0

Guest

Courses New

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
10 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
5 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
3 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

Think You Can Provide A Better Answer ?

Other Related Questions on General Physics

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More