A particle is droped from a height H. The de-broglie wavelength of the particle as a function of height is proportional to

Options

A. H
B. `H^(1//2)`
C. `H^(0)`
D. `H^(-1//2)`

Explanation

To solve the problem of how the de Broglie wavelength of a particle changes as it falls from a height \( H \), we can follow these steps:

### Step 1: Understand the relationship between height and velocity
When a particle is dropped from a height \( H \), it accelerates due to gravity. The velocity \( v \) of the particle at any height \( h \) can be expressed using the equation:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity.

**Hint:** Remember that the velocity of a freely falling object increases as it falls due to gravitational acceleration.

### Step 2: Write the expression for de Broglie wavelength
The de Broglie wavelength \( \lambda \) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) can be expressed as:
\[ p = mv \]
where \( m \) is the mass of the particle.

**Hint:** The de Broglie wavelength depends on the momentum of the particle, which is related to its velocity.

### Step 3: Substitute the expression for momentum
Substituting the expression for momentum into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{mv} \]
Now, substituting \( v = \sqrt{2gh} \) into this equation, we get:
\[ \lambda = \frac{h}{m \sqrt{2gh}} \]

**Hint:** This substitution helps us relate the wavelength directly to the height from which the particle is dropped.

### Step 4: Simplify the expression
Now, we can simplify the expression for \( \lambda \):
\[ \lambda = \frac{h}{m \sqrt{2g}} \cdot \frac{1}{\sqrt{h}} = \frac{h}{m \sqrt{2g}} \cdot h^{-1/2} \]
This can be rewritten as:
\[ \lambda \propto h^{-1/2} \]

**Hint:** Recognizing that \( h^{-1/2} \) indicates an inverse relationship helps us understand how the wavelength changes with height.

### Step 5: Conclusion
From the simplification, we conclude that the de Broglie wavelength \( \lambda \) is inversely proportional to the square root of the height \( H \):
\[ \lambda \propto \frac{1}{\sqrt{H}} \]

Thus, the de Broglie wavelength of the particle as a function of height is proportional to \( H^{-1/2} \).

**Final Answer:** The de Broglie wavelength of the particle is proportional to \( H^{-1/2} \).