Question 44: Atoms & Nuclei

Question

The closest approach of an alpha particle when it make a head on collision with a gold nucleus is $$10 	imes 10^{-14} \mathrm{~m}$$, then the kinetic energy of the alpha particle is :

Accepted Answer

To determine the kinetic energy of the alpha particle when it makes a head-on collision with a gold nucleus, we can use the concept of electrostatic potential energy at the closest approach. The kinetic energy of the alpha particle will be completely converted into electrostatic potential energy at the point of closest approach.
The formula for the electrostatic potential energy between two charges is given by:
$$ U = \frac{1}{4 \pi \epsilon_0} \frac{q_1 q_2}{r} $$
where:
$$q_1$$ and $$q_2$$ are the charges of the alpha particle and gold nucleus, respectively
$$r$$ is the distance of closest approach
$$\epsilon_0$$ is the permittivity of free space, which is approximately $$8.85 	imes 10^{-12} \mathrm{~F/m}$$
For an alpha particle, $$q_1$$ is the charge of 2 protons, so:
$$ q_1 = 2e = 2 	imes 1.6 	imes 10^{-19} \mathrm{~C} $$
For a gold nucleus (with atomic number 79), $$q_2$$ is the charge of 79 protons, so:
$$ q_2 = 79e = 79 	imes 1.6 	imes 10^{-19} \mathrm{~C} $$
The distance of closest approach, $$r$$, is given as:
$$ r = 10 	imes 10^{-14} \mathrm{~m} $$
Now, substituting these values into the formula for electrostatic potential energy:
$$ U = \frac{1}{4 \pi \epsilon_0} \frac{(2 	imes 1.6 	imes 10^{-19}) (79 	imes 1.6 	imes 10^{-19})}{10 	imes 10^{-14}} $$
Simplifying the equation:
$$ U = \frac{9 	imes 10^9}{10 	imes 10^{-14}} (2 	imes 79 	imes 1.6^2 	imes 10^{-38}) $$
Calculating each component:
$$ U = \frac{9 	imes 10^9}{10^{-13}} (2 	imes 79 	imes 2.56 	imes 10^{-38}) $$
Or:
$$ U = 9 	imes 10^{22} 	imes 404.48 	imes 10^{-38} $$
Simplifying further:
$$ U = 3.64 	imes 10^{-13} \mathrm{~J} $$
Therefore, the kinetic energy of the alpha particle is:
Option D: $$3.64 	imes 10^{-13} \mathrm{~J}$$

Answer

3640 J

Answer

3.64 J

Answer

$$
3.64 	imes 10^{-16} \mathrm{~J}
$$

Answer

$$
3.64 	imes 10^{-13} \mathrm{~J}
$$

Atoms & Nuclei

The closest approach of an alpha particle when it make a head on collision with a gold nucleus is $10 \times 10^{-14} \mathrm{~m}$, then the kinetic energy of the alpha particle is :

Choose the correct answer: