The de Broglie wavelength of an electron accelerated by an electric field of V volt is given by:

Options

A. `lamda=(1.23)/(sqrt(m))`
B. `lamda=(1.23m)/(sqrt(h))`
C. `(1.23)/(sqrt(V))nm`
D. `lamda=(1.23)/(V)`.

Answer

To find the de Broglie wavelength of an electron accelerated by an electric field of V volts, we can use the de Broglie wavelength formula. Here’s a step-by-step solution:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \( h \) is Planck's constant and \( p \) is the momentum of the particle.

### Step 2: Relate momentum to kinetic energy
For an electron accelerated through a potential difference \( V \), the kinetic energy (KE) gained by the electron is given by:
\[ KE = eV \]
where \( e \) is the charge of the electron. The momentum \( p \) can be expressed in terms of kinetic energy:
\[ p = \sqrt{2m \cdot KE} = \sqrt{2m \cdot eV} \]
where \( m \) is the mass of the electron.

### Step 3: Substitute momentum into the de Broglie wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{\sqrt{2m \cdot eV}} \]

### Step 4: Use known values
Using known constants:
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- Mass of the electron \( m = 9.11 \times 10^{-31} \, \text{kg} \)
- Charge of the electron \( e = 1.602 \times 10^{-19} \, \text{C} \)

### Step 5: Simplify the expression
We can simplify the expression for the de Broglie wavelength:
\[ \lambda = \frac{h}{\sqrt{2m \cdot eV}} = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 1.602 \times 10^{-19} \times V}} \]
This can be further simplified to:
\[ \lambda = \frac{12.27}{\sqrt{V}} \, \text{Å} \]
(Here, the factor of 12.27 is derived from the constants used.)

### Step 6: Convert to nanometers
To convert from angstroms (Å) to nanometers (nm), we use the conversion factor:
\[ 1 \, \text{Å} = 0.1 \, \text{nm} \]
Thus, we have:
\[ \lambda = 12.27 \cdot 0.1 \cdot \frac{1}{\sqrt{V}} \, \text{nm} = 1.227 \cdot \frac{1}{\sqrt{V}} \, \text{nm} \]

### Final Result
The de Broglie wavelength of an electron accelerated by an electric field of \( V \) volts is given by:
\[ \lambda \approx \frac{1.227}{\sqrt{V}} \, \text{nm} \]