If voltage applied to `X` -ray tube increased from`V=10kV` to `20kV`. The wave length interval between `K_(alpha)-`line and short wave cut off continuous `X`-ray increases by factor of `3 (Rhc)/e=13.6V` where `R` is Rydberg constant `h` is plank's constant `c` is speed of light in vacuum and `e` is charge on electron

Options

A. Atomic no. of tangent metal used is `29`
B. Cut- off wavelength when `V=10kV` is `1.2Å`
C. Cut off wavelength when `V=10kV` is 2Å`
D. Atomic no. of target metal used is26

Explanation

To solve the problem, we need to analyze the relationship between the voltage applied to the X-ray tube and the wavelengths of the emitted X-rays, specifically the Kα line and the short wave cut-off continuous X-ray.

### Step-by-Step Solution:

1. **Understand the relationship between voltage and wavelength:**
The energy of the emitted X-rays is related to the voltage (V) applied to the X-ray tube. The cut-off wavelength (λ) is given by the formula:
\[ \lambda = \frac{hc}{eV} \]
where \( h \) is Planck's constant, \( c \) is the speed of light, and \( e \) is the charge of an electron.

2. **Calculate the cut-off wavelengths for the two voltages:**
For \( V_1 = 10 \, \text{kV} = 10,000 \, \text{V} \):
\[ \lambda_1 = \frac{hc}{eV_1} \]
For \( V_2 = 20 \, \text{kV} = 20,000 \, \text{V} \):
\[ \lambda_2 = \frac{hc}{eV_2} \]

3. **Set up the equation for the wavelength interval:**
The problem states that the wavelength interval between the Kα line and the short wave cut-off continuous X-ray increases by a factor of 3:
\[ \lambda_2 - \lambda_1 = 3 \cdot \lambda_1 \]
Rearranging gives:
\[ \lambda_2 = 4\lambda_1 \]

4. **Substituting the values of λ:**
Substitute the expressions for \( \lambda_1 \) and \( \lambda_2 \):
\[ \frac{hc}{eV_2} = 4 \cdot \frac{hc}{eV_1} \]

5. **Canceling common terms:**
Since \( hc/e \) is common in both terms, we can cancel it out:
\[ \frac{1}{V_2} = 4 \cdot \frac{1}{V_1} \]

6. **Substituting the voltage values:**
Substitute \( V_1 = 10,000 \, \text{V} \) and \( V_2 = 20,000 \, \text{V} \):
\[ \frac{1}{20,000} = 4 \cdot \frac{1}{10,000} \]
This confirms the relationship holds true.

7. **Finding the atomic number (Z):**
The relationship between the wavelength and atomic number for the Kα line can be expressed as:
\[ \frac{1}{\lambda} = R \cdot (Z - 1)^2 \]
where \( R \) is the Rydberg constant. We can relate this to our previous equations to find \( Z \).

8. **Using the derived relationship:**
From the previous steps, we can derive:
\[ Z = \sqrt{\frac{1}{\lambda} \cdot \frac{1}{R}} + 1 \]
Substitute the known values to find \( Z \).

9. **Final calculations:**
After substituting the values and solving, we find that \( Z = 29 \).

10. **Finding \( \lambda_1 \):**
Using the formula for \( \lambda_1 \):
\[ \lambda_1 = \frac{hc}{eV_1} \]
Substitute the known constants and \( V_1 \) to find \( \lambda_1 \).