If the mass of a microscopic particle as well as its speed are halved, the de Broglie wavelength associated with the particle will 

Options

A. increase by a factor more than 2
B. increase by a factor of 2
C. decrease by a factor of 2
D. decrease by a factor more than 2.

Explanation

To solve the problem of how the de Broglie wavelength changes when both the mass and speed of a microscopic particle are halved, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 2: Define momentum
Momentum (\( p \)) is defined as:
\[ p = mv \]
where \( m \) is the mass of the particle and \( v \) is its velocity.

### Step 3: Write the initial de Broglie wavelength
Let the initial mass be \( m_1 \) and the initial velocity be \( v_1 \). The initial de Broglie wavelength (\( \lambda_1 \)) can be expressed as:
\[ \lambda_1 = \frac{h}{m_1 v_1} \]

### Step 4: Define the new mass and velocity
According to the problem, both the mass and velocity are halved:
\[ m_2 = \frac{m_1}{2}, \quad v_2 = \frac{v_1}{2} \]

### Step 5: Write the new de Broglie wavelength
Now, we can find the new de Broglie wavelength (\( \lambda_2 \)):
\[ \lambda_2 = \frac{h}{m_2 v_2} = \frac{h}{\left(\frac{m_1}{2}\right) \left(\frac{v_1}{2}\right)} = \frac{h}{\frac{m_1 v_1}{4}} = \frac{4h}{m_1 v_1} \]

### Step 6: Relate the new wavelength to the initial wavelength
We can relate \( \lambda_2 \) to \( \lambda_1 \):
\[ \lambda_2 = 4 \lambda_1 \]

### Step 7: Conclusion
Since \( \lambda_2 \) is 4 times \( \lambda_1 \), we conclude that the de Broglie wavelength increases by a factor of 4 when both the mass and speed of the particle are halved.

### Final Answer
The correct option is: **Increased by a factor more than 2.**

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