Consider an electron, a neutron and a proton initially at rest and placed along a straight line such that the neutron is exactly at the centre of the line joining the electron and proton. At t = 0 , the particles are released but are constrained to move along the same straight line. Which of these will collide first?

Understanding the Initial Setup

We have three particles: an electron (e⁻), a neutron (n), and a proton (p⁺). They are initially at rest and placed along a straight line. The neutron is positioned exactly at the midpoint between the electron and the proton. Let's denote the position of the electron as $x_e=-d$, the neutron as $x_n=0$, and the proton as $x_p=+d$. The distance between the electron and the proton is $2d$. At time $t=0$, they are released and can only move along this line.

Analyzing Forces and Acceleration

We need to determine which pair collides first. This depends on the forces acting on the particles and their resulting accelerations.

- Neutron: A neutron is electrically neutral. It does not experience any electrostatic force from the electron or the proton. Assuming gravity is negligible, the net force on the neutron is zero. Therefore, the neutron remains stationary at its initial position ($x_n=0$).
- Electron: An electron has a negative charge ($-e$) and a small mass ($m_e$). It is attracted towards the positively charged proton ($+e$) by the electrostatic force ($F_{ep}$). It is also attracted towards the proton, and repelled by the electron. Wait, the electron is attracted by the proton. The force on the electron due to the proton is towards the right (positive x-direction). The magnitude of this force is given by Coulomb's law: $F_{ep}=k\frac{|q_e{q}_p|}{r^2}=k\frac{e^2}{(x_p-x_e{)}^2}$, where $k$ is Coulomb's constant. Since the distance between the electron and proton changes, the force is not constant. The acceleration of the electron is $a_e=\frac{F_{ep}}{m_e}$. Since the force is attractive towards the proton, the electron accelerates towards the right.
- Proton: A proton has a positive charge ($+e$) and a larger mass ($m_p$) compared to the electron ($m_p\gg m_e$). It is attracted towards the negatively charged electron ($-e$) by the electrostatic force ($F_{pe}$). The force on the proton due to the electron is towards the left (negative x-direction). The magnitude is the same: $F_{pe}=k\frac{|q_p{q}_e|}{r^2}=k\frac{e^2}{(x_p-x_e{)}^2}$. The acceleration of the proton is $a_p=\frac{F_{pe}}{m_p}$. Since the force is attractive towards the electron, the proton accelerates towards the left.

Crucially, the magnitudes of the forces are equal ($F_{ep}=F_{pe}=F$), but the masses are different. Since acceleration is force divided by mass ($a=F/m$), and $m_e<m_p$, the acceleration of the electron is greater than the acceleration of the proton ($a_e=F/m_e>F/m_p=a_p$).

Comparing Collision Times

Now let's analyze the time it takes for each potential collision:

1. **Electron-Neutron Collision ($t_{en}$):** The electron starts at $x_e=-d$ and moves towards the right. The neutron is stationary at $x_n=0$. The electron needs to cover a distance $d$. Since the electron accelerates ($a_e>0$), it will eventually reach the neutron's position. The time taken depends on the electron's acceleration.
2. **Proton-Neutron Collision ($t_{pn}$):** The proton starts at $x_p=+d$ and moves towards the left. The neutron is stationary at $x_n=0$. The proton needs to cover a distance $d$. The proton accelerates ($a_p>0$, towards the left). The time taken depends on the proton's acceleration.
3. **Electron-Proton Collision ($t_{ep}$):** The electron starts at $x_e=-d$ moving right, and the proton starts at $x_p=+d$ moving left. They are moving towards each other. They will collide somewhere between $-d$ and $+d$. The total distance covered by both particles combined until they collide is $2d$.

Comparison:

- Both the electron and the proton need to travel the same distance ($d$) to reach the stationary neutron.
- The electron has a significantly larger acceleration ($a_e$) than the proton ($a_p$) because $m_e<m_p$.
- Since acceleration determines how quickly speed is gained, the electron will cover the distance $d$ faster than the proton. Therefore, $t_{en}<t_{pn}$.
- Now, let's compare $t_{en}$ with $t_{ep}$. The electron moves towards the proton, and the proton moves towards the electron. They are effectively accelerating towards each other. The time for collision $t_{ep}$ can be related to the combined accelerations. Intuitively, since the electron has a much larger acceleration and needs to cover only distance $d$ to hit the neutron, while the electron-proton collision involves their combined motion over $2d$ (with the proton's slower acceleration playing a role), it's likely the electron-neutron collision happens first.
- More formally, let's approximate using average accelerations. Time $t_{en}\approx \sqrt{2d/a_e}$. Time $t_{ep}\approx \sqrt{2\times (\text{distance})/(\text{average relative acceleration})}$. The effective distance is $2d$. The relative acceleration is $a_e+a_p$. So, $t_{ep}\approx \sqrt{4d/(a_e+a_p)}$. Since $a_p>0$, we have $a_e+a_p>a_e$. Thus, $\sqrt{4d/(a_e+a_p)}<\sqrt{4d/a_e}=2\sqrt{d/a_e}$. Also, $\sqrt{2d/a_e}<\sqrt{4d/(a_e+a_p)}$ because $a_p<a_e$. This confirms $t_{en}<t_{ep}$.

Combining the inequalities, $t_{en}<t_{pn}$ and $t_{en}<t_{ep}$. This means the electron will collide with the neutron before the proton collides with the neutron, and also before the electron and proton collide with each other.

Conclusion

The electron, having the smallest mass, experiences the greatest acceleration due to the electrostatic force from the proton. As the neutron remains stationary at the center, the electron reaches it first. Therefore, the electron and neutron will collide first.