If the binding energy per nucleon in 73Li and 42He nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction p+73Li→242He energy of proton must be

28.24 MeV

17.28 MeV

1.46MeV

39.2MeV

Views: 1,034 students

Found 3 tutors discussing this question

Owen Discussed

If the binding energy per nucleon in 73Li and 42He nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction p+73Li→242He energy of proton must be

7 mins ago

Discuss this question LIVE

7 mins ago

Text solutionVerified

[b] Let E be the energy of proton, then E+7×5.6=2×[4×7.06] ⇒E=56.48−39.2=17.28MeV

Was this solution helpful?

150

Report

One destination for complete JEE/NEET preparation

One destination to cover all your homework and assignment needs

Learn Practice Revision Succeed

Instant 1:1 help, 24x7

60, 000+ Expert tutors

Textbook solutions

Big idea maths, McGraw-Hill Education etc

Essay review

Get expert feedback on your essay

Schedule classes

High dosage tutoring from Dedicated 3 experts

Trusted by 4 million+ students

Stuck on the question or explanation?

Connect with our physics tutors online and get step by step solution of this question.

231 students are taking LIVE classes

Question Text	If the binding energy per nucleon in 73Li and 42He nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction p+73Li→242He energy of proton must be
Answer Type	Text solution:1
Upvotes	150

If the binding energy per nucleon in 73Li and 42He nuclei are 5.60 MeV and 7.06 MeV respectively, then in the reaction p+73Li→242He energy of proton must be

Text solutionVerified

Practice questions from similar books