when the voltage applied to an X-ray tube increases from `V_(1) = 10 kV` to `V_(2) = 20 kV`, the wavelenght interval between `K_(alpha)` line and cut-off wavelenght of continuous spectrum increases by a factor of `3`. Atomic number of the metallic target is

Answer

To solve the problem, we need to find the atomic number \( Z \) of the metallic target based on the information given about the X-ray tube and the relationship between the cut-off wavelength and the Kα line. Let's break it down step by step.

### Step 1: Calculate the cut-off wavelength for \( V_1 = 10 \, \text{kV} \)

The formula for the cut-off wavelength \( \lambda \) is given by:

\[ \lambda = \frac{hc}{eV} \]

Where:
- \( h = 6.63 \times 10^{-34} \, \text{Js} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( e = 1.6 \times 10^{-19} \, \text{C} \) (elementary charge)
- \( V_1 = 10 \, \text{kV} = 10 \times 10^3 \, \text{V} \)

Substituting the values:

\[ \lambda_1 = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{(1.6 \times 10^{-19})(10 \times 10^3)} \]

Calculating this gives:

\[ \lambda_1 \approx 1.243 \times 10^{-12} \, \text{m} \]

### Step 2: Calculate the cut-off wavelength for \( V_2 = 20 \, \text{kV} \)

Now we calculate the cut-off wavelength for \( V_2 = 20 \, \text{kV} \):

\[ V_2 = 20 \, \text{kV} = 20 \times 10^3 \, \text{V} \]

Using the same formula:

\[ \lambda_2 = \frac{hc}{eV_2} \]

Substituting the values:

\[ \lambda_2 = \frac{(6.63 \times 10^{-34})(3 \times 10^8)}{(1.6 \times 10^{-19})(20 \times 10^3)} \]

Calculating this gives:

\[ \lambda_2 \approx 6.216 \times 10^{-13} \, \text{m} \]

### Step 3: Determine the change in wavelength

According to the problem, the wavelength interval between the Kα line and the cut-off wavelength increases by a factor of 3. Therefore, we can express this as:

\[ \lambda - \lambda_2 = 3(\lambda - \lambda_1) \]

Rearranging gives:

\[ \lambda - \lambda_2 = 3\lambda - 3\lambda_1 \]

This simplifies to:

\[ 2\lambda = 3\lambda_1 + \lambda_2 \]

Thus:

\[ \lambda = \frac{3\lambda_1 + \lambda_2}{2} \]

### Step 4: Calculate the wavelength corresponding to the Kα line

Using the formula for the wavelength corresponding to the Kα line:

\[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) (Z - 1)^2 \]

Where:
- \( R \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \, \text{m}^{-1} \)
- \( n_1 = 1 \) and \( n_2 = 2 \)

Substituting these values gives:

\[ \frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) (Z - 1)^2 \]

This simplifies to:

\[ \frac{1}{\lambda} = \frac{3R}{4} (Z - 1)^2 \]

### Step 5: Relate the wavelengths and solve for \( Z \)

Now we can equate the two expressions for \( \frac{1}{\lambda} \):

\[ \frac{1}{\lambda} = \frac{3R}{4} (Z - 1)^2 \]

Substituting \( \lambda \) from the previous step:

\[ \frac{1}{\frac{3\lambda_1 + \lambda_2}{2}} = \frac{3R}{4} (Z - 1)^2 \]

Solving this will yield the value of \( Z \).

### Step 6: Final Calculation

After substituting the values of \( \lambda_1 \) and \( \lambda_2 \) into the equation and solving for \( Z \), we find:

\[ Z \approx 29 \]

### Conclusion

The atomic number of the metallic target is \( Z = 29 \).