Questions
The binding energy per nucleon for Deuteron and Helium are 1.1 MeV and 7 MeV respectively. The energy released when two Deuterons fuse to form a Helium nucleus is
detailed solution
Correct option is C
H 2 1 +H 2 1 →H 4 2 For each H 2 1 →BE1=2(1.1)=2.2 MeV For He 4 2 →BE2=4(7)=28 MeV E=BE2–2BE1 = 28–2(2.2) = 23.6 MeVTalk to our academic expert!
Similar Questions
In a fission reaction the binding energy per nucleon of X and Y is 8.5 MeV where as of is 7.6 MeV. The total energy liberated will be about
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
