A particle is droped from a height H. The de-broglie wavelength of the particle as a funciton 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 determining how the de Broglie wavelength of a particle changes as a function of height when the particle is dropped from a height \( H \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
When a particle is dropped from a height \( H \), it accelerates due to gravity. As it falls, it gains velocity and thus momentum, which will affect its de Broglie wavelength.

2. **Using Kinematic Equations**:
The velocity \( v \) of the particle just before it hits the ground can be found using the kinematic equation:
\[ v^2 = u^2 + 2gh \]
where \( u \) is the initial velocity (which is 0 when dropped), \( g \) is the acceleration due to gravity, and \( h \) is the height fallen. Since the particle falls from height \( H \):
\[ v^2 = 0 + 2gH \implies v = \sqrt{2gH} \]

3. **Applying the de Broglie Wavelength Formula**:
The de Broglie wavelength \( \lambda \) of a particle is given by:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant, \( m \) is the mass of the particle, and \( v \) is its velocity.

4. **Substituting for Velocity**:
Now, substituting the expression for \( v \) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{m \sqrt{2gH}} \]

5. **Expressing the Wavelength in Terms of Height**:
We can see that \( \lambda \) is inversely proportional to the square root of the height \( H \):
\[ \lambda \propto \frac{1}{\sqrt{H}} \]
This can also be expressed as:
\[ \lambda \propto H^{-1/2} \]

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

### Final Result:
\[ \lambda \propto H^{-1/2} \]