The wavelength of `K_(a)` line copper is `1.5Å`. The ionisation energy of K electron in copper is

Options

A. `11.2xx10^(-17)J`
B. `12.9xx10^(-16)J`
C. `1.7xx10^(-15)J`
D. `10xx10^(-16)J`

Explanation

To find the ionization energy of the K electron in copper given the wavelength of the Kα line, we can use the formula that relates energy to wavelength:

### Step-by-Step Solution:

1. **Identify the Formula**:
The energy (E) can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy,
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)),
- \(\lambda\) is the wavelength in meters.

2. **Convert Wavelength to Meters**:
The given wavelength is \(1.5 \, \text{Å}\) (angstroms). We need to convert this to meters:
\[ 1 \, \text{Å} = 10^{-10} \, \text{m} \]
Therefore:
\[ \lambda = 1.5 \, \text{Å} = 1.5 \times 10^{-10} \, \text{m} \]

3. **Substitute the Values into the Formula**:
Now we can substitute the values of \(h\), \(c\), and \(\lambda\) into the energy formula:
\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s})(3 \times 10^8 \, \text{m/s})}{1.5 \times 10^{-10} \, \text{m}} \]

4. **Calculate the Energy**:
Performing the calculation:
\[ E = \frac{(6.626 \times 3) \times 10^{-34 + 8}}{1.5 \times 10^{-10}} \]
\[ E = \frac{19.878 \times 10^{-26}}{1.5 \times 10^{-10}} = 13.252 \times 10^{-16} \, \text{J} \]
Simplifying gives:
\[ E \approx 1.3252 \times 10^{-15} \, \text{J} \]

5. **Final Result**:
The ionization energy of the K electron in copper is approximately:
\[ E \approx 1.3252 \times 10^{-15} \, \text{J} \text{ or } 12.9 \times 10^{-16} \, \text{J} \]

### Conclusion:
The ionization energy of the K electron in copper is \(12.9 \times 10^{-16} \, \text{J}\).