How much mass is converted into energy in a hydrogen bomb.
Answer
527.1k+ views
Hint: The working principle of a hydrogen bomb is nuclear fusion. In the nuclear fusion process, four hydrogen atoms (small one) are combined to generate one helium atom (larger one). In order to calculate the amount of mass converted into energy, we have to take the difference between mass of reactants and products.
Complete answer:
The reaction takes place in a hydrogen bomb is as follows:
$ 4 {_1}{^1}H \to {_2}{^4}He + 2 {_1}{^0}e $
According to the above reaction, we conclude that four hydrogen atoms are combined to form one molecule of helium. As we know that the atomic mass of hydrogen $ \left( H \right)=1.007824u $
And the atomic mass of helium $ \left( He \right)=4.002602u $
The mass difference $ \left( \Delta m \right) $ between the reactants (hydrogen atoms) and product (helium) is given as:
$ \Delta m=4(1.007824)-4.002602 $
$ \Rightarrow \Delta m=4.031296-4.002602 $
Hence, $ \Delta m=0.028694u $
It means that $ 0.028694u $ amount of mass is released as energy in a hydrogen bomb.
Therefore, $ 0.028694u $ mass is converted into energy in a hydrogen bomb.
Additional Information:
In a hydrogen bomb, $ 0.028694u $ mass is converted into energy according to the Einstein’s equation which is represented as follows:
$ E=m{{c}^{2}} $
Where $ E $ and $ m $ refers to the energy and mass, respectively. $ c $ represents the speed of light which is constant and equals to $ 3\times {{10}^{8}}m{{s}^{-1}} $ .
Note:
It is important to note that the process of nuclear fusion occurs in a hydrogen bomb. According to this process, four hydrogen atoms are combined to generate one helium atom in a hydrogen bomb. During this fusion, the mass difference $ \left( \Delta m \right) $ between four hydrogen atoms and one helium atom comes out to be $ 0.028694u $ . This is the amount of mass i.e. $ 0.028694u $ which is released as energy in the hydrogen bomb.
Complete answer:
The reaction takes place in a hydrogen bomb is as follows:
$ 4 {_1}{^1}H \to {_2}{^4}He + 2 {_1}{^0}e $
According to the above reaction, we conclude that four hydrogen atoms are combined to form one molecule of helium. As we know that the atomic mass of hydrogen $ \left( H \right)=1.007824u $
And the atomic mass of helium $ \left( He \right)=4.002602u $
The mass difference $ \left( \Delta m \right) $ between the reactants (hydrogen atoms) and product (helium) is given as:
$ \Delta m=4(1.007824)-4.002602 $
$ \Rightarrow \Delta m=4.031296-4.002602 $
Hence, $ \Delta m=0.028694u $
It means that $ 0.028694u $ amount of mass is released as energy in a hydrogen bomb.
Therefore, $ 0.028694u $ mass is converted into energy in a hydrogen bomb.
Additional Information:
In a hydrogen bomb, $ 0.028694u $ mass is converted into energy according to the Einstein’s equation which is represented as follows:
$ E=m{{c}^{2}} $
Where $ E $ and $ m $ refers to the energy and mass, respectively. $ c $ represents the speed of light which is constant and equals to $ 3\times {{10}^{8}}m{{s}^{-1}} $ .
Note:
It is important to note that the process of nuclear fusion occurs in a hydrogen bomb. According to this process, four hydrogen atoms are combined to generate one helium atom in a hydrogen bomb. During this fusion, the mass difference $ \left( \Delta m \right) $ between four hydrogen atoms and one helium atom comes out to be $ 0.028694u $ . This is the amount of mass i.e. $ 0.028694u $ which is released as energy in the hydrogen bomb.
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