Statement-1 : Among the particles of same kinetic energy, lighter particle has greater de brogile wavelength.
and
Statement-2 : The de-Brogile wavelength of a particle depends only on the charge of the particle.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-11
B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-11
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

To solve the given question, we will analyze both statements one by one and determine their validity.

### Step 1: Analyze Statement 1
**Statement 1:** Among the particles of the same kinetic energy, a lighter particle has a greater de Broglie wavelength.

**Solution:**
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is the Planck constant and \(p\) is the momentum of the particle. The momentum \(p\) can be expressed as:
\[ p = mv \]
where \(m\) is the mass and \(v\) is the velocity of the particle.

The kinetic energy (KE) of a particle is given by:
\[ KE = \frac{1}{2}mv^2 \]
For particles with the same kinetic energy, we can express the velocity in terms of kinetic energy:
\[ v = \sqrt{\frac{2KE}{m}} \]
Substituting this into the momentum formula gives:
\[ p = m \sqrt{\frac{2KE}{m}} = \sqrt{2m \cdot KE} \]
Now substituting this back into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]
From this equation, we can see that the de Broglie wavelength is inversely proportional to the square root of the mass \(m\). Therefore, for particles with the same kinetic energy, a lighter particle (smaller \(m\)) will have a greater de Broglie wavelength (larger \(\lambda\)).

Thus, **Statement 1 is true.**

### Step 2: Analyze Statement 2
**Statement 2:** The de Broglie wavelength of a particle depends only on the charge of the particle.

**Solution:**
From the de Broglie wavelength formula:
\[ \lambda = \frac{h}{p} \]
we see that the wavelength depends on the momentum \(p\), which is determined by the mass and velocity of the particle, not its charge.

The charge of the particle does not appear in the formula for de Broglie wavelength, indicating that the wavelength is independent of the charge. Therefore, **Statement 2 is false.**

### Conclusion
- **Statement 1** is true.
- **Statement 2** is false.

Thus, the correct answer is that Statement 1 is true and Statement 2 is false.

### Final Answer:
C: Statement 1 is true, Statement 2 is false.

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