Nuclear Physics and RadioactivityMedium
Question
The binding energy per nucleon of $\ _{83}^{209}Bi$ is
$\_\_\_\_$ MeV .
[Take $m\left( \ _{83}^{209}Bi \right) = 208.980388u,m_{p} = 1.007825u$, $m_{n} = 1.008665u,1u = 931MeV/c^{2}$ ]
Options
A.7.48
B.7.84
C.8.79
D.6.94
Solution
Sol. [Key Concept: $\Delta\ m = {Zm}_{p} + {Nm}_{n} - M$; then BE$/A = \Delta\ m \cdot 931/A$.]
$${\ 83\ _{1}P^{1} + 126\ _{0}n^{1} \rightarrow \ _{83}{Bi}^{209} }{\Delta m = 83{\text{ }m}_{p} + 126{\text{ }m}_{n} - M_{Bi} }{= 83 \times 1.007825 + 126 \times 1.008665 - 208.980388 }{= 1.760877u }{BE = 1.760877 \times 931MeV = 1639.3764 }{\frac{BE}{A} = \frac{1639.3764}{209}MeV = 7.8439MeV }$$
Create a free account to view solution
View Solution FreeMore Nuclear Physics and Radioactivity Questions
A nuclear transformation is denoted by X(n, α) Which of the following is the nucleus of element X ?...As the mass number increases, binding energy per nucleon :-...Identify the possible nuclear fusion reactions from the following:...The radioactivity sample of a sample is R1 at time T1 and R2 at time 2. the half life of the specimen is than number of ...When radioactive substance emits an α - particle, then its position in the periodic in the periodic table is lowere...