Question

Given m_p = 1.007876 amu and m_n = 1.008665 amu. Calculate the binding energy of an α -particle if 1 amu = 931.3 MeV. [Mass of the α-particle is 4.0028 amu]
1. 28.29 MeV
2. 29.28 MeV
3. 14.15 MeV
4. 28.29 GeV

Answer 1

Correct Answer - Option 1 : 28.29 MeV

Concept:

Alpha Particles: 

• The alpha particles are obtained during radioactive decay. 
• It is basically a Helium nucleus. 
• The helium Nucleus consists of 2 neutrons and 2 protons. 

Binding Energy:

• The binding energy is the minimum energy required to remove nucleons (Protons or neutrons) to an infinite distance from the nucleus.
• It can be expressed as 

ΔE =  Δm c2

Δm is the mass defect that is the difference of mass between a particle and nucleons present in it. c is the speed of light

Calculation:

Given mass of α-particle = 4.0028 amu

mass of nucleons present in alpha particle = 2 (m_p + m_n ) = 2 (1.007876 + 1.008665 ) amu = 4.033 amu

Mass defect = Δm = 4.033 - 4.0028 = 0,0302 amu

Now it is given energy equivalent to 1 amu = 931.3 MeV

The energy equivalent to 0,0302 amu = 0.0302 × 931.3 MeV ≈ 28.29 MeV

So, the correct option is 28.29 MeV.