Question 75: Atoms & Nuclei

Question

Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass numbers 3,4 and 10 are $5.6 \mathrm{MeV}, 7.4 \mathrm{MeV}$ and 6.1 MeV , respectively, then in the process, $\Delta \mathrm{Mc}^2=$ $\_\_\_\_$ MeV .

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

6.9

Answer

7.9

Answer

2.2

Answer

4.3

Atoms & Nuclei

Choose the correct answer: