In which of the following cases the heavier of the two particles has a smaller de-Broglie wavelength ? The two particles

Options

A. move with the same speed
B. move with the same linear momentum
C. move with the same kinetic energy
D. have fallen through the same height

Answer

To determine in which case the heavier of the two particles has a smaller de-Broglie wavelength, we will analyze each of the given scenarios step by step.

### Given Options:
1. Particles move with the same speed.
2. Particles move with the same linear momentum.
3. Particles move with the same kinetic energy.
4. Particles have fallen to the same height.

### Step-by-Step Solution:

#### Step 1: De-Broglie Wavelength Formula
The de-Broglie wavelength (\(\lambda\)) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \(h\) is Planck's constant, \(m\) is the mass of the particle, and \(v\) is its velocity.

#### Step 2: Analyze Each Case

**Case 1: Particles move with the same speed**
- Let the mass of particle X (heavier) be \(m_x\) and the mass of particle Y (lighter) be \(m_y\).
- Since both particles move with the same speed \(v\):
\[ \lambda_x = \frac{h}{m_x v} \quad \text{and} \quad \lambda_y = \frac{h}{m_y v} \]
- Since \(m_x > m_y\), it follows that:
\[ \lambda_x < \lambda_y \]
Thus, the heavier particle (X) has a smaller de-Broglie wavelength. **(This case is correct)**

**Case 2: Particles move with the same linear momentum**
- Linear momentum \(p\) is given by \(p = mv\).
- If both particles have the same momentum:
\[ m_x v_x = m_y v_y \]
- Rearranging gives:
\[ v_x = \frac{m_y}{m_x} v_y \]
- Substituting \(v_x\) and \(v_y\) into the de-Broglie wavelength equations:
\[ \lambda_x = \frac{h}{m_x v_x} \quad \text{and} \quad \lambda_y = \frac{h}{m_y v_y} \]
- Since \(m_x v_x = m_y v_y\), we find:
\[ \lambda_x = \lambda_y \]
Thus, the heavier particle does not have a smaller wavelength in this case. **(This case is incorrect)**

**Case 3: Particles move with the same kinetic energy**
- Kinetic energy \(K\) is given by:
\[ K = \frac{1}{2} mv^2 \]
- If both particles have the same kinetic energy:
\[ \frac{1}{2} m_x v_x^2 = \frac{1}{2} m_y v_y^2 \]
- Rearranging gives:
\[ m_x v_x^2 = m_y v_y^2 \]
- Using the de-Broglie wavelength formula:
\[ \lambda_x = \frac{h}{mv} \quad \text{and} \quad \lambda_y = \frac{h}{mv} \]
- We can derive that the heavier particle will have a smaller wavelength:
\[ \lambda_x < \lambda_y \]
Thus, the heavier particle (X) has a smaller de-Broglie wavelength. **(This case is correct)**

**Case 4: Particles have fallen to the same height**
- When falling from the same height \(h\), both particles gain the same potential energy which converts to kinetic energy:
\[ mgh = \frac{1}{2} mv^2 \]
- Since both particles fall the same height, they will have the same velocity:
\[ v_x = v_y \]
- Therefore, using the de-Broglie wavelength formula:
\[ \lambda_x = \frac{h}{m_x v} \quad \text{and} \quad \lambda_y = \frac{h}{m_y v} \]
- Since \(m_x > m_y\):
\[ \lambda_x < \lambda_y \]
Thus, the heavier particle (X) has a smaller de-Broglie wavelength. **(This case is correct)**

### Conclusion:
The correct options where the heavier particle has a smaller de-Broglie wavelength are:
- Case 1: Particles move with the same speed.
- Case 3: Particles move with the same kinetic energy.
- Case 4: Particles have fallen to the same height.

### Final Answer:
The heavier of the two particles has a smaller de-Broglie wavelength in cases 1, 3, and 4.

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