The binding energy per nucleon of 209 Bi is MeV.

Question

The binding energy per nucleon of ${}_{83}^{209}B i$ is ______ MeV.

[Take $m ({}_{83}^{209}B i) = 208.980388 u, m_{p} = 1.007825 u, m_{n} = 1.008665 u, 1 u = 931 {MeV/c}^{2}$ ] [2026]

Smart Achievers · Accepted Answer

(2)

$83 {{}_{1}p}^{1} + 126 {{}_{0}n}^{1} \to {}_{83}B i^{209}$

$Δ m = 83 m_{p} + 126 m_{n} - M_{B i}$

$= 83 \times 1.007825 + 126 \times 1.008665 - 208.980388$

$= 1.760877 u$

$B E = 1.760877 \times 931 MeV = 1639.3764 MeV$

$\frac{B E}{A} = \frac{1639.3764}{209} MeV = 7.8439 MeV$

Want Us to Email you About Special Offers & Updates?