In the fission of U-235, the percentage of mass converted into energy is:
Option A: 0.01%
Option B: 0.1%
Option C: 1%
Option D: 10%
Answer
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Hint: When nuclear fission takes place some mass of the parent atom is given out as energy. The released energy can be used to calculate the reduction of mass.
Complete solution:
New possibilities emerge when we go beyond natural radioactive decays and study nuclear reactions by bombarding nuclei with other nuclear particles such as protons, neutrons, $\alpha $ particles, etc.
A most important neutron-induced nuclear reaction is fission. In nuclear fission thermal neutrons or protons or other low mass highly energized particles are bombarded on the nucleus of heavy metals. This bombarding of neutrons results in breaking of the heavy metal into two or more less mass elements.
This process is based on the fact that heavy nuclei are relatively less stable than low mass nuclei. In this process there takes place reduction of atomic mass unit (amu) i.e. sometimes the sum of the atomic masses of the daughter nuclei is less than that of the parent nucleus. This is because, small amount of mass (mostly neutrons) escapes the parent nucleus and some mass is converted into energy. This energy is called Q value and is related to amu as given below;
931MeV = 1amu.
The unit for energy in such cases is MeV.
Now, if we talk about U-235, its fission is done by bombarding its nucleus with thermal neutrons. It breaks into Ba-141 and Kr-92 along with three neutral neutrons. The reaction can be written as;
${}_{92}^{235}U \to {}_{56}^{141}Ba + {}_{36}^{92}Kr + 3{}_0^1n + 200MeV$.
Here the q-value for the reaction is 200MeV.
Also, we know that, $931MeV = 1amu$
Hence, $200MeV = \dfrac{{200}}{{931}} = 0.215amu$
The amount of mass converted into energy is 0.215 amu.
Also atomic mass of U-235 is 235 amu.
Hence percentage mass converted into energy is $ = \dfrac{{0.215}}{{235}} \times 100 = 0.09 = 0.1\% $.
Therefore option B is correct.
Note:
(1) The mass lost as neutrons and the mass lost as energy are not the same.
(2) The neutrons in the product side can be the neurons that were thrown or the neutrons from the parent nucleus.
Complete solution:
New possibilities emerge when we go beyond natural radioactive decays and study nuclear reactions by bombarding nuclei with other nuclear particles such as protons, neutrons, $\alpha $ particles, etc.
A most important neutron-induced nuclear reaction is fission. In nuclear fission thermal neutrons or protons or other low mass highly energized particles are bombarded on the nucleus of heavy metals. This bombarding of neutrons results in breaking of the heavy metal into two or more less mass elements.
This process is based on the fact that heavy nuclei are relatively less stable than low mass nuclei. In this process there takes place reduction of atomic mass unit (amu) i.e. sometimes the sum of the atomic masses of the daughter nuclei is less than that of the parent nucleus. This is because, small amount of mass (mostly neutrons) escapes the parent nucleus and some mass is converted into energy. This energy is called Q value and is related to amu as given below;
931MeV = 1amu.
The unit for energy in such cases is MeV.
Now, if we talk about U-235, its fission is done by bombarding its nucleus with thermal neutrons. It breaks into Ba-141 and Kr-92 along with three neutral neutrons. The reaction can be written as;
${}_{92}^{235}U \to {}_{56}^{141}Ba + {}_{36}^{92}Kr + 3{}_0^1n + 200MeV$.
Here the q-value for the reaction is 200MeV.
Also, we know that, $931MeV = 1amu$
Hence, $200MeV = \dfrac{{200}}{{931}} = 0.215amu$
The amount of mass converted into energy is 0.215 amu.
Also atomic mass of U-235 is 235 amu.
Hence percentage mass converted into energy is $ = \dfrac{{0.215}}{{235}} \times 100 = 0.09 = 0.1\% $.
Therefore option B is correct.
Note:
(1) The mass lost as neutrons and the mass lost as energy are not the same.
(2) The neutrons in the product side can be the neurons that were thrown or the neutrons from the parent nucleus.
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