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
254.1k+ 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
Mass vs Weight: Key Differences Explained for Students

Uniform Acceleration Explained: Formula, Examples & Graphs

JEE Main 2022 (June 25th Shift 2) Chemistry Question Paper with Answer Key

Average Atomic Mass - Important Concepts and Tips for JEE

JEE Main 2023 (April 15th Shift 1) Physics Question Paper with Answer Key

JEE Main 2022 (June 27th Shift 2) Chemistry Question Paper with Answer Key

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

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

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Understanding Atomic Structure for Beginners

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

