If the de Broglie wavelength of the electron in $$n^{th}$$ Bohr orbit in a hydrogenic atom is equal to $$1.5\pi a_0$$ ($$a_0$$ is Bohr radius), then the value of $$\frac{n}{z}$$ is:

We begin by recalling the two standard Bohr‐model relations for a hydrogenic (one-electron) atom with nuclear charge $$Z$$.

First, the radius of the $$n^{\text{th}}$$ orbit is given by the well-known formula

$$r_n \;=\; \dfrac{n^{2}a_0}{Z},$$

where $$a_0$$ is the Bohr radius for the hydrogen atom.

Second, de Broglie’s standing-wave condition states that an integral number of wavelengths must fit exactly into the circumference of the circular orbit. Mathematically,

$$2\pi r_n \;=\; n\lambda,$$

where $$\lambda$$ is the de Broglie wavelength of the electron in that orbit.

We are told that this wavelength has the numerical value

$$\lambda \;=\; 1.5\pi a_0 \;=\; \dfrac{3}{2}\pi a_0.$$

Our task is to find the ratio $$\dfrac{n}{Z}$$.

From the standing-wave condition we first isolate $$\lambda$$:

$$\lambda \;=\; \dfrac{2\pi r_n}{n}.$$

Now we substitute the explicit expression for $$r_n$$ from the radius formula:

$$\lambda \;=\; \dfrac{2\pi}{n}\,\Bigl(\dfrac{n^{2}a_0}{Z}\Bigr) \;=\; 2\pi\,\dfrac{n a_0}{Z}.$$

This theoretical value of $$\lambda$$ must equal the given value. Hence we set

$$2\pi\,\dfrac{n a_0}{Z} \;=\; \dfrac{3}{2}\pi a_0.$$

We can now simplify step by step. First cancel the common factors $$\pi$$ and $$a_0$$ on both sides:

$$2\,\dfrac{n}{Z} \;=\; \dfrac{3}{2}.$$

Multiplying both sides by $$Z$$ and, simultaneously, by $$2$$ to clear the fraction, we have

$$4n \;=\; 3Z.$$

Finally, dividing both sides by $$4Z$$ gives the desired ratio:

$$\dfrac{n}{Z} \;=\; \dfrac{3}{4} \;=\; 0.75.$$

Hence, the correct answer is Option 4.