All Courses
Explore Courses UPSC Exam UPSC Test Series Free Notes EduRev Infinity Get the App Login Join for Free
Sign in Open App
JEE Exam  >  JEE Questions  >  Two deuterons undergo nuclear fusion to form ... Start Learning for Free
SaveJoin the Discussion on the App
Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)
  • a)
    30.2 MeV
  • b)
    32.4 MeV
  • c)
    23.6 MeV
  • d)
    25.8 MeV
Correct answer is option 'C'. Can you explain this answer?
FREE This question is part of
Download PDF Attempt Test
Download PDF Attempt this Test
Most Upvoted Answer
Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy ...
Given Information:
- Binding energy per nucleon for deuteron = 1.1 MeV
- Binding energy per nucleon for helium = 7.0 MeV

Objective:
To calculate the energy released in the nuclear fusion process of two deuterons to form a helium nucleus.

Explanation:
Nuclear fusion is a process in which two light atomic nuclei combine to form a heavier nucleus. In this case, two deuterons (nuclei of heavy hydrogen, each containing one proton and one neutron) combine to form a helium nucleus (containing two protons and two neutrons). The reaction can be represented as:

2H + 2H → 4He

To calculate the energy released in this process, we need to consider the difference in binding energy between the reactants (deuterons) and the product (helium nucleus).

Step 1:
Calculate the total binding energy of the reactants (deuterons):
Binding energy per nucleon for deuteron = 1.1 MeV
Number of nucleons in a deuteron = 2 (1 proton + 1 neutron)
Total binding energy of 2 deuterons = 2 x 2 x 1.1 MeV = 4.4 MeV

Step 2:
Calculate the total binding energy of the product (helium nucleus):
Binding energy per nucleon for helium = 7.0 MeV
Number of nucleons in helium nucleus = 4 (2 protons + 2 neutrons)
Total binding energy of helium nucleus = 4 x 7.0 MeV = 28.0 MeV

Step 3:
Calculate the energy released:
Energy released = (Total binding energy of reactants) - (Total binding energy of product)
Energy released = 4.4 MeV - 28.0 MeV = -23.6 MeV

The negative sign indicates that energy is released in the fusion process.

Final Answer:
The energy released in the nuclear fusion process of two deuterons to form a helium nucleus is 23.6 MeV. Hence, the correct answer is option C.
< Previous Question Next Question >
Explore Courses for JEE exam

Similar JEE Doubts

PASSAGE-2Scientists are working hard to develop nuclear fusion reactor.Nuclei of heavy hydrogen,known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D r eaction isenergy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection ofnuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma.Special techniques are used which confine the plasma for a time t0 before the particles fly away form the core. If n is the density (number/volum e) of deuterons, the product nt0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 × 1014 s/cm3.It may be helpful to use the following: Boltzmann constantQ.Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion?

PASSAGE-2Scientists are working hard to develop nuclear fusion reactor.Nuclei of heavy hydrogen,known as deuteron and denoted by D, can be thought of as a candidate for fusion reactor. The D-D r eaction isenergy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection ofnuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma.Special techniques are used which confine the plasma for a time t0 before the particles fly away form the core. If n is the density (number/volum e) of deuterons, the product nt0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 × 1014 s/cm3.It may be helpful to use the following: Boltzmann constantQ.Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 × 10–15 m is in the range

Top Courses for JEEView all

Top Courses for JEE

Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)a)30.2 MeVb)32.4 MeVc)23.6 MeVd)25.8 MeVCorrect answer is option 'C'. Can you explain this answer?
Question Description
Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)a)30.2 MeVb)32.4 MeVc)23.6 MeVd)25.8 MeVCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)a)30.2 MeVb)32.4 MeVc)23.6 MeVd)25.8 MeVCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)a)30.2 MeVb)32.4 MeVc)23.6 MeVd)25.8 MeVCorrect answer is option 'C'. Can you explain this answer?.
Solutions for Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)a)30.2 MeVb)32.4 MeVc)23.6 MeVd)25.8 MeVCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)a)30.2 MeVb)32.4 MeVc)23.6 MeVd)25.8 MeVCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)a)30.2 MeVb)32.4 MeVc)23.6 MeVd)25.8 MeVCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)a)30.2 MeVb)32.4 MeVc)23.6 MeVd)25.8 MeVCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)a)30.2 MeVb)32.4 MeVc)23.6 MeVd)25.8 MeVCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Two deuterons undergo nuclear fusion to form a Helium nucleus. Energy released in this process is : (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)a)30.2 MeVb)32.4 MeVc)23.6 MeVd)25.8 MeVCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses

Suggested Free Tests

Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Get App
© EduRev
Education Revolution
Follow Us
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup