Suppose we think of fission of a \(^{56}_{26}Fe\) nucleus into two equal fragments, \(^{28}_{13} Al.\) Is the fission energetically possible? Argue by working out Q of the process. Given \(m\)(\(^{56}_{26}Fe\)) = 55.93494 u and \(m\)(\(^{28}_{13} Al\)) = 27.98191 u.
Suppose we think of fission of a \(^{56}_{26}Fe\) nucleus into two equal fragments, \(^{28}_{13} Al.\) Is the fission energetically possible? Argue by working out Q of the process. Given \(m\)(\(^{56}_{26}Fe\)) = 55.93494 u and \(m\)(\(^{28}_{13} Al\)) = 27.98191 u.
Solution and Explanation
The fission of \( ^{56}_ {26}Fe \) can be given as:
\(^{56}_{13} Fe → 2\space ^{28}_{13} Al\)
It is given that:
Atomic mass of \(m(^{56}_ {26}Fe) = 55.93494 u\)
Atomic mass of \(m ( ^{28}_{13} Al ) = 27.98191 u\)
The Q-value of this nuclear reaction is given as:
Q = \([m(^{56}_{26}Fe) - 2m(^{28}_{13} Al)]c^2\)
\(Q = [55.93494 - 2 \times 27.98191]c^2\)
\(Q = (-0.02888 \space c^2)u\)
But 1u = 931.5 \(\frac{MeV}{c^{2}}\)
Q = -0.02888 x 931.5
Q = -26.902 MeV
The Q-value of the fission is negative. Therefore, the fission is not possible energetically. For an energetically possible fission reaction, the Q-value must be positive.
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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.