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The Binding energy per nucleon of \frac{7}{3}\text{Li} and \frac{4}{2}He nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction

\frac{7}{3}\text{Li}+\frac{1}{1}\text{H}\rightarrow \frac{4}{2}\text{He}+\frac{4}{2}\text{He}+\text{Q}, the value of energy Q released is:

  • Option 1)

    19.6 MeV

  • Option 2)

    -2.4 MeV

  • Option 3)

    8.4 MeV

  • Option 4)

    17.3 MeV

 

Answers (1)

best_answer

_{3}^{7}\textrm{Li}+_{1}^{1}\textrm{H}\rightarrow _{2}^{4}\textrm{He}+_{2}^{4}\textrm{He}+Q

Q value of reaction

=2\times E_{He}-\left [ E_{Li} \right ]

=2\times \left [ 7.06\times4 \right ]-[7\times5.60]\:MeV

=56.48 - 39.20 MeV

= 17.3 MeV 


Option 1)

19.6 MeV

This option is incorrect.

Option 2)

-2.4 MeV

This option is incorrect.

Option 3)

8.4 MeV

This option is incorrect.

Option 4)

17.3 MeV

This option is correct.

Posted by

divya.saini

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