The `K_(alpha)` wavelength of an element with atomic number z is `lambda_z`. The `k_(alpha)` wavelength of an element with atomic number 2z is `lambda_(2z)` . How are `lambda_z and lambda_(2z)` related ?

Options

A. `lambda_z gt 4lambda_(2z)`
B. `lambda_z=4lambda_(2z)`
C. `lambda_z lt 4 lambda_(2z)`
D. Depending on z,`lambda_z` could be greater than or less than `4lambda_(2z)`

Answer

To find the relationship between the K-alpha wavelengths \( \lambda_z \) and \( \lambda_{2z} \) for elements with atomic numbers \( z \) and \( 2z \), respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding K-alpha Transition**:
The K-alpha wavelength is emitted when an electron transitions from the L shell (n=2) to the K shell (n=1).

2. **Using the Rydberg Formula**:
The formula for the wavelength of the emitted photon during this transition is given by:
\[ \frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
where \( R \) is the Rydberg constant, \( Z \) is the atomic number, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the shells involved in the transition.

3. **Applying the Formula for Atomic Number \( z \)**:
For the element with atomic number \( z \):
\[ \frac{1}{\lambda_z} = R \cdot z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \cdot z^2 \left( 1 - \frac{1}{4} \right) = R \cdot z^2 \cdot \frac{3}{4} \]
Let’s denote \( K = R \cdot \frac{3}{4} \). Thus, we have:
\[ \frac{1}{\lambda_z} = K \cdot z^2 \]

4. **Applying the Formula for Atomic Number \( 2z \)**:
For the element with atomic number \( 2z \):
\[ \frac{1}{\lambda_{2z}} = R \cdot (2z)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \cdot 4z^2 \cdot \frac{3}{4} = 3R \cdot z^2 \]
This can be expressed as:
\[ \frac{1}{\lambda_{2z}} = 4K \cdot z^2 \]

5. **Relating \( \lambda_z \) and \( \lambda_{2z} \)**:
Now, we can relate \( \lambda_z \) and \( \lambda_{2z} \):
\[ \frac{1}{\lambda_{2z}} = 4 \cdot \frac{1}{\lambda_z} \]
Taking the reciprocal gives:
\[ \lambda_{2z} = \frac{\lambda_z}{4} \]

6. **Final Relationship**:
Rearranging the above equation, we find:
\[ \lambda_z = 4 \cdot \lambda_{2z} \]

### Conclusion:
The relationship between the K-alpha wavelengths of the two elements is:
\[ \lambda_z = 4 \cdot \lambda_{2z} \]