Calculate the binding energy per nucleon of the $^{40}_{20}\mathrm{Ca}$ nucleus. [Given: $m(^{40}_{20}\mathrm{Ca}) = 40.078\ u$; $m_n = 1.008665\ u$; $m_p = 1.007825\ u$.]

$\text{BE}/A \approx 8.55\ \text{MeV}$ (using nuclidic mass $39.96\ u$).

CBSE Class 12 · 2004 · Modern PhysicsShort Answer

Calculate the binding energy per nucleon of the

_{20}^{40} Ca

nucleus. [Given:

m (_{20}^{40} Ca) = 40.078 u

;

m_{n} = 1.008665 u

;

m_{p} = 1.007825 u

Answer: $BE / A \approx 8.55 MeV$ (using nuclidic mass $39.96 u$ ).

For

_{20}^{40} Ca

Z = 20

protons and

A - Z = 20

neutrons.
Mass defect

Δ m = [Z m_{p} + (A - Z) m_{n}] - M

.
Using the tabulated nuclidic mass

M (^{40} Ca) = 39.9626 u

(the value

40.078 u

quoted is the chemical atomic weight, i.e. the natural average, and is not the mass of a single

^{40}

Ca nucleus):

Δ m = [20 (1.007825) + 20 (1.008665)] - 39.9626 = 40.3298 - 39.9626 = 0.3672 u .

BE = 0.3672 \times 931.5 = 342.1 MeV, \frac{BE}{A} = \frac{342.1}{40} \approx 8.55 MeV .

BE / A \approx 8.55 MeV/nucleon

CBSE Class 12 Board · 2004 · Subjective