Stopping potential of 24, 100, 110 and 115 k V are measured for photoelectrons emitted from a certain element when it is radiated with monochromatic X-ray . If this element is used as a target in an X-ray tube,what will be the wavelength of `K_a` - line?

Explanation

To solve the problem, we need to find the wavelength of the K_alpha line for an element based on the stopping potentials provided. The stopping potential gives us information about the maximum kinetic energy of the emitted photoelectrons, which we can use to determine the energy transition associated with the K_alpha line.

### Step-by-Step Solution:

1. **Identify the Stopping Potentials:**
The stopping potentials given are:
- 24 kV
- 100 kV
- 110 kV
- 115 kV

2. **Convert Stopping Potentials to Energy:**
The stopping potential (V) in kilovolts can be converted to energy (E) in electron volts (eV) using the formula:
\[ E = V \times 1000 \text{ eV} \]
Therefore:
- For 24 kV: \( E_{K} = 24 \times 10^3 \text{ eV} = 24000 \text{ eV} \)
- For 100 kV: \( E_{L} = 100 \times 10^3 \text{ eV} = 100000 \text{ eV} \)
- For 110 kV: \( E_{M} = 110 \times 10^3 \text{ eV} = 110000 \text{ eV} \)
- For 115 kV: \( E_{N} = 115 \times 10^3 \text{ eV} = 115000 \text{ eV} \)

3. **Determine the Energy Transition for K_alpha:**
The K_alpha line corresponds to the transition of an electron from the L shell to the K shell. The energy difference (ΔE) for this transition can be calculated as:
\[ \Delta E = E_{L} - E_{K} \]
Substituting the values:
\[ \Delta E = 100000 \text{ eV} - 24000 \text{ eV} = 76000 \text{ eV} = 76 \times 10^3 \text{ eV} \]

4. **Calculate the Wavelength of K_alpha Line:**
The relationship between energy and wavelength is given by:
\[ E = \frac{hc}{\lambda} \]
Rearranging for wavelength (λ):
\[ \lambda = \frac{hc}{E} \]
Where:
- \( h \) (Planck's constant) = \( 4.1357 \times 10^{-15} \text{ eV.s} \)
- \( c \) (speed of light) = \( 3 \times 10^{10} \text{ cm/s} = 3 \times 10^{18} \text{ Å/s} \)

Now substituting the values:
\[ \lambda = \frac{(4.1357 \times 10^{-15} \text{ eV.s})(3 \times 10^{18} \text{ Å/s})}{76 \times 10^3 \text{ eV}} \]

5. **Calculate the Wavelength:**
Performing the calculation:
\[ \lambda = \frac{1.240 \times 10^{-6} \text{ Å.eV.s}}{76 \times 10^3 \text{ eV}} \approx 0.163 \text{ Å} \]

### Final Answer:
The wavelength of the K_alpha line is approximately **0.163 Å**.