A particle of mass m is projected form ground with velocity u making angle `theta` with the vertical. The de Broglie wavelength of the particle at the highest point is

Options

A. `infty`
B. `(h)/(m u sin theta)`
C. `(h)/(m u costheta)`
D. `(h)/(m u)`

Explanation

To find the de Broglie wavelength of a particle projected from the ground with velocity \( u \) at an angle \( \theta \) with the vertical, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components of Velocity**:
The particle is projected with velocity \( u \) at an angle \( \theta \) with the vertical. We need to find the horizontal and vertical components of this velocity.
- The horizontal component of the velocity \( V_x \) can be expressed as:
\[ V_x = u \sin(\theta) \]
- The vertical component of the velocity \( V_y \) is:
\[ V_y = u \cos(\theta) \]

2. **Determine the Velocity at the Highest Point**:
At the highest point of the projectile's motion, the vertical component of the velocity becomes zero, while the horizontal component remains unchanged. Therefore, the velocity at the highest point is:
\[ V = V_x = u \sin(\theta) \]

3. **Calculate the de Broglie Wavelength**:
The de Broglie wavelength \( \lambda_d \) of a particle is given by the formula:
\[ \lambda_d = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle. The momentum \( p \) can be expressed as:
\[ p = mV \]
Substituting the expression for \( V \) at the highest point:
\[ p = m (u \sin(\theta)) \]
Now substituting this into the de Broglie wavelength formula:
\[ \lambda_d = \frac{h}{m (u \sin(\theta))} \]

4. **Final Expression**:
Therefore, the de Broglie wavelength of the particle at the highest point is:
\[ \lambda_d = \frac{h}{m u \sin(\theta)} \]

### Summary:
The de Broglie wavelength of the particle at the highest point is given by:
\[ \lambda_d = \frac{h}{m u \sin(\theta)} \]