A particle is projected horizantally with a velocity `10m/s`. What will be the ratio of de-Broglie wavelengths of the particle, when the velocity vector makes an angle `30^(0)` and `60^(0)` with the horizontal

Options

A. `sqrt3:1`
B. `1:sqrt3`
C. `2:sqrt3`
D. `sqrt3:2`

Explanation

To solve the problem of finding the ratio of de-Broglie wavelengths of a particle projected horizontally at a velocity of `10 m/s` when the velocity vector makes angles of `30°` and `60°` with the horizontal, we can follow these steps:

### Step-by-Step Solution:

**Step 1: Understand the Problem**
- The particle is projected horizontally with an initial velocity \( u = 10 \, \text{m/s} \).
- We need to find the de-Broglie wavelengths when the velocity vector makes angles of \( 30^\circ \) and \( 60^\circ \) with the horizontal.

**Step 2: Find the Horizontal Component of Velocity**
- The horizontal component of the velocity remains constant since there is no horizontal acceleration.
- The horizontal component of the velocity when the angle is \( \theta \) is given by:
\[ V_x = V \cos(\theta) \]
- For \( \theta = 30^\circ \):
\[ V_{30} = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s} \]
- For \( \theta = 60^\circ \):
\[ V_{60} = 10 \cos(60^\circ) = 10 \times \frac{1}{2} = 5 \, \text{m/s} \]

**Step 3: Use the de-Broglie Wavelength Formula**
- The de-Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} = \frac{h}{mv} \]
where \( p \) is the momentum, \( m \) is the mass of the particle, and \( v \) is the velocity.
- Since the mass \( m \) is constant for the same particle, we can express the ratio of the wavelengths as:
\[ \frac{\lambda_1}{\lambda_2} = \frac{v_2}{v_1} \]

**Step 4: Calculate the Ratio of Wavelengths**
- Let \( \lambda_{30} \) be the wavelength at \( 30^\circ \) and \( \lambda_{60} \) be the wavelength at \( 60^\circ \).
- Thus, we have:
\[ \frac{\lambda_{30}}{\lambda_{60}} = \frac{V_{60}}{V_{30}} = \frac{5}{5\sqrt{3}} = \frac{1}{\sqrt{3}} \]

**Step 5: Final Result**
- The ratio of the de-Broglie wavelengths is:
\[ \frac{\lambda_{30}}{\lambda_{60}} = \frac{1}{\sqrt{3}} \]