The expression for de-Broglie wavelength of an electron moving under a potential difference of V volt is.............. .

Answer

To derive the expression for the de-Broglie wavelength of an electron moving under a potential difference of V volts, we can follow these steps:

### Step 1: Understand the Work Done on the Electron
When an electron is accelerated through a potential difference \( V \), the work done on the electron is given by:
\[ W = Q \cdot V \]
where \( Q \) is the charge of the electron. For an electron, \( Q = e \), where \( e \) is the elementary charge (\( e \approx 1.6 \times 10^{-19} \) coulombs). Thus, the work done becomes:
\[ W = eV \]

### Step 2: Relate Work Done to Kinetic Energy
The work done on the electron is converted into kinetic energy (KE). Therefore, we can write:
\[ KE = eV \]

### Step 3: Express Kinetic Energy in Terms of Momentum
The kinetic energy can also be expressed in terms of momentum \( P \) as:
\[ KE = \frac{P^2}{2m} \]
where \( m \) is the mass of the electron.

### Step 4: Set the Two Expressions for Kinetic Energy Equal
Equating the two expressions for kinetic energy, we have:
\[ eV = \frac{P^2}{2m} \]

### Step 5: Solve for Momentum \( P \)
Rearranging the equation to solve for momentum \( P \):
\[ P^2 = 2m \cdot eV \]
Taking the square root gives:
\[ P = \sqrt{2m \cdot eV} \]

### Step 6: Use 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 (\( h \approx 6.63 \times 10^{-34} \) Js).

### Step 7: Substitute for Momentum \( P \)
Substituting the expression for \( P \) into the de-Broglie wavelength formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot eV}} \]

### Step 8: Final Expression for de-Broglie Wavelength
Thus, the expression for the de-Broglie wavelength of an electron moving under a potential difference \( V \) is:
\[ \lambda = \frac{h}{\sqrt{2meV}} \]

### Step 9: Numerical Approximation
If we want to express this in terms of numerical values, we can plug in the constants:
\[ \lambda \approx \frac{6.63 \times 10^{-34}}{\sqrt{2 \cdot 9.1 \times 10^{-31} \cdot 1.6 \times 10^{-19} \cdot V}} \]
This can be simplified further to:
\[ \lambda \approx \frac{12.27 \times 10^{-10}}{\sqrt{V}} \text{ meters} \quad \text{or} \quad \frac{12.27}{\sqrt{V}} \text{ angstroms} \]