de-Broglie wavelength associated with an electron at V potential difference is :

Options

A. `(12.27)/sqrt(V)Å`
B. `sqrt((12.27)/V) Å`
C. `(1.277)/sqrt(V) Å`
D. `sqrt((1.227)/V) Å`

Answer

To find the de-Broglie wavelength associated with an electron accelerated through a potential difference \( V \), we can follow these steps:

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

### Step 2: Express Momentum in Terms of Kinetic Energy
The momentum \( p \) of an electron can be expressed in terms of its kinetic energy \( KE \):
\[ p = mv \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

### Step 3: Relate Kinetic Energy to Potential Difference
When an electron is accelerated through a potential difference \( V \), its kinetic energy \( KE \) can be expressed as:
\[ KE = eV \]
where \( e \) is the charge of the electron.

### Step 4: Relate Kinetic Energy to Momentum
The kinetic energy can also be expressed in terms of momentum:
\[ KE = \frac{p^2}{2m} \]
From this, we can rearrange to find momentum:
\[ p = \sqrt{2m \cdot KE} \]

### Step 5: Substitute Kinetic Energy into Momentum Equation
Substituting \( KE = eV \) into the momentum equation gives:
\[ p = \sqrt{2m \cdot eV} \]

### Step 6: Substitute Momentum into de-Broglie Wavelength Formula
Now, substituting this expression for momentum back into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot eV}} \]

### Step 7: Calculate the Wavelength
Using the known values:
- Planck's constant \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- Mass of the electron \( m = 9.11 \times 10^{-31} \, \text{kg} \)
- Charge of the electron \( e = 1.6 \times 10^{-19} \, \text{C} \)

We can substitute these values into the equation:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \cdot 9.11 \times 10^{-31} \cdot 1.6 \times 10^{-19} \cdot V}} \]

### Step 8: Simplify the Expression
This simplifies to:
\[ \lambda = \frac{12.27 \times 10^{-10}}{\sqrt{V}} \, \text{meters} \]

### Step 9: Convert to Angstroms
To convert meters to angstroms (1 angstrom = \( 10^{-10} \) meters):
\[ \lambda = \frac{12.27}{\sqrt{V}} \, \text{angstroms} \]

### Final Result
Thus, the de-Broglie wavelength associated with an electron at a potential difference \( V \) is:
\[ \lambda = \frac{12.27}{\sqrt{V}} \, \text{angstroms} \]

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