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If the binding energy per nucleon in 3 Li 7 and 2 He 4 nuclei are respectively 5.60 MeV and 7.06 Me V, then the energy of proton in the reaction 3Li 7 + p ---> 2 2 He 4 is (a) 19.6 MeV (b) 2.4 MeV (c) 8.4 MeV (d) 17.3 MeV



If the binding energy per nucleon in 3Li7 and 2He4 nuclei are respectively 5.60 MeV and 7.06 Me V, then the energy of proton in the reaction 3Li7 + p --->2 2He4 is


        (a)    19.6 MeV          


(b)    2.4 MeV


        (c)    8.4 MeV


        (d)    17.3 MeV


Grade:upto college level

3 Answers

Amit Saxena
35 Points
8 years ago

(d)

Applying principle of energy conservation, Energy of proton

=      total B.E. of 2a - energy ofLi7

=      8 × 7.06 - 7 × 5.6

=      56.48 - 39.2 = 17.28 MeV

Anurag Thamke
15 Points
3 years ago
By applying principle ofel energy conservation,   
Total binding energy of product -energy of lithium 
=8×7.60 - 7×5.6
=56.48-) - 39.2 =17.28
 
 
Rishi Sharma
askIITians Faculty 646 Points
2 years ago
Dear Student,
Please find below the solution to your problem.

Q=2(4×7.06)−7×(5.60)
=(56.48−39.2)MeV
=17.28MeV
=17.3MeV

Thanks and Regards

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