1. Use the de Broglie condition for Bohr orbit

For an electron in the $n^{\text{th}}$ Bohr orbit,

$2\pi r_n = n\lambda$

So,

$\lambda = \frac{2\pi r_n}{n}$

2. Radius of the $n^{\text{th}}$ orbit in a hydrogenic atom

For a hydrogenic species,

$r_n = \frac{n^2 a_0}{Z}$

Substitute into the wavelength expression:

$\lambda = \frac{2\pi}{n}\cdot \frac{n^2 a_0}{Z}$

$\lambda = \frac{2\pi n a_0}{Z}$

3. Given wavelength

It is given that

$\lambda = 1.5\pi a_0$

Therefore,

$\frac{2\pi n a_0}{Z} = 1.5\pi a_0$

Cancel $\pi a_0$ from both sides:

$\frac{2n}{Z} = 1.5$

$\frac{n}{Z} = \frac{1.5}{2} = 0.75$

4. Match with the options

$\frac{n}{Z} = 0.75$

So the correct option is A.

5. Comparison with stored answer

Stored correct answer: A

This matches our derived answer.