We wish to see inside an atom. Assuming the atom to have a diameter of `100` pm, this means that one must be able to resolve a width of say `10` pm. If an electron microscope is used, the minimum electron energy required is about

Options

A. 1.5 keV
B. 15 KeV
C. 150 keV
D. 1.5 MeV.

Explanation

To solve the problem of determining the minimum electron energy required to resolve a width of 10 pm using an electron microscope, we can follow these steps:

### Step 1: Understand the relationship between wavelength and resolution
The resolving power of an electron microscope is related to the wavelength of the electrons used. To resolve a feature of size \( d \), the wavelength \( \lambda \) must be comparable to or smaller than \( d \). In this case, we want to resolve a width of 10 pm (picometers).

### Step 2: Convert units
Convert the width from picometers to meters for calculation:
\[ 10 \text{ pm} = 10 \times 10^{-12} \text{ m} = 10^{-11} \text{ m} \]

### Step 3: Use the de Broglie wavelength formula
The de Broglie wavelength \( \lambda \) of an electron is given by:
\[ \lambda = \frac{h}{\sqrt{2 m_e E}} \]
where:
- \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \, \text{Js} \)),
- \( m_e \) is the mass of the electron (\( 9.11 \times 10^{-31} \, \text{kg} \)),
- \( E \) is the energy of the electron.

### Step 4: Rearrange the formula to find energy
Rearranging the formula to solve for energy \( E \):
\[ E = \frac{h^2}{2 m_e \lambda^2} \]

### Step 5: Substitute values into the equation
Substituting the known values into the equation:
\[ E = \frac{(6.63 \times 10^{-34})^2}{2 \times (9.11 \times 10^{-31}) \times (10^{-11})^2} \]

### Step 6: Calculate the energy
Calculating the numerator:
\[ (6.63 \times 10^{-34})^2 = 4.39 \times 10^{-67} \text{ J}^2 \]
Calculating the denominator:
\[ 2 \times (9.11 \times 10^{-31}) \times (10^{-11})^2 = 2 \times 9.11 \times 10^{-31} \times 10^{-22} = 1.82 \times 10^{-52} \text{ J} \]
Now substituting these values back into the energy equation:
\[ E = \frac{4.39 \times 10^{-67}}{1.82 \times 10^{-52}} \approx 2.41 \times 10^{-15} \text{ J} \]

### Step 7: Convert energy to electron volts
To convert joules to electron volts, use the conversion factor \( 1 \text{ eV} = 1.6 \times 10^{-19} \text{ J} \):
\[ E \approx \frac{2.41 \times 10^{-15}}{1.6 \times 10^{-19}} \approx 15.06 \text{ keV} \]

### Conclusion
The minimum electron energy required to resolve a width of 10 pm is approximately **15 keV**.

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