The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu. The binding energy of the helium nucleus will be [1 amu= 931 MeV]
A
28.4 MeV
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
20.8 MeV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
27.3 MeV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
14.2 MeV
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A 28.4 MeV Helium nucleus consist of two neutrons and two protons. So binding energy E=Δm×931MeV ⇒E=(2×mp+2mn−M)×931MeV=(2×1.0073+2×1.0087−4.0015)×931=28.4MeV