If the binding energy per nucleon in _{3}^{7}Li and _{2}^{4}He nuclei are 5.60 MeV and 7.06 MeV  respectively, then in the reaction :

p+_{3}^{7}Li\rightarrow 2_{2}^{4}He energy of proton must be :

  • Option 1)

    39.2 MeV

  • Option 2)

    28.24 MeV]

  • Option 3)

    17.28 MeV

  • Option 4)

    1.46 MeV

 

Answers (1)

As we learnt in

Q value -

X+Y\rightarrow Z+Q

Q=(M_{x}+M_{y}-M_{z})C^{2}

     

- wherein

M_{x} and M_{y} are mass of reactant

M_{z}  is mass of product

 

 

 

Binding energy of    _{3}^{7}\textrm{Li}=7\times 5.60=39.2\; Me\, V

Binding energy of    _{2}^{4}\textrm{He}=4\times 7.06=28.24\, Me\, V

\therefore \;    Energy of proton = Energy of \left [ 2(_{2}^{4}\textrm{He})-_{3}^{7}\textrm{Li} \right ]

=2\times 28.24-39.2

=17.28\; Me\, V

Correct option is 3.


Option 1)

39.2 MeV

This is an incorrect option.

Option 2)

28.24 MeV]

This is an incorrect option.

Option 3)

17.28 MeV

This is the correct option.

Option 4)

1.46 MeV

This is an incorrect option.

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