Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments ‘B’ and ‘C’ of mass numbers 105 and 115. The binding energy of nucleons in ‘B’ and ‘C’ is 6.4 MeV per nucleon. The energy Q released per fission will be :
Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments ‘B’ and ‘C’ of mass numbers 105 and 115. The binding energy of nucleons in ‘B’ and ‘C’ is 6.4 MeV per nucleon. The energy Q released per fission will be :
- 0.8 MeV
- 275 MeV
- 220 MeV
- 176 MeV
The Correct Option is D
Solution and Explanation
220A → 105B+115C
⇒ Q = [105 × 6.4 + 115×6.4] – [220 × 5.6] MeV
⇒ Q = 176 MeV
The correct option is (D) : 176 MeV
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Concepts Used:
Nuclear Physics
Nuclear physics is the field of physics that studies atomic nuclei and their constituents and interactions, in addition to the study of other forms of nuclear matter. Nuclear physics should not be confused with atomic physics, which studies the atom as a whole, including its electrons
Radius of Nucleus
‘R’ represents the radius of the nucleus. R = RoA1/3
Where,
- Ro is the proportionality constant
- A is the mass number of the element
Total Number of Protons and Neutrons in a Nucleus
The mass number (A), also known as the nucleon number, is the total number of neutrons and protons in a nucleus.
A = Z + N
Where, N is the neutron number, A is the mass number, Z is the proton number
Mass Defect
Mass defect is the difference between the sum of masses of the nucleons (neutrons + protons) constituting a nucleus and the rest mass of the nucleus and is given as:
Δm = Zmp + (A - Z) mn - M
Where Z = atomic number, A = mass number, mp = mass of 1 proton, mn = mass of 1 neutron and M = mass of nucleus.