A particle initially at rest having charge `q` coulomb, `&` mass `m kg` is accelerated by a potential difference of `V` volts. What would be its `K.E. & ` de broglie wavelength respectively after acceleration.

Options

A. `qV,(h)/(sqrt(2qVm))`
B. `(h)/(sqrt(2qVm)),qV`
C. `qV,(h)/(mV)`
D. `(h)/(mV),qV`

Answer

To solve the problem step by step, we need to find the kinetic energy (K.E.) and the de Broglie wavelength of a particle that is accelerated by a potential difference.

### Step 1: Calculate Kinetic Energy
When a charged particle is accelerated through a potential difference \( V \), the work done on the particle is equal to the kinetic energy gained by the particle. The relationship can be expressed as:

\[ \text{K.E.} = qV \]

Where:
- \( q \) = charge of the particle (in coulombs)
- \( V \) = potential difference (in volts)

### Step 2: Calculate Velocity
The kinetic energy can also be expressed in terms of mass and velocity:

\[ \text{K.E.} = \frac{1}{2} mv^2 \]

Setting the two expressions for kinetic energy equal gives:

\[ qV = \frac{1}{2} mv^2 \]

From this, we can solve for the velocity \( v \):

\[ v^2 = \frac{2qV}{m} \]

Taking the square root:

\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 3: Calculate de Broglie Wavelength
The de Broglie wavelength \( \lambda \) of a particle is given by the formula:

\[ \lambda = \frac{h}{mv} \]

Where:
- \( h \) = Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \))
- \( m \) = mass of the particle (in kg)
- \( v \) = velocity of the particle (calculated in Step 2)

Substituting the expression for \( v \) into the de Broglie wavelength formula gives:

\[ \lambda = \frac{h}{m \sqrt{\frac{2qV}{m}}} \]

This simplifies to:

\[ \lambda = \frac{h}{\sqrt{2qmV}} \]

### Summary of Results
1. **Kinetic Energy (K.E.)**:
\[ \text{K.E.} = qV \]

2. **de Broglie Wavelength (\( \lambda \))**:
\[ \lambda = \frac{h}{\sqrt{2qmV}} \]