Obtain the binding energy of the nuclei $_{26}^{56} F e$ and $_{83}^{209} B i$ in units of MeV from the following data:
(a) $m (_{26}^{56} F e) = 55.934939 u$
(b) $m (_{83}^{209} B i) = 208.980388 u$

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Solution

(a) $_{26}^{56} F e$ contains 26 protons and 30 neutrons.
Mass of 26 protons= $26 \times 1.007825 u = 26.26345 u$
Mass of 30 neutrons= $30 \times 1.008665 u = 30.25995 u$
Mass of $_{26}^{56} F e$ nucleus= $55.934939 u$
Thus, mass defect $Δ m = 26.26345 u + 30.25995 u - 55.934939 u = 0.528461 u$ .
Thus, binding energy of $_{26}^{56} F e$ nucleus is $0.528461 \times 931.5 M e V = 492.26 M e V$ .
Binding energy per nucleon= $\frac{492.26 M e V}{56} = 8.790 M e V$

(b) $_{83}^{209} B i$ contains 83 protons and 126 neutrons.
Mass of 83 protons= $83 \times 1.007825 u = 83.649475 u$
Mass of 126 neutrons= $126 \times 1.008665 u = 127.09170 u$
Mass of $_{83}^{209} B i$ nucleus= $208.986388 u$
Thus, mass defect $Δ m = 83.649475 u + 127.09170 u - 208.986388 u = 1.760872 u$ .
Thu,s binding energy of $_{83}^{209} B i$ nucleus is $1.760872 \times 931.5 M e V = 1640.26 M e V$ .
Binding energy per nucleon= $\frac{1640.26 M e V}{209} = 7.848 M e V$

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Similar questions

_{8}^{17} O \to 131 M e V

_{26}^{56} F e \to 493 M e V

_{92}^{238} U \to 1804 M e V

(Δ = \frac{B}{A})

1._{8}^{17} O \to 131 M e V

2._{26}^{56} F e \to 493 M e V

3._{92}^{238} U \to 1804 M e V

(Δ = \frac{B}{A})

_{26} F e^{56}

_{83} B i^{209}

= 1.007825

= 1.08665

_{26} F e^{56}

= 55.934939

_{83} B i^{209}

= 208.980388

_{26}^{56} F e

^{56} F e

^{1} H

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