Decoding the Quantum State
Imagine you are a quantum detective, and you've just been handed a dossier on a mysterious electronic state, Ψ.
This state belongs to a helium ion, specifically He+.
Why is this important? Because He+ has only one electron, making it a hydrogen-like species.
This means we can use the elegant and predictable mathematics of the Bohr model and the quantum mechanical model to uncover its secrets.
The dossier gives us three crucial clues: the energy is −3.4 eV, the azimuthal quantum number l is 2, and the magnetic quantum number m is 0.
The Master Equation
Energy of an Orbit
To find out exactly where this electron lives, we need to determine its principal quantum number, n.
For any hydrogen-like species, the energy of an electron in the n-th orbit is governed by a beautiful master equation.
Here, Z represents the atomic number.
Since we are dealing with helium, the nucleus has two protons, so Z=2.
The negative sign is a reminder that the electron is bound to the nucleus; it's trapped in the electrostatic potential well.
Finding the Principal Quantum Number
Now, let's plug our known values into the master equation.
We know the energy E is −3.4 eV, and Z is 2.
Let's simplify the numerator. Two squared is four.
Now, we rearrange the equation to isolate n2.
If you look closely, the negative signs cancel out.
Dividing 13.6 by 3.4 gives exactly 4.
Multiplying that by the 4 from our Z2 term gives us 16.
Taking the square root, we find our principal quantum number.
The electron resides in the fourth principal shell!
Unveiling the Orbital Identity
We have n=4, but what about the shape of the orbital?
This is where the azimuthal quantum number, l, comes into play.
The problem states that l=2.
Recall the standard spectroscopic notation: l=0 is an s-orbital, l=1 is a p-orbital, and l=2 corresponds to a d-orbital.
Combining the principal shell and the subshell, we can definitively say that the state Ψ is a 4d state.
This makes statement (C) absolutely correct.
The Geometry of Probability
Nodes
Next, let's investigate the nodes of this orbital.
A node is a region in space where the probability of finding the electron is exactly zero.
There are two types of nodes: angular and radial.
The number of angular nodes is simply equal to the azimuthal quantum number, l.
Since l=2, there are exactly 2 angular nodes.
This confirms that statement (A) is correct.
What about the radial nodes?
The formula for the number of radial nodes is n−l−1.
Let's substitute our values.
There is only 1 radial node in a 4d orbital.
Statement (B) claims there are 3 radial nodes, which is a trap! So, statement (B) is incorrect.
The Pull of the Nucleus
Effective Nuclear Charge
Finally, let's evaluate the nuclear charge experienced by the electron.
In multi-electron atoms, inner electrons repel outer electrons, shielding them from the full attractive force of the nucleus.
However, He+ is a single-electron species.
There are no other electrons to get in the way!
The shielding constant, σ, is exactly zero.
Therefore, the effective nuclear charge Zeff is simply the actual nuclear charge.
The electron experiences the full, unadulterated pull of the two protons, which is exactly 2e.
Statement (D) suggests the charge experienced is less than 2e, which is false.
Final Verdict
By systematically decoding the quantum numbers and applying the principles of atomic structure, we have exposed the truth.
The state has 2 angular nodes and is indeed a 4d state.
Therefore, the only true statements are (A) and (C).
