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  • I have a doubt, kindly clarify. - Atoms And Nuclei - JEE Main-2

The binding energy per nucleon of deuteron   \left ( _{1}^{2} H\right )   and helium nucleus   \left ( _{2}^{4} He\right ) is   1.1MeV\: and\: \: 7MeV respectively. If two deuteron nuclei react to form a single helium nucleus, then the energy released is

  • Option 1)

    13.9\: \: MeV

  • Option 2)

    26.9\: \: MeV

  • Option 3)

    23.6\: \: MeV

  • Option 4)

    19.2\: \: MeV

 

Answers (2)

best_answer

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

 

 

 Energy released =E \left(^{4}_{2}He \right )-2E\left(^{2}_{1}H \right )=(7\times4-4\times1.1)MeV=23.6\ MeV

Correct option is 3.


Option 1)

13.9\: \: MeV

This is an incorrect option.

Option 2)

26.9\: \: MeV

This is an incorrect option.

Option 3)

23.6\: \: MeV

This is the correct option.

Option 4)

19.2\: \: MeV

This is an incorrect option.

Posted by

solutionqc

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JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

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

 

 

 Energy released =E \left(^{4}_{2}He \right )-2E\left(^{2}_{1}H \right )=(7\times4-4\times1.1)MeV=23.6\ MeV

Correct option is 3.


Option 1)

13.9\: \: MeV

This is an incorrect option.

Option 2)

26.9\: \: MeV

This is an incorrect option.

Option 3)

23.6\: \: MeV

This is the correct option.

Option 4)

19.2\: \: MeV

This is an incorrect option.

Posted by

perimeter

View full answer

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