A potential of 10000 V is applied across an x-ray tube. Find the ratio of de-Broglie wavelength associated with incident electrons to the minimum wavelength associated with x-rays.

Options

A. `10`
B. `20`
C. `1//10`
D. `1//20`

Explanation

To solve the problem, we need to find the ratio of the de Broglie wavelength associated with incident electrons to the minimum wavelength associated with x-rays when a potential of 10,000 V is applied across an x-ray tube.

### Step-by-Step Solution:

1. **Determine the de Broglie wavelength of the electrons:**
The de Broglie wavelength (\( \lambda_{dB} \)) of an electron can be calculated using the formula:
\[ \lambda_{dB} = \frac{12.27}{V} \text{ (in Å)} \]
where \( V \) is the potential difference in volts.

Given \( V = 10,000 \, V \):
\[ \lambda_{dB} = \frac{12.27}{10,000} = 0.001227 \, \text{Å} \]

2. **Determine the minimum wavelength of the x-rays:**
The minimum wavelength (\( \lambda_{min} \)) associated with x-rays can be calculated using the formula:
\[ \lambda_{min} = \frac{123.7}{V} \text{ (in pm)} \]
where \( V \) is the potential difference in volts.

Again, using \( V = 10,000 \, V \):
\[ \lambda_{min} = \frac{123.7}{10,000} = 0.01237 \, \text{pm} \]

3. **Convert units if necessary:**
Since \( \lambda_{dB} \) is in Ångstroms and \( \lambda_{min} \) is in picometers, we can convert \( \lambda_{dB} \) to picometers:
\[ 1 \, \text{Å} = 100 \, \text{pm} \implies \lambda_{dB} = 0.001227 \, \text{Å} = 0.1227 \, \text{pm} \]

4. **Calculate the ratio of the wavelengths:**
Now, we can find the ratio of the de Broglie wavelength to the minimum wavelength:
\[ \text{Ratio} = \frac{\lambda_{dB}}{\lambda_{min}} = \frac{0.1227 \, \text{pm}}{0.01237 \, \text{pm}} \]

Performing the division:
\[ \text{Ratio} = \frac{0.1227}{0.01237} \approx 9.91 \approx 10 \]

5. **Final Result:**
The ratio of the de Broglie wavelength associated with incident electrons to the minimum wavelength associated with x-rays is approximately:
\[ \text{Ratio} \approx 10 \]