If an X-ray tube operates at the voltage of 10kV, find the ratio of the de-broglie wavelength of the incident electrons to the shortest wavelength of X-ray producted. The specific charge of electron is `1.8xx10^11 C/kg`.

Explanation

To solve the problem, we need to find the ratio of the de Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced in an X-ray tube operating at a voltage of 10 kV.

### Step-by-Step Solution:

1. **Identify the Shortest Wavelength of X-rays**:
The shortest wavelength (λ_min) of the X-rays produced can be calculated using the formula:
\[ \lambda_{\text{min}} = \frac{hc}{eV} \]
where:
- \( h \) = Planck's constant \( = 6.626 \times 10^{-34} \, \text{Js} \)
- \( c \) = speed of light \( = 3 \times 10^8 \, \text{m/s} \)
- \( e \) = charge of an electron \( = 1.6 \times 10^{-19} \, \text{C} \)
- \( V \) = operating voltage \( = 10 \, \text{kV} = 10 \times 10^3 \, \text{V} \)

2. **Calculate the Shortest Wavelength**:
Substituting the values into the formula:
\[ \lambda_{\text{min}} = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{(1.6 \times 10^{-19})(10 \times 10^3)} \]

3. **Calculate the De Broglie Wavelength of the Incident Electrons**:
The de Broglie wavelength (λ) of the electrons can be calculated using:
\[ \lambda = \frac{h}{mv} \]
where \( v \) is the velocity of the electrons. The velocity can be derived from the kinetic energy acquired by the electrons when accelerated through the potential \( V \):
\[ KE = eV = \frac{1}{2} mv^2 \]
Rearranging gives:
\[ v = \sqrt{\frac{2eV}{m}} \]
Substituting this into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{m\sqrt{\frac{2eV}{m}}} = \frac{h}{\sqrt{2meV}} \]

4. **Calculate the Ratio of the Wavelengths**:
We want the ratio of the de Broglie wavelength to the shortest wavelength of X-rays:
\[ \text{Ratio} = \frac{\lambda}{\lambda_{\text{min}}} = \frac{\frac{h}{\sqrt{2meV}}}{\frac{hc}{eV}} = \frac{1}{c} \cdot \frac{h}{\sqrt{2meV}} \cdot \frac{eV}{hc} \]
Simplifying this gives:
\[ \text{Ratio} = \frac{1}{c} \cdot \frac{1}{\sqrt{2m}} \cdot \sqrt{\frac{e}{V}} \]

5. **Substituting the Values**:
Given the specific charge of the electron \( \frac{e}{m} = 1.8 \times 10^{11} \, \text{C/kg} \), we can substitute this into the equation:
\[ \text{Ratio} = \sqrt{\frac{e/m}{2}} \cdot \frac{1}{c} \cdot \sqrt{V} \]
With \( V = 10^4 \, \text{V} \) and \( c = 3 \times 10^8 \, \text{m/s} \):
\[ \text{Ratio} = \sqrt{\frac{1.8 \times 10^{11}}{2}} \cdot \frac{1}{3 \times 10^8} \cdot \sqrt{10^4} \]

6. **Final Calculation**:
Calculate the numerical value to find the ratio.