Answer to Question #198670 in Atomic and Nuclear Physics for Hamna

Question

Find mass defect and binding energy for helium nucleus

Answer 1

$1\;u=931.5 \;MeV/c^2$

Mass Number of He, A= 4

Atomic Number of He, Z=2

$m_p=1.0078 \;u \\ m_n=1.0086 \;u \\ m_{nuc}= 4.0026 \;u$

Formation of the nucleus by combining neutron and proton releases energy. This energy is called as Binding energy and it is calculated by:

$E_b=Δmc^2$

where, Δm=mass defect

c= Light velocity

Mass of nucleus is less than the mass of individual constituent in a nucleus, this difference in a mass is called a mass defect.

Mass defect is calculated by:

$Δm=Zm_p+(A-Z)m_n-m_{nuc}$

where, $Zm_p$ = Total mass of proton

$(A-Z)m_n$ = Total mass of neutron

$m_{nuc}$ = Mass of nucleus.

A mass defect for Helium is,

$Δm=(2 \times 1.0079\;u)+((4-2) \times 1.0086\;u)-4.0026u \\ Δm=2.0156\;u+2.0173\;u-4.0026\;u \\ Δm= 0.03037\;u$

The energy released from Helium-4 is:

$E_b=Δmc^2 \\ = 0.03037\;u \times c^2 \\ = (0.03037 \times 931.5) \;MeV/c^2 \times c^2 \\ E_b=28.3\;MeV$

Hence, the binding energy is the energy released by the formation of nucleus and mass defect is the difference between the mass of nucleus and masses of individual content in the nucleus.

Mass defect for 4He is 0.03037 u

Binding energy for 4He is 28.3 MeV