A proton and an electron are released from an infinite distance apart and they get attracted towars each other. Which of the following statement about their kinetic energy is true?

Options

A. Kinetic energy of electron is more than that of proton
B. Kinetic energy of electron is less than that of proton
C. Kinetic energy of electron = kinetic energy of proton
D. None of the above is true as it depends on the distance between the particles

Answer

To solve the problem regarding the kinetic energy of a proton and an electron that are released from an infinite distance apart and attracted towards each other, we can follow these steps:

### Step 1: Understand the System
When a proton and an electron are released from an infinite distance, they experience an attractive force due to their opposite charges. This force causes them to accelerate towards each other.

**Hint:** Consider the nature of the forces acting on the particles and their initial conditions.

### Step 2: Conservation of Momentum
Since there are no external forces acting on the system, the total momentum of the system must be conserved. If we denote the mass of the proton as \( m_p \) and the mass of the electron as \( m_e \), and their velocities as \( v_p \) and \( v_e \) respectively, we can express the conservation of momentum as:

\[ m_p v_p + m_e v_e = 0 \]

This implies that:

\[ m_p v_p = -m_e v_e \]

**Hint:** Remember that momentum is a vector quantity, and the direction of velocities matters.

### Step 3: Relate Kinetic Energy to Momentum
The kinetic energy \( K \) of an object is given by the formula:

\[ K = \frac{1}{2} m v^2 \]

For both particles, we can write their kinetic energies as:

\[ K_p = \frac{1}{2} m_p v_p^2 \quad \text{(for the proton)} \]
\[ K_e = \frac{1}{2} m_e v_e^2 \quad \text{(for the electron)} \]

**Hint:** Use the relationship between momentum and velocity to express kinetic energy in terms of momentum.

### Step 4: Express Kinetic Energy in Terms of Momentum
Using the relationship \( p = mv \), we can express the velocities in terms of momentum:

\[ v_p = \frac{p_p}{m_p}, \quad v_e = \frac{p_e}{m_e} \]

Substituting these into the kinetic energy formulas gives:

\[ K_p = \frac{p_p^2}{2 m_p}, \quad K_e = \frac{p_e^2}{2 m_e} \]

Since \( p_p = p_e \) (from conservation of momentum), we can denote \( p = p_p = p_e \):

\[ K_p = \frac{p^2}{2 m_p}, \quad K_e = \frac{p^2}{2 m_e} \]

**Hint:** Consider how the masses of the particles affect their kinetic energies.

### Step 5: Compare Kinetic Energies
Since \( m_p > m_e \), it follows that:

\[ K_p < K_e \]

This means the kinetic energy of the electron is greater than that of the proton.

**Hint:** Think about how mass affects kinetic energy when momentum is constant.

### Conclusion
The correct statement regarding their kinetic energies is that the kinetic energy of the electron is greater than that of the proton.

**Final Answer:** The kinetic energy of the electron is more than that of the proton.