If `lamda_0` stands for mid-wavelength in the visible region, the de Broglie wavelength for 100 V electrons is nearest to

Options

A. `(lamda_0)/(5)`
B. `(lamda_0)/(50)`
C. `(lamda_0)/(500)`
D. `(lamda_0)/(5000)`

Explanation

To find the de Broglie wavelength for electrons accelerated through a potential difference of 100 V, we will follow these steps:

### Step 1: Understand 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, \( m \) is the mass of the electron, and \( v \) is its velocity.

### Step 2: Relate kinetic energy to potential difference
The kinetic energy (\( KE \)) of an electron accelerated through a potential difference \( V \) is given by:
\[ KE = qV \]
where \( q \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) C) and \( V \) is the potential difference (100 V in this case).

### Step 3: Express kinetic energy in terms of mass and velocity
The kinetic energy can also be expressed as:
\[ KE = \frac{1}{2} mv^2 \]
Setting these two expressions for kinetic energy equal gives:
\[ qV = \frac{1}{2} mv^2 \]

### Step 4: Solve for velocity
Rearranging the equation to solve for \( v \):
\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 5: Substitute \( v \) into the de Broglie wavelength formula
Substituting the expression for \( v \) into the de Broglie wavelength formula:
\[ \lambda = \frac{h}{m \sqrt{\frac{2qV}{m}}} = \frac{h}{\sqrt{2mqV}} \]

### Step 6: Substitute known values
Now, substituting the known values:
- \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- \( q = 1.6 \times 10^{-19} \, \text{C} \)
- \( V = 100 \, \text{V} \)
- \( m = 9.1 \times 10^{-31} \, \text{kg} \)

The equation becomes:
\[ \lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times 100}} \]

### Step 7: Calculate the value of \( \lambda \)
Calculating the denominator:
\[ \sqrt{2 \times (9.1 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times 100} \approx \sqrt{2.912 \times 10^{-48}} \approx 5.39 \times 10^{-24} \]
Now substituting back to find \( \lambda \):
\[ \lambda \approx \frac{6.63 \times 10^{-34}}{5.39 \times 10^{-24}} \approx 1.23 \times 10^{-10} \, \text{m} \]

### Step 8: Relate to \( \lambda_0 \)
Given that \( \lambda_0 \) (the mid-wavelength in the visible region) is approximately \( 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \):
\[ \lambda \approx \frac{1.23 \times 10^{-10}}{5 \times 10^{-7}} = \frac{1.23}{5} \times 10^{-7} \approx 0.246 \times 10^{-7} \approx 0.246 \, \lambda_0 \]

### Final Answer
The de Broglie wavelength for 100 V electrons is approximately \( \frac{\lambda_0}{5000} \).