De Broglie Wavelength of Electron in Lithium Ion Excited State

According to Bohr's quantization condition, the orbital angular momentum $L$ of an electron in the $n$-th orbit is given by: $L = \frac{nh}{2\pi}$

Given that $L = \frac{3h}{2\pi}$, we find the principal quantum number: $n = 3$

For a hydrogen-like ion with atomic number $Z$, the radius of the $n$-th orbit is: $r_n = \frac{n^2}{Z} a_0$

For a lithium ion ($\text{Li}^{2+}$), the atomic number is $Z = 3$. Substituting $n = 3$ and $Z = 3$: $r_3 = \frac{3^2}{3} a_0 = 3a_0$

According to de Broglie's hypothesis for stationary Bohr orbits, the circumference of the orbit must contain an integral number of de Broglie wavelengths: $2\pi r_n = n\lambda$

Rearranging for the de Broglie wavelength $\lambda$: $\lambda = \frac{2\pi r_n}{n} = \frac{2\pi (3a_0)}{3} = 2\pi a_0$

Comparing this with the given expression $\lambda = p\pi a_0$: $p = 2$