Relativistic corrections become necessary when the expression for the kinetic energy `1/2mv^(2)`, becomes comparable with `mc^(2)`, where m is the mass of the particle. At what de-broglie wavelength will relativistic corrections become important for an electron?

Options

A. `lamda=1nm`
B. `lamda=10nm`
C. `lamda=10^(-1)`
D. `lamda=10^(-4)`

Explanation

To determine at what de Broglie wavelength relativistic corrections become important for an electron, we follow these steps:

### Step 1: Understand the condition for relativistic corrections
Relativistic corrections become necessary when the kinetic energy \( K \) of the particle becomes comparable to its rest energy \( mc^2 \). The kinetic energy of a particle is given by:
\[ K = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the particle and \( v \) is its velocity. The rest energy is given by:
\[ E_0 = mc^2 \]

### Step 2: Set the condition for relativistic corrections
We set the kinetic energy equal to the rest energy:
\[ \frac{1}{2} mv^2 \approx mc^2 \]
This simplifies to:
\[ v^2 \approx 2c^2 \]
Taking the square root gives:
\[ v \approx \sqrt{2}c \]

### Step 3: Relate velocity to de Broglie wavelength
The de Broglie wavelength \( \lambda \) is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the particle. The momentum \( p \) can also be expressed as:
\[ p = mv \]
Thus, we can rewrite the de Broglie wavelength as:
\[ \lambda = \frac{h}{mv} \]

### Step 4: Substitute the velocity condition into the de Broglie wavelength
Substituting \( v \approx \sqrt{2}c \) into the de Broglie wavelength equation gives:
\[ \lambda \approx \frac{h}{m\sqrt{2}c} \]

### Step 5: Insert known values for Planck's constant and the mass of the electron
Using the values:
- \( h = 6.626 \times 10^{-34} \, \text{J s} \)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)
- \( c = 3 \times 10^8 \, \text{m/s} \)

We can calculate:
\[ \lambda \approx \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \cdot \sqrt{2} \cdot 3 \times 10^8} \]

### Step 6: Calculate the numerical value
Calculating the denominator:
\[ m\sqrt{2}c \approx 9.1 \times 10^{-31} \cdot 1.414 \cdot 3 \times 10^8 \approx 3.85 \times 10^{-22} \]
Now substituting this back into the wavelength equation:
\[ \lambda \approx \frac{6.626 \times 10^{-34}}{3.85 \times 10^{-22}} \approx 1.72 \times 10^{-12} \, \text{m} \text{ or } 1.72 \, \text{pm} \]

### Step 7: Conclusion
Relativistic corrections become important for an electron when the de Broglie wavelength is approximately \( 1.72 \, \text{pm} \).