The `K_(alpha)` `X` - rays arising from a cobalt `(z = 27)` target have a wavelength of 179 pm. The `K_(alpha)` `X` - rays arising from a nickel target `(z = 28)` is

Options

A. `gt179"pm"`
B. `lt179 "pm"`
C. `=179 "pm"`
D. None of these

Answer

To solve the problem of finding the wavelength of the K_alpha X-rays arising from a nickel target (Z = 28), we will follow these steps:

### Step 1: Understand the K_alpha Transition
The K_alpha transition occurs when an electron transitions from the n = 2 energy level to the n = 1 energy level in an atom. The effective nuclear charge (Z_effective) for this transition can be expressed as:
\[ Z_{\text{effective}} = Z - 1 \]
where Z is the atomic number of the element.

### Step 2: Use the Wavelength Formula
The wavelength (λ) of the emitted photon during this transition can be related to the effective nuclear charge and the energy levels using the formula:
\[ \frac{1}{\lambda} \propto Z_{\text{effective}}^2 \]
This means that:
\[ \frac{1}{\lambda_1} = k \cdot Z_{\text{effective}_1}^2 \]
\[ \frac{1}{\lambda_2} = k \cdot Z_{\text{effective}_2}^2 \]
where k is a constant.

### Step 3: Set Up the Ratio of Wavelengths
From the above relationships, we can derive:
\[ \frac{\lambda_1}{\lambda_2} = \frac{Z_{\text{effective}_2}^2}{Z_{\text{effective}_1}^2} \]

### Step 4: Calculate Z_effective for Cobalt and Nickel
For cobalt (Z = 27):
\[ Z_{\text{effective}_1} = 27 - 1 = 26 \]

For nickel (Z = 28):
\[ Z_{\text{effective}_2} = 28 - 1 = 27 \]

### Step 5: Substitute the Known Values
We know from the problem that:
\[ \lambda_1 = 179 \text{ pm} \]
Now substituting the values into the ratio:
\[ \frac{\lambda_1}{\lambda_2} = \frac{27^2}{26^2} \]
This can be rearranged to find λ2:
\[ \lambda_2 = \lambda_1 \cdot \frac{26^2}{27^2} \]

### Step 6: Calculate λ2
Now substituting the known value of λ1:
\[ \lambda_2 = 179 \cdot \frac{26^2}{27^2} \]
Calculating the squares:
\[ 26^2 = 676 \]
\[ 27^2 = 729 \]
Thus:
\[ \lambda_2 = 179 \cdot \frac{676}{729} \]

### Step 7: Perform the Calculation
Now we can calculate λ2:
\[ \lambda_2 = 179 \cdot \frac{676}{729} \approx 179 \cdot 0.925 = 165.075 \text{ pm} \]

### Conclusion
The wavelength of the K_alpha X-rays arising from a nickel target (Z = 28) is approximately **165.1 pm**.