For the same objective, what is the ratio of the least separation between two points to be distinguished by a microscope for light of 5000 Ã… and electrons accelerated through 100 V used as an illuminating substance?

Options

A. 3075
B. 3575
C. 4075
D. 5075

Explanation

To solve the problem, we need to find the ratio of the least separation between two points that can be distinguished by a microscope using light of wavelength 5000 Å (angstroms) and electrons accelerated through 100 V.

### Step-by-Step Solution:

1. **Understanding the Limit of Resolution**:
The limit of resolution (d) for a microscope is given by the formula:
\[ d = \frac{\lambda}{2 \sin \theta} \]
Since we are using the same objective, the angle \(\theta\) is the same for both cases, which means we can express the ratio of the least separations (d1 and d2) in terms of their wavelengths:
\[ \frac{d_1}{d_2} = \frac{\lambda_1}{\lambda_2} \]

2. **Identifying the Wavelengths**:
- For the light microscope, the wavelength \(\lambda_1\) is given as 5000 Å.
- For the electrons, we need to calculate the de Broglie wavelength \(\lambda_2\).

3. **Calculating the de Broglie Wavelength**:
The de Broglie wavelength \(\lambda\) of a particle is given by:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle. The momentum \(p\) can be expressed in terms of kinetic energy \(K\):
\[ p = \sqrt{2mK} \]
For electrons accelerated through a potential \(V\), the kinetic energy \(K\) is given by:
\[ K = qV \]
where \(q\) is the charge of the electron (approximately \(1.6 \times 10^{-19} C\)) and \(V\) is the accelerating voltage (100 V).

4. **Substituting Values**:
The kinetic energy for the electron is:
\[ K = (1.6 \times 10^{-19} C)(100 V) = 1.6 \times 10^{-17} J \]
Now substituting this into the momentum equation:
\[ p = \sqrt{2m(1.6 \times 10^{-17})} \]
where the mass \(m\) of an electron is approximately \(9.11 \times 10^{-31} kg\).

5. **Calculating the Momentum**:
\[ p = \sqrt{2 \times (9.11 \times 10^{-31}) \times (1.6 \times 10^{-17})} \approx \sqrt{2.91 \times 10^{-47}} \approx 5.39 \times 10^{-24} kg \cdot m/s \]

6. **Calculating the de Broglie Wavelength**:
Using \(h \approx 6.63 \times 10^{-34} J \cdot s\):
\[ \lambda_2 = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{5.39 \times 10^{-24}} \approx 1.23 \times 10^{-10} m = 1.23 \times 10^{10} Å \]

7. **Finding the Ratio of Least Separations**:
Now we can find the ratio:
\[ \frac{d_1}{d_2} = \frac{\lambda_1}{\lambda_2} = \frac{5000 Å}{1.23 \times 10^{10} Å} \approx 4075 \]

### Final Answer:
The ratio of the least separation between two points to be distinguished by a microscope for light of 5000 Å and electrons accelerated through 100 V is approximately **4075**.