JEE Main 2017PhysicsNuclear PhysicsMediumMCQ

JEE Main 2017 — Nuclear Physics Question with Solution

JEE Main 2017 (08 Apr Online)

Question

Two deuterons undergo nuclear fusion to form a Helium nucleus. The energy released in this process is (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)

Choose an option

Show full solutionCorrect option: A
Correct answer
A23.6 MeV

Step-by-step explanation

Q=∑B.E. of products- ∑B.E. of reactants

The equation for two deuterons combining to form Helium nucleus is given as H2+1H2→2H4 

Energy released=Q=4B.E.2H4 -4B.E.1H2  
 =4×7−4×1.1

=28−4.4

The energy released in this process is =23.6 MeV

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Nuclear Physics chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2017, covering the Nuclear Physics chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.