Step 1: Understanding the relationship.
The wavelength of \( K_{\alpha} \)-X-ray emitted by an element is inversely proportional to the square of its atomic number. Using the relationship \( \lambda \propto \frac{1}{Z^2} \), we can solve for the atomic number emitting a wavelength of \( 4\lambda \).
Step 2: Calculations.
Since \( Z_1 = 11 \) and the wavelength changes by a factor of 4, we find the new atomic number is 6.