An electron microscope is used to probe the atomic arrangement to a resolution of `5 Å`. What should be the electric potential to which the electrons need to be accelerated

Options

A. 2.5 V
B. 6 V
C. 2.5 kV
D. 5 kV

Explanation

To find the electric potential to which the electrons need to be accelerated in an electron microscope that probes atomic arrangements to a resolution of 5 Å, we can follow these steps:

### Step 1: Understand the relationship between wavelength and kinetic energy
The wavelength (\( \lambda \)) of the electrons is related to their kinetic energy (\( KE \)) by the formula:
\[ \lambda = \frac{h}{\sqrt{2m \cdot KE}} \]
where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \, \text{kg} \)),
- \( KE \) is the kinetic energy of the electron.

### Step 2: Relate kinetic energy to electric potential
The kinetic energy of the electron when accelerated through a potential \( V \) is given by:
\[ KE = e \cdot V \]
where \( e \) is the charge of the electron (\( 1.6 \times 10^{-19} \, \text{C} \)).

### Step 3: Substitute the expression for kinetic energy into the wavelength formula
Substituting \( KE = e \cdot V \) into the wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2m \cdot e \cdot V}} \]

### Step 4: Rearrange the equation to solve for \( V \)
Rearranging the equation to solve for \( V \):
\[ V = \frac{h^2}{2m \cdot e \cdot \lambda^2} \]

### Step 5: Substitute known values
Now, we can substitute the known values into the equation:
- \( \lambda = 5 \, \text{Å} = 5 \times 10^{-10} \, \text{m} \)
- \( h = 6.63 \times 10^{-34} \, \text{Js} \)
- \( m = 9.11 \times 10^{-31} \, \text{kg} \)
- \( e = 1.6 \times 10^{-19} \, \text{C} \)

Substituting these values into the equation:
\[ V = \frac{(6.63 \times 10^{-34})^2}{2 \cdot (9.11 \times 10^{-31}) \cdot (1.6 \times 10^{-19}) \cdot (5 \times 10^{-10})^2} \]

### Step 6: Calculate the value of \( V \)
Calculating the numerator:
\[ (6.63 \times 10^{-34})^2 = 4.39 \times 10^{-67} \]

Calculating the denominator:
\[ 2 \cdot (9.11 \times 10^{-31}) \cdot (1.6 \times 10^{-19}) \cdot (5 \times 10^{-10})^2 = 2 \cdot (9.11 \times 10^{-31}) \cdot (1.6 \times 10^{-19}) \cdot (25 \times 10^{-20}) = 7.27 \times 10^{-68} \]

Now substituting these values into the equation for \( V \):
\[ V = \frac{4.39 \times 10^{-67}}{7.27 \times 10^{-68}} \approx 6.03 \, \text{V} \]

### Step 7: Final answer
Thus, the electric potential to which the electrons need to be accelerated is approximately:
\[ \boxed{6 \, \text{V}} \]