A potential difference of `10^(3)` V is applied across an X-ray tube. The ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of X-rays products is -
`(e//m=1.8xx10^(14)C//kg`for an electron)

Options

A. `1//20`
B. `1//100`
C. 1
D. None of these

Explanation

To solve the problem, we need to find the ratio of the de-Broglie wavelength of the incident electrons (λ1) to the shortest wavelength of the X-rays produced (λ2) when a potential difference of \(10^3\) V is applied across an X-ray tube.

### Step-by-Step Solution:

1. **Understanding the de-Broglie Wavelength**:
The de-Broglie wavelength (λ) of a particle can be calculated using the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

2. **Momentum of the Electron**:
The momentum \(p\) of an electron accelerated through a potential difference \(V\) is given by:
\[ p = \sqrt{2m_e eV} \]
where \(m_e\) is the mass of the electron, \(e\) is the charge of the electron, and \(V\) is the potential difference.

3. **Substituting for λ1 (de-Broglie wavelength of the electron)**:
Thus, the de-Broglie wavelength of the electron (λ1) can be expressed as:
\[ \lambda_1 = \frac{h}{\sqrt{2m_e eV}} \]

4. **Shortest Wavelength of X-rays**:
The shortest wavelength (λ2) of the X-rays produced is given by:
\[ \lambda_2 = \frac{h c}{eV} \]
where \(c\) is the speed of light.

5. **Finding the Ratio**:
We need to find the ratio of the two wavelengths:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\frac{h}{\sqrt{2m_e eV}}}{\frac{h c}{eV}} = \frac{eV}{h c} \cdot \frac{h}{\sqrt{2m_e eV}} = \frac{eV}{c \sqrt{2m_e eV}} \]
Simplifying this gives:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\sqrt{eV}}{c \sqrt{2m_e}} \]

6. **Substituting Values**:
Given:
- \(e/m = 1.8 \times 10^{14} \, \text{C/kg}\)
- \(V = 10^3 \, \text{V}\)
- \(c = 3 \times 10^8 \, \text{m/s}\)

We can express \(m_e\) in terms of \(e/m\):
\[ m_e = \frac{e}{(e/m)} = \frac{e}{1.8 \times 10^{14}} \]

7. **Calculating the Ratio**:
Plugging in the values:
\[ \frac{\lambda_1}{\lambda_2} = \frac{\sqrt{(1.6 \times 10^{-19})(10^3)}}{(3 \times 10^8) \sqrt{2 \left(\frac{1.6 \times 10^{-19}}{1.8 \times 10^{14}}\right)}} \]

8. **Final Calculation**:
After performing the calculations, we find the numerical value of the ratio.

### Final Answer:
The ratio of the de-Broglie wavelength of the incident electrons to the shortest wavelength of the X-rays produced is found to be approximately \( \text{Option 3} \).