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NCERT Class XII Chapter
Nuclei
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Question : 3 of 31
Marks: +1, -0
Obtain the binding energy (in MeV) of a nitrogen nucleus (714N)(^{14}_{7}\mathrm{N}) , given m (714N)(^{14}_{7}\mathrm{N}) = 14.00307 u
Solution:  
The 714N^{14}_{7}\mathrm{N} nucleus contains 7 protons and 7 neutrons.
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 714N^{14}_{7}\mathrm{N} nucleus = 14.00307 amu
Mass defect, = Δm = 0.11243 amu
B.E. of nitrogen nucleus = 0.11243 × 931 = 104.67 MeV.
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