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  • Tell me? - Consider the nuclear fission Given that the binding energy/ nucleon of are respectively, 8.03 MeV, - Atoms And Nuclei - JEE Main

Consider the nuclear fission

Ne^{20}\rightarrow 2He^{4}+C^{12}

Given that the binding energy/ nucleon of Ne^{20}, He^{4}\: and \: C^{12}are respectively,

8.03 MeV, 7.07MeV and 7.86 Me V , identify the correct statement:

 

  • Option 1)

    energy of 3.6 MeV will be released

  • Option 2)

    energy of 11.9 MeV has to supplied

  • Option 3)

    energy of 12.4 MeV will be supplied

  • Option 4)

    8.3 MeV energy will be released

Answers (1)

best_answer

 

Binding energy -

B.E=\left [ Zm_{p}+(A-Z)m_{n}-M_{x} \right ]c^{2} 

amount of energy released when nucleons  come together to form a nuclei.

M_{x} = mass of nuclei formed

 

- wherein

m_{p}- mass\ of\ proton

m_{n}- mass\ of \ neutron

Z- atomic\ number

A- atomic\ mass

 

 

 

Binding energy per nucleon -

 

- wherein

This graph shows the stability of nuclei many nuclear phenomena can be explained by this graph

 

 

Ne^{20}\rightarrow 4He^{4}+C^{12}

E_{B} = (BE)_{reactant} - (BE)_{product}

E_{B} = 8.03\times 20 - \left ( 2\times 7.07\times 4+7.86\times 12 \right ) = 9.72MeV


Option 1)

energy of 3.6 MeV will be released

Option 2)

energy of 11.9 MeV has to supplied

Option 3)

energy of 12.4 MeV will be supplied

Option 4)

8.3 MeV energy will be released

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