de Broglie wavelength of an electron after being accelerated by a potential difference of V volt from rest is :

Options

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

Answer

To find the de Broglie wavelength of an electron that has been accelerated by a potential difference of \( V \) volts from rest, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and potential difference
When an electron is accelerated through a potential difference \( V \), it gains kinetic energy equal to the work done on it by the electric field. The kinetic energy (\( KE \)) gained by the electron can be expressed as:
\[ KE = eV \]
where \( e \) is the charge of the electron.

### Step 2: Relate kinetic energy to velocity
The kinetic energy of the electron can also be expressed in terms of its mass (\( m \)) and velocity (\( v \)):
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives us:
\[ eV = \frac{1}{2} mv^2 \]

### Step 3: Solve for velocity
Rearranging the equation to solve for \( v \):
\[ v^2 = \frac{2eV}{m} \]
Taking the square root of both sides, we find:
\[ v = \sqrt{\frac{2eV}{m}} \]

### Step 4: Use the de Broglie wavelength formula
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where \( h \) is Planck's constant.

### Step 5: Substitute the expression for velocity into the de Broglie wavelength formula
Substituting \( v \) from Step 3 into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{m \sqrt{\frac{2eV}{m}}} \]
This simplifies to:
\[ \lambda = \frac{h}{\sqrt{2emV}} \]

### Step 6: Substitute known values for constants
Using the known values:
- Planck's constant \( h = 6.626 \times 10^{-34} \, \text{Js} \)
- Charge of the electron \( e = 1.6 \times 10^{-19} \, \text{C} \)
- Mass of the electron \( m = 9.1 \times 10^{-31} \, \text{kg} \)

### Step 7: Final expression for de Broglie wavelength
Substituting these values into the equation, we can express the de Broglie wavelength in terms of \( V \):
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.6 \times 10^{-19} \times 9.1 \times 10^{-31} \times V}} \]
This can be simplified further to yield:
\[ \lambda \approx \frac{1.23 \times 10^{-9}}{\sqrt{V}} \text{ meters} \]
Converting to nanometers:
\[ \lambda \approx 1.23 \sqrt{V} \text{ nanometers} \]

### Conclusion
Thus, the de Broglie wavelength of an electron after being accelerated by a potential difference of \( V \) volts is:
\[ \lambda = 1.23 \sqrt{V} \text{ nanometers} \]