Question

What is the mass defect and the binding energy of ${}_{27}{Co}^{59}$ which has a nucleus of mass of $58.933u$ ? $(m_p=1.0078u,m_n=1.0087u)$

• A. 517.914 MeV
• B. 617.914 MeV
• C. 17.914 MeV
• D. 717.914 MeV

Answer 1

The correct option is A 517.914 MeV

In ${}_{27}{Co}^{59}$ number of protons $=27$.
Number of neutrons $=59-27=32$.
$\therefore$ The total mass of the nucleus $=27\times 1.0078u+32\times 1.0087u=27.2106u+32.2784u=59.489u$.
Given, the actual mass of the nucleus $=58.933u$.
$\therefore$ The mass defect $=59.489u-58.933u=0.556u$.
$\therefore$ The binding energy $=\Delta m\times 931.5MeV=0.556u\times 931.5MeV=517.914MeV$.