1. Use Bohr’s quantization condition

   For an electron in the $n^{\text{th}}$ Bohr orbit, $2\pi r_n = n\lambda$ where $\lambda$ is the de-Broglie wavelength.

2. Radius of the $n^{\text{th}}$ orbit

   In Bohr model, $r_n = n^2 a_0$

   For the $4^{\text{th}}$ orbit: $r_4 = 4^2 a_0 = 16a_0$

3. Apply the standing wave condition

   $2\pi r_4 = 4\lambda$

   Substituting $r_4 = 16a_0$: $2\pi(16a_0) = 4\lambda$ $32\pi a_0 = 4\lambda$ $\lambda = 8\pi a_0$

4. Compare with the required form

   The question asks: $\lambda = \underline{\hspace{2cm}}\,\pi a_0$

   Hence the blank is: $8$

Final Answer: $8$