NCERT Class XII Chapter
Nuclei
Questions With Solutions
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Question : 3
Total: 31
Obtain the binding energy (in MeV) of a nitrogen nucleus (
N ) (
N )
Solution:
The
N
Mass of 7 protons = 7 × 1.00783 = 7.05481 amu
Mass of 7 neutrons = 7 × 1.00867 = 7.06069 amu
Total mass = 14.11550 amu
Mass of
N
Mass defect, = Δm = 0.11243 amu
B.E. of nitrogen nucleus = 0.11243 × 931 = 104.67 MeV.
Mass of 7 protons = 7 × 1.00783 = 7.05481 amu
Mass of 7 neutrons = 7 × 1.00867 = 7.06069 amu
Total mass = 14.11550 amu
Mass of
Mass defect, = Δm = 0.11243 amu
B.E. of nitrogen nucleus = 0.11243 × 931 = 104.67 MeV.
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