Chemistry - Atomic Structure Question with Solution | TestHub

The given wave function is $\Psi =\frac{1}{9\sqrt{8\pi }}{\left[\frac{Z}{a_0}\right]}^{\frac{3}{2}}{e}^{-\frac{\sigma }{2}}\sigma \left(4-\sigma \right)\sin \theta \cos \varphi$.

The probability density is $|\Psi {|}^2$. For the probability density to be zero at $r=3a_0$ irrespective of $\theta$ and $\varphi$, the radial part must be zero.

This means $\sigma (4-\sigma )=0$. Since $\sigma \ne 0$ (as $r\ne 0$), we must have $4-\sigma =0$, so $\sigma =4$.

We are given $\sigma =\frac{2Zr}{n{a}_0}$.

Substituting $\sigma =4$ and $r=3a_0$:

$4=\frac{2Z(3a_0)}{n{a}_0}=\frac{6Z}{n}$

So, $4n=6Z$, or $2n=3Z$.

From the angular part of the wave function, $\sin \theta \cos \varphi$, we can identify the orbital.

The general angular part for p-orbitals is proportional to $\sin \theta \cos \varphi$ (for $p_x$) or $\sin \theta \sin \varphi$ (for $p_y$) or $\cos \theta$ (for $p_z$).

Thus, the orbital is a p-orbital, which means the azimuthal quantum number $x=\ell =1$.

For a p-orbital, the principal quantum number $n$ must be at least 2.

The radial part $\sigma (4-\sigma )$ has a radial node when $4-\sigma =0$.

The number of radial nodes is $n-\ell -1$.

Here, there is one radial node (at $\sigma =4$).

So, $n-\ell -1=1$.

Substituting $\ell =1$:

$n-1-1=1⟹n-2=1⟹n=3$.

Now we have $n=3$ and $2n=3Z$.

$2(3)=3Z⟹6=3Z⟹Z=2$.

So, $x=\ell =1$ and $y=Z=2$.

The required value is $x+y=1+2=3$.