A single slit diffraction pattern is formed with light of wavelength $6384 Å$. The second secondary maximum for this wavelength coincides with the third secondary maximum in the pattern for light of wavelength ' $\lambda_0^{\prime}$. The value of ' $\lambda_0^{\prime}$ is

The condition for a secondary maximum in single-slit diffraction is $a \sin \theta = (m + \frac{1}{2})\lambda$, where $a$ is the slit width and $m$ is the order starting from 1.

For wavelength $\lambda = 6384$ Å at $m=2$:
$a \sin \theta_1 = \frac{5}{2}\lambda$

For wavelength $\lambda_0$ at $m=3$:
$a \sin \theta_2 = \frac{7}{2}\lambda_0$

Since $\theta_1 = \theta_2$, the expressions are equal:
$\frac{5}{2}\lambda = \frac{7}{2}\lambda_0$

Solving yields $\lambda_0 = \frac{5}{7}\lambda = \frac{5}{7} \times 6384 = 4560$ Å.

Final answer: $\boxed{4560 \text{ Å}}$