Question 76: Atoms & Nuclei

Question

The binding energy per nucleon of $^{209}_{83}Bi$ is _______ MeV.
[Take $m(^{209}_{83}Bi) = 208.980388\ 	ext{u}$, $m_p = 1.007825\ 	ext{u}$, $m_n = 1.008665\ 	ext{u}$, $1\ 	ext{u} = 931\ 	ext{MeV}/c^2$]

Accepted Answer

The nucleus is assumed as a sphere where the volume is proportional to the total number of nucleons (mass number, A).
Because the density of nuclear matter is approximately constant, the radius R of a nucleus is related to its mass number A by the following formula :
$$ \mathrm{R}=\mathrm{R}_0 \mathrm{~A}^{1 / 3} $$
Where R is the nuclear radius.
$R_0$ is a constant (approximately $1.2 	imes 10^{-15} \mathrm{~m}$ or 1.2 fm ).
A is the mass number.
For the first nucleus : $\mathrm{R}_\alpha=\mathrm{R}_0(\alpha)^{1 / 3}$
For the second nucleus : $R_\beta=R_0(\beta)^{1 / 3}$
Taking the ratio $\frac{\mathrm{R}_\alpha}{\mathrm{R}_\beta}$ :
$$ \frac{\mathrm{R}_\alpha}{\mathrm{R}_\beta}=\frac{\mathrm{R}_0 \alpha^{1 / 3}}{\mathrm{R}_0 \beta^{1 / 3}}=\left(\frac{\alpha}{\beta}ight)^{1 / 3} $$
It is given that $\beta=8 \alpha$.
$$ \frac{\mathrm{R}_\alpha}{\mathrm{R}_\beta}=\left(\frac{\alpha}{8 \alpha}ight)^{1 / 3}=\frac{1}{2}=0.5 $$
Therefore, the ratio of the radius of the first nucleus to the second is 0.5 . Hence, the correct option is (D).

Answer

7.48

Answer

7.84

Answer

8.79

Answer

6.94

Atoms & Nuclei

The binding energy per nucleon of ^{209}_{83}Bi is _______ MeV. [Take m(^{209}_{83}Bi) = 208.980388\ \text{u}, m_p = 1.007825\ \text{u}, m_n = 1.008665\ \text{u}, 1\ \text{u} = 931\ \text{MeV}/c^2]

Choose the correct answer: