For the same objective, the ratio of least separation between two points to be distiguised by a microscope for light of `5000 Å` and electrons acclerated through 100V used as illuminating substance is __________ (neraly)

Options

A. `2xx10^(-4)`
B. `2xx10^(-3)`
C. `2xx10^(-2)`
D. `4xx10^(-4)`

Explanation

To solve the problem of finding the ratio of the least separation between two points to be distinguished by a microscope for light of wavelength \(5000 \, \text{Å}\) and electrons accelerated through \(100 \, \text{V}\), we will follow these steps:

### Step 1: Understand the relationship between wavelength and minimum separation
The minimum separation (\(\Delta x\)) that can be distinguished by a microscope is directly proportional to the wavelength (\(\lambda\)). Therefore, we can write:
\[ \Delta x \propto \lambda \]

### Step 2: Identify the wavelengths
1. **For light**: The wavelength (\(\lambda_1\)) is given as:
\[ \lambda_1 = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \]
2. **For electrons**: We need to calculate the de Broglie wavelength (\(\lambda_2\)) of electrons accelerated through \(100 \, \text{V}\). The formula for the de Broglie wavelength is:
\[ \lambda = \frac{h}{p} \]
where \(p\) is the momentum of the electron. The momentum can be expressed as:
\[ p = mv = \sqrt{2m \cdot KE} \]
where \(KE\) (kinetic energy) is given by:
\[ KE = eV \]
Here, \(e\) is the charge of the electron and \(V\) is the accelerating voltage.

### Step 3: Calculate the de Broglie wavelength for electrons
Using the known values:
- Charge of electron, \(e \approx 1.6 \times 10^{-19} \, \text{C}\)
- Mass of electron, \(m \approx 9.11 \times 10^{-31} \, \text{kg}\)
- Voltage, \(V = 100 \, \text{V}\)

The kinetic energy \(KE\) is:
\[ KE = eV = (1.6 \times 10^{-19} \, \text{C})(100 \, \text{V}) = 1.6 \times 10^{-17} \, \text{J} \]

Now, substituting into the momentum equation:
\[ p = \sqrt{2m \cdot KE} = \sqrt{2 \cdot (9.11 \times 10^{-31} \, \text{kg}) \cdot (1.6 \times 10^{-17} \, \text{J})} \]
Calculating this gives:
\[ p \approx \sqrt{2.91 \times 10^{-47}} \approx 5.39 \times 10^{-24} \, \text{kg m/s} \]

Now, substituting \(p\) into the de Broglie wavelength formula:
\[ \lambda_2 = \frac{h}{p} = \frac{6.63 \times 10^{-34} \, \text{Js}}{5.39 \times 10^{-24} \, \text{kg m/s}} \approx 1.23 \times 10^{-10} \, \text{m} = 1.23 \, \text{Å} \]

### Step 4: Calculate the ratio of least separations
Now we have:
- \(\lambda_1 = 5000 \, \text{Å}\)
- \(\lambda_2 \approx 1.23 \, \text{Å}\)

The ratio of the least separations is given by:
\[ \frac{\Delta x_1}{\Delta x_2} = \frac{\lambda_1}{\lambda_2} \]
Substituting the values:
\[ \frac{\Delta x_1}{\Delta x_2} = \frac{5000 \, \text{Å}}{1.23 \, \text{Å}} \approx 4065.04 \]

### Step 5: Final ratio
The final ratio of least separation between two points to be distinguished by a microscope for light and electrons is approximately:
\[ \Delta x_2 : \Delta x_1 \approx 1 : 4065 \]

### Final Answer
The ratio of least separation is nearly \(2 \times 10^{-4}\).