Question 53: Atoms & Nuclei

Question

In the head-on collision of two alpha particles $$\alpha_1$$ and $$\alpha_2$$ with the gold nucleus, the closest approaches are 31.4 fermi and 94.2 fermi respectively. Then the ratio of the energy possessed by the alpha particles $$\alpha_2 / \alpha_1$$ is:

Accepted Answer

When we consider a head-on collision of two alpha particles with a gold nucleus, the closest approaches directly relate to the kinetic energies of the alpha particles. This relationship can be understood from the perspective of electrostatic repulsion and conservation of energy. The kinetic energy of each alpha particle is transformed into potential energy at the point of closest approach, since at that point, their velocities become zero, assuming a non-relativistic scenario.
For an alpha particle approaching a gold nucleus, the potential energy at the closest approach is given by the Coulomb potential energy formula:
$$
PE = \frac{k \cdot q_1 \cdot q_2}{r}
$$
where:
$k$ is Coulomb's constant,
$q_1$ and $q_2$ are the charges of the alpha particle and gold nucleus, respectively,
$r$ is the distance of the closest approach, which corresponds to the distance at which the entire kinetic energy of the alpha particle has been converted into electrostatic potential energy.
Given that the charge of the alpha particle and the nucleus remain constant and considering $k$ is a constant as well, the potential energy (and hence, the initial kinetic energy) varies inversely with the distance of closest approach.
Thus, the ratio of the kinetic energies ($KE$) of the alpha particles can be expressed in terms of their closest approaches, $r_2$ for $\alpha_2$ and $r_1$ for $\alpha_1$, as follows:
$$
\frac{KE_2}{KE_1} = \frac{r_1}{r_2}
$$
Substituting the given values of the closest approaches, 31.4 fermi for $\alpha_1$ and 94.2 fermi for $\alpha_2$, into the equation:
$$
\frac{KE_2}{KE_1} = \frac{31.4}{94.2} = \frac{1}{3}
$$
Therefore, the ratio of the energy possessed by the alpha particles $\alpha_2 / \alpha_1$ is $1: 3$, corresponding to Option A.

Answer

$$1: 3$$

Answer

$$9: 1$$

Answer

$$3: 1$$

Answer

$$1: 9$$

Atoms & Nuclei

In the head-on collision of two alpha particles $\alpha_1 and \alpha_2 with the gold nucleus, the closest approaches are 31.4 fermi and 94.2 fermi respectively. Then the ratio of the energy possessed by the alpha particles \alpha_2 / \alpha_1$ is:

Choose the correct answer: