
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
216.6k+ views
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.
Recently Updated Pages
Wheatstone Bridge Explained: Working, Formula & Uses

Young’s Double Slit Experiment Derivation Explained

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Electricity and Magnetism Explained: Key Concepts & Applications

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Understanding Atomic Structure for Beginners

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Understanding the Electric Field of a Uniformly Charged Ring

