To solve the problem, we will follow these steps:
### Step 1: Calculate the cut-off wavelength at V1
The cut-off wavelength (λ2) for the X-ray spectrum can be calculated using the formula:
\[
\lambda_2 = \frac{hc}{eV_1}
\]
Where:
- \( h \) is Planck's constant (given as \( hc = 1240 \, \text{eV nm} \))
- \( e \) is the charge of an electron (1 eV = 1.6 x \( 10^{-19} \) J)
- \( V_1 = 15.5 \, \text{kV} = 15.5 \times 10^3 \, \text{V} \)
Substituting the values:
\[
\lambda_2 = \frac{1240 \, \text{eV nm}}{(1.6 \times 10^{-19} \, \text{C})(15.5 \times 10^3 \, \text{V})}
\]
Calculating this gives:
\[
\lambda_2 \approx 0.08 \, \text{nm}
\]
### Step 2: Determine the new cut-off wavelength at V2
When the voltage is increased to \( V_2 = 31 \, \text{kV} \), the new cut-off wavelength (λ2') can be calculated as:
\[
\lambda_2' = \frac{hc}{eV_2}
\]
Substituting \( V_2 = 31 \, \text{kV} = 31 \times 10^3 \, \text{V} \):
\[
\lambda_2' = \frac{1240 \, \text{eV nm}}{(1.6 \times 10^{-19} \, \text{C})(31 \times 10^3 \, \text{V})}
\]
Calculating this gives:
\[
\lambda_2' \approx 0.04 \, \text{nm}
\]
### Step 3: Calculate the change in wavelength
The problem states that the wavelength interval between the K-alpha line and the cut-off wavelength increases by a factor of 1.3. Let \( \lambda_1 \) be the wavelength of the K-alpha line. The initial difference is:
\[
\Delta \lambda = \lambda_1 - \lambda_2
\]
After the voltage increase, the new difference is:
\[
\Delta \lambda' = \lambda_1 - \lambda_2' = 1.3 \Delta \lambda
\]
### Step 4: Set up the equation
From the above, we have:
\[
\lambda_1 - \lambda_2' = 1.3(\lambda_1 - \lambda_2)
\]
Substituting \( \lambda_2' = 0.04 \, \text{nm} \) and \( \lambda_2 = 0.08 \, \text{nm} \):
\[
\lambda_1 - 0.04 = 1.3(\lambda_1 - 0.08)
\]
### Step 5: Solve for λ1
Expanding the equation:
\[
\lambda_1 - 0.04 = 1.3\lambda_1 - 1.04
\]
Rearranging gives:
\[
1.3\lambda_1 - \lambda_1 = 1.04 - 0.04
\]
\[
0.3\lambda_1 = 1.0
\]
\[
\lambda_1 = \frac{1.0}{0.3} \approx 3.33 \, \text{nm}
\]
### Step 6: Relate λ1 to atomic number Z
Using the formula for the K-alpha line:
\[
\frac{1}{\lambda_1} = R(Z - 1)^2
\]
Where \( R = 1 \times 10^7 \, \text{m}^{-1} \) or \( R = 1 \times 10^{9} \, \text{nm}^{-1} \).
Substituting \( \lambda_1 \):
\[
\frac{1}{3.33} = R(Z - 1)^2
\]
Calculating gives:
\[
Z - 1 = \sqrt{\frac{1}{3.33 \times 10^9}}
\]
Calculating \( Z \):
\[
Z \approx 26
\]
### Step 7: Calculate \( \frac{Z}{13} \)
Finally, we find:
\[
\frac{Z}{13} = \frac{26}{13} = 2
\]
### Final Answer
Thus, the value of \( \frac{Z}{13} \) is:
\[
\boxed{2}
\]
