The wavelength of very fast moving electron `(v ~~ c)` is :

Options

A. `lamda = (h)/(m_0 v)`
B. `lamda = (h)/(sqrt(2mE) )`
C. `lamda^2 = (h^2)/(sqrt(2mE) )`
D. `lamda = (h)/(m_0v // sqrt(1 - v^2/c^2 ) )`

Answer

To find the wavelength of a very fast-moving electron (where its velocity \( v \) is comparable to the speed of light \( c \)), 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 \( \lambda \) 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: Define Momentum
For a particle, the momentum \( p \) is defined as:
\[ p = mv \]
where \( m \) is the inertial mass of the particle and \( v \) is its velocity.

### Step 3: Consider Relativistic Effects
When the velocity \( v \) of the electron is very close to the speed of light \( c \), we need to account for relativistic effects. The inertial mass \( m \) in this case is given by:
\[ m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}} \]
where \( m_0 \) is the rest mass of the electron.

### Step 4: Substitute Momentum in the Wavelength Formula
Substituting the expression for momentum into the de Broglie wavelength formula, we get:
\[ \lambda = \frac{h}{mv} = \frac{h}{\left(\frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}\right)v} \]

### Step 5: Simplify the Expression
This can be simplified to:
\[ \lambda = \frac{h \sqrt{1 - \frac{v^2}{c^2}}}{m_0 v} \]

### Step 6: Conclusion
Thus, the wavelength of a very fast-moving electron (where \( v \) is close to \( c \)) is given by:
\[ \lambda = \frac{h}{m_0 v \sqrt{1 - \frac{v^2}{c^2}}} \]