Q29 / 199 · NTA Abhyas (NEET) 2020 · Physics · Nuclear Physics

Using the following data
Mass hydrogen atom = 1.00783 u
Mass of neutron = 1.00867 u
Mass of nitrogen atom ( 7N14)= 14.00307 u
The calculated value of the binding energy of the nucleus of the nitrogen atom ( 7N14) is close to

  • A
    56 MeV
     
  • B
    98 MeV
     
  • Correct answer
    C
    104 MeV
     
  • D
    112 MeV
     

Solution

The binding energy of nucleus may be defined as the energy equivalent to the mass defect of the nucleus.

If ∆m is mass defect then according to Einstein's mass-energy relation.

Binding Energy

=∆ mc2 = Zmp + A-Z mn -M c2

=7 × 1.00783 + 7 × 1.00867-14.00307c2

or  BE=0.1124 × 931.5 MeV

or BE = 104.6 MeV

Hence, the correct answer is option: 'C'.

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