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Normal 0false false falseEN-US X-NONE X-NONE/* Style Definitions */table.MsoNormalTable{mso-style-name:Table Normal;mso-tstyle-rowband-size:0;mso-tstyle-colband-size:0;mso-style-noshow:yes;mso-style-priority:99;mso-style-parent:;mso-padding-alt:0in 5.4pt 0in 5.4pt;mso-para-margin-top:0in;mso-para-margin-right:0in;mso-para-margin-bottom:10.0pt;mso-para-margin-left:0in;line-height:115%;mso-pagination:widow-orphan;font-size:11.0pt;font-family:Calibri,sans-serif;mso-ascii-font-family:Calibri;mso-ascii-theme-font:minor-latin;mso-hansi-font-family:Calibri;mso-hansi-theme-font:minor-latin;mso-bidi-font-family:Times New Roman;mso-bidi-theme-font:minor-bidi;}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

Amit Saxena , 11 Years ago
Grade upto college level
anser 3 Answers
Amit Saxena

Last Activity: 11 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

Last Activity: 6 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

Last Activity: 4 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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