Question

A proton and electron are lying in a box having impenetrable walls, the ratio of uncertainty in their velocities are \(\ldots \ldots\) \(\left(\mathrm{m}_{\mathrm{e}}=\right.\) mass of electron and \(\mathrm{m}_{\mathrm{p}}=\) mass of proton. (A) \(\left(\mathrm{m}_{\mathrm{e}} / \mathrm{m}_{\mathrm{p}}\right)\) (B) \(\mathrm{m}_{\mathrm{e}} \cdot \mathrm{m}_{\mathrm{p}}\) (C) \(\left.\sqrt{\left(m_{e}\right.} \cdot m_{p}\right)\) (D) \(\sqrt{\left(m_{e} / m_{p}\right)}\)

Short answer

The short answer to the question is: (A) \(\left(\frac{m_p}{m_e}\right)\)

Step-by-step solution

Define the Heisenberg's Uncertainty Principle for position and momentum

According to the Heisenberg's Uncertainty Principle, we cannot simultaneously measure the position (x) and momentum (p) of a particle with absolute certainty. Mathematically, this is expressed as: \[\Delta x \cdot \Delta p \ge \frac{\hbar}{2}\] where \(\Delta x\) is the uncertainty in position, \(\Delta p\) is the uncertainty in momentum, and \(\hbar\) is the reduced Planck's constant.

Express the uncertainty in momentum in terms of velocity

The momentum of a particle is given by the product of its mass and velocity: \[p = m \cdot v\] Here, \(m\) is the particle mass, and \(v\) is its velocity. Knowing this, we can express the uncertainty in momentum as: \[\Delta p = m \cdot \Delta v\] where \(\Delta v\) is the uncertainty in velocity.

Rewrite the Heisenberg's Uncertainty Principle with the uncertainty in velocity

Substitute \(\Delta p = m \cdot \Delta v\) in the Heisenberg's Uncertainty Principle equation: \[\Delta x \cdot m \cdot \Delta v \ge \frac{\hbar}{2}\] Now we can rearrange the equation to find the uncertainty in velocity: \[\Delta v \ge \frac{\hbar}{2m \cdot \Delta x}\]

Compare the uncertainties in velocities of the proton and electron

The mass of the electron is denoted by \(m_e\) and the mass of the proton is denoted by \(m_p\). Using the formula for the uncertainty in velocity, we can write the uncertainties in velocities for the electron and the proton as: \[\Delta v_e \ge \frac{\hbar}{2m_e \cdot \Delta x}\] \[\Delta v_p \ge \frac{\hbar}{2m_p \cdot \Delta x}\] Now we want to find the ratio of the uncertainty in their velocities: \[\frac{\Delta v_e}{\Delta v_p} = \frac{\frac{\hbar}{2m_e \cdot \Delta x}}{\frac{\hbar}{2m_p \cdot \Delta x}}\]

Simplify the ratio of uncertainties in velocities

Notice that the terms \(\frac{\hbar}{2 \cdot \Delta x}\) cancel out in the numerator and the denominator of the fraction: \[\frac{\Delta v_e}{\Delta v_p} = \frac{m_p}{m_e}\] Since \(m_p\) and \(m_e\) are the masses of the proton and electron, respectively, the ratio of the uncertainty in their velocities is: \[\frac{\Delta v_e}{\Delta v_p} = \frac{m_p}{m_e}\] Thus, the correct answer is: (A) \(\left(\frac{m_e}{m_p}\right)\)