The wavelength of a charged particle `…………..` the square root of the potential difference through which it is accelerated .

Options

A. is inversely proportional to
B. is directly proportional to
C. is independent of
D. is unrelated with

Answer

To solve the question regarding the relationship between the wavelength of a charged particle and the square root of the potential difference through which it is accelerated, we can follow these steps:

### Step 1: Understand the de Broglie Wavelength
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is Planck's constant,
- \( m \) is the mass of the particle,
- \( v \) is the velocity of the particle.

### Step 2: Relate Velocity to Potential Difference
When a charged particle with charge \( q \) is accelerated through a potential difference \( V \), the work done on the particle is equal to the kinetic energy gained by it. The work done can be expressed as:
\[ W = qV \]
This work done is equal to the kinetic energy (KE) of the particle, which is given by:
\[ KE = \frac{1}{2} mv^2 \]
Setting these equal gives:
\[ qV = \frac{1}{2} mv^2 \]

### Step 3: Solve for Velocity
Rearranging the equation for velocity \( v \):
\[ mv^2 = 2qV \]
\[ v^2 = \frac{2qV}{m} \]
Taking the square root of both sides:
\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 4: Substitute Velocity into the de Broglie Equation
Now, substitute this expression for \( v \) back into the de Broglie wavelength equation:
\[ \lambda = \frac{h}{m\sqrt{\frac{2qV}{m}}} \]
This simplifies to:
\[ \lambda = \frac{h}{\sqrt{2qmV}} \]

### Step 5: Analyze the Relationship
From the final equation, we can see that:
\[ \lambda \propto \frac{1}{\sqrt{V}} \]
This indicates that the wavelength \( \lambda \) is inversely proportional to the square root of the potential difference \( V \).

### Final Conclusion
Thus, the wavelength of a charged particle is inversely proportional to the square root of the potential difference through which it is accelerated.

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