For an electron microscope that uses electrons accelerated throught 120 kV.

Options

A. Resolving power is approximately 5 times larger than that of an ordinary microscope using white light.
B. Smallest object that can be seen is approximately 1/5 times smaller that of an ordinary microscope with white light
C. Resolving power increases if opticaly denser medium is used on one side of lens of an ordinary microscope with white light.
D. Resolving power decreases if optically denser medium is used on one side of lens of an ordinary microscope with white light.

Answer

To solve the problem regarding the resolving power of an electron microscope that uses electrons accelerated through a potential difference of 120 kV, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Resolving Power Formula**:
The resolving power (R) of a microscope is given by the formula:
\[ R = \frac{2n \sin \theta}{\lambda} \]
where:
- \( n \) is the refractive index of the medium,
- \( \theta \) is the half-angle of the maximum cone of light that can enter the lens,
- \( \lambda \) is the wavelength of the electrons used in the microscope.

2. **Calculating the Wavelength of Electrons**:
When electrons are accelerated through a potential difference (V), their wavelength (\( \lambda \)) can be calculated using the formula:
\[ \lambda = \sqrt{\frac{150}{V}} \text{ angstroms} \]
For this problem, \( V = 120 \text{ kV} = 120,000 \text{ V} \).

3. **Substituting the Value of V**:
Substitute \( V \) into the wavelength formula:
\[ \lambda = \sqrt{\frac{150}{120,000}} \text{ angstroms} \]

4. **Calculating the Wavelength**:
\[ \lambda = \sqrt{\frac{150}{120,000}} = \sqrt{\frac{15}{12,000}} \text{ angstroms} \]
Now, calculate:
\[ \lambda = \sqrt{0.00125} \text{ angstroms} \approx 0.035 \text{ angstroms} \]

5. **Comparing with Wavelength of White Light**:
The wavelength of white light is approximately \( 5500 \text{ angstroms} \). To find the factor by which the wavelength of electrons is smaller:
\[ \text{Factor} = \frac{5500}{0.035} \approx 157,142.86 \]
This shows that the wavelength of electrons is significantly smaller than that of ordinary light.

6. **Conclusion on Resolving Power**:
Since the wavelength of electrons is much smaller than that of light, the resolving power of the electron microscope is much greater than that of optical microscopes. Additionally, if an optically denser medium is used, the refractive index \( n \) increases, which further increases the resolving power.

### Final Answer:
- The resolving power of the electron microscope is significantly higher than that of optical microscopes due to the much smaller wavelength of electrons compared to visible light.