When the voltage applied to an X-ray tube increases from `10` to `20 kV` the wavelength difference between the ka line and the short wave cut-off of continuous X-ray spectrum increases by a factor of `3.0`. Identify the target element. Take `4/3R = 1200 Å` where R is Rydberg’s contant and `hc = 12.4 keV Å`

Answer

To solve the problem, we need to analyze the relationship between the wavelength of the characteristic X-ray (Ka line) and the short-wave cut-off of the continuous X-ray spectrum as the voltage applied to the X-ray tube changes.

### Step-by-Step Solution:

1. **Understanding the Wavelengths**:
- The wavelength of the characteristic X-ray (Ka line) is given by:
\[ \lambda_k = \frac{4}{3R} \cdot \frac{1}{(Z - 1)^2} \]
- The short-wave cut-off wavelength (continuous spectrum) is given by:
\[ \lambda_c = \frac{hc}{eV} \]
- Where \( hc = 12.4 \, \text{keV} \cdot \text{Å} \) and \( eV \) is the accelerating voltage.

2. **Calculating Wavelengths for Different Voltages**:
- For \( V = 10 \, \text{kV} \):
\[ \lambda_c(10) = \frac{12.4}{10} = 1.24 \, \text{Å} \]
- For \( V = 20 \, \text{kV} \):
\[ \lambda_c(20) = \frac{12.4}{20} = 0.62 \, \text{Å} \]

3. **Finding the Wavelength Difference**:
- The difference in wavelengths for \( V = 10 \, \text{kV} \):
\[ \Delta \lambda_{10} = \lambda_k - \lambda_c(10) = \frac{4}{3R} \cdot \frac{1}{(Z - 1)^2} - 1.24 \]
- The difference in wavelengths for \( V = 20 \, \text{kV} \):
\[ \Delta \lambda_{20} = \lambda_k - \lambda_c(20) = \frac{4}{3R} \cdot \frac{1}{(Z - 1)^2} - 0.62 \]

4. **Setting Up the Relationship**:
- According to the problem, the difference increases by a factor of 3:
\[ \Delta \lambda_{20} = 3 \Delta \lambda_{10} \]
- Substituting the expressions:
\[ \left(\frac{4}{3R} \cdot \frac{1}{(Z - 1)^2} - 0.62\right) = 3 \left(\frac{4}{3R} \cdot \frac{1}{(Z - 1)^2} - 1.24\right) \]

5. **Solving for Z**:
- Rearranging the equation:
\[ \frac{4}{3R} \cdot \frac{1}{(Z - 1)^2} - 0.62 = 3 \cdot \frac{4}{3R} \cdot \frac{1}{(Z - 1)^2} - 3.72 \]
- Simplifying gives:
\[ 2.62 = 2 \cdot \frac{4}{3R} \cdot \frac{1}{(Z - 1)^2} \]
- Solving for \( Z \):
\[ (Z - 1)^2 = \frac{2 \cdot \frac{4}{3R}}{2.62} \]
- Substitute \( 4/3R = 1200 \, \text{Å} \):
\[ (Z - 1)^2 = \frac{2 \cdot 1200}{2.62} \]
- Calculate \( Z \):
\[ Z - 1 = \sqrt{\frac{2400}{2.62}} \approx 30.0 \implies Z \approx 31 \]

6. **Identifying the Target Element**:
- The atomic number \( Z = 31 \) corresponds to Gallium (Ga).

### Final Answer:
The target element is Gallium (Ga).