Obtain the binding energy of a nitrogen nucleus `14_7` N. Given m `^14_7` N = 14.00307 u.

Find the binding energy per nucleon of (Ca^40_20) nucleus given m^40(_20) Ca = 39.962589u , m_p = 1.00783u , m_n=1.00876u , TAKE 1amu = 931MeV/C^2

Number of atoms is 1.4 g nitrogen is,

Calculate the percentage of nitrogen in NH_3 (Atomic mass of N = 14, H = 1amu)

The energy released in the fission reaction ""U^(236) to ""X^(117) + Y^(117) + 2 ""_(0)n^(1) , given that the binding energy per nucleon of X and Y is 8.5 MeV and that of ""_(92)U^(236) is 7.6 MeV, is nearly.