Question

Find the total binding energy of the mass C = 12.000000 u. nucleus (Z =6), given the atomic A. 127.6 MeV B. 104.6 MeV C. 931.5 MeV D. 92.2 MeV E. 89.1 MeV

Answer 1

Answer: The correct answer is Option D.

Explanation:

We are given a nucleus having representation:  $_6^{12}\textrm{C}$

Number of protons = 6

Number of neutrons = 12 - 6 = 6

Number of electrons = 6

To calculate the mass defect of the nucleus, we use the equation:

$\Delta m=[(n_p\times m_p)+(n_n\times m_n)+(n_e\times m_e)]-M$

where,

$n_p$ = number of protons  = 6

$m_p$ = mass of one proton  = 1.00728 u

$n_n$ = number of neutrons  = 6

$m_n$ = mass of one neutron = 1.00866 u

$n_e$ = number of electrons  = 6

$m_e$ = mass of one electron = 0.00054858 u

M = Mass number = 12

Putting values in above equation, we get:

$\Delta m=[(6\times 1.00728)+(6\times 1.00866)+(6\times 0.00054858)]-12\\\\\Delta m=0.098931u$

To calculate the binding energy of the nucleus, we use the equation:

$E=\Delta mc^2\\E=(0.098931u)\times c^2$

$E=(0.098931u)\times (931.5MeV)$    (Conversion factor:  $1u=931.5MeV/c^2$  )

$E=92.2MeV$

Hence, the correct answer is Option D.