed. Each of these steps releases a significant amount of energy.
Complete step-by-step solution:
Given,
Mass of Hydrogen \[\left( {{m_1}} \right) = 1gm = 1 \times {10^{ - 3}}kg\]
Mass of Helium \[\left( {{m_2}} \right) = 0.993{\text{ }}gm = 0.993 \times {10^{ - 3}}kg\]
For energy released,
\[E = \left( {{m_1} - {m_2}} \right){c^2}\]
\[ \Rightarrow E = \Delta m{c^2}\]
\[ \Rightarrow E = \left\{ {\left. {\left( {1 \times {{10}^{ - 3}}} \right) - \left( {0.993 \times {{10}^{ - 3}}} \right)} \right\} \times {{\left( {3 \times {{10}^8}} \right)}^2}} \right.\]
\[ \Rightarrow E = 7 \times {10^6} \times 9 \times {10^6} = 63 \times {10^{10}}J\]
Note:Because the helium nucleus has a somewhat lower mass than the original four protons, energy is released. Each conversion of four hydrogen nuclei into a helium nucleus releases approximately 10 million times more energy than the chemical reaction that produces water when hydrogen interacts with oxygen.
In the sun, nuclear fusion transforms hydrogen to helium while releasing energy.
In a solar fusion, some mass is transformed into energy.
Helium nuclei are formed when hydrogen nuclei unite. The energy associated with the mass defect is released.
The proton-proton cycle and the carbon-nitrogen-oxygen cycle are two chains of processes that are used to convert hydrogen to helium.
Two hydrogen nuclei (protons) are fused in the proton-proton cycle, and one of these protons is transformed to a neutron by beta decay (see radioactivity) to form a deuterium nucleus (one proton and one neutron). The light isotope of helium, helium-3, is created by adding a third proton to deuterium. When two helium-3 nuclei collide, a nucleus of conventional helium, helium-4 (two protons and one neutron), is formed. Each of these steps releases a significant amount of energy.
Complete step-by-step solution:
Given,
Mass of Hydrogen \[\left( {{m_1}} \right) = 1gm = 1 \times {10^{ - 3}}kg\]
Mass of Helium \[\left( {{m_2}} \right) = 0.993{\text{ }}gm = 0.993 \times {10^{ - 3}}kg\]
For energy released,
\[E = \left( {{m_1} - {m_2}} \right){c^2}\]
\[ \Rightarrow E = \Delta m{c^2}\]
\[ \Rightarrow E = \left\{ {\left. {\left( {1 \times {{10}^{ - 3}}} \right) - \left( {0.993 \times {{10}^{ - 3}}} \right)} \right\} \times {{\left( {3 \times {{10}^8}} \right)}^2}} \right.\]
\[ \Rightarrow E = 7 \times {10^6} \times 9 \times {10^6} = 63 \times {10^{10}}J\]
Note:Because the helium nucleus has a somewhat lower mass than the original four protons, energy is released. Each conversion of four hydrogen nuclei into a helium nucleus releases approximately 10 million times more energy than the chemical reaction that produces water when hydrogen interacts with oxygen.
In the sun, nuclear fusion transforms hydrogen to helium while releasing energy.
In a solar fusion, some mass is transformed into energy.
Helium nuclei are formed when hydrogen nuclei unite. The energy associated with the mass defect is released.
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