Structure of Atom JEE 2019: He+ Nodes and 4d Solution

What is the answer to the Structure of Atom JEE 2019 He+ question?

A and B only are correct in this Structure of Atom JEE 2019 question: the electron occupies a 4d state with two angular nodes and one radial node. The source is JEE Advanced 2019, Paper 2, Chemistry, and the format is multiple-correct.

The question bank rates it medium and suggests 120 seconds for an attempt. That is a suggested time, not an observed student average.

Hydrogen’s ground-state energy and the given helium-ion state are:

E_{H, g r o u n d} = - 13.6 e V

{H e}^{+} : E = - 3.4 e V, l = 2, m_{l} = 0

Decide which claims hold:

The elementary charge is defined as: $e = magnitude of the electron's charge$

How do you use the energy to find the principal quantum number?

The energy gives a principal quantum number of four. Combining that with the supplied azimuthal quantum number identifies the state as 4d, so A is correct. Start with the hydrogen-like energy formula, not with the magnetic quantum number.

For a hydrogen-like species containing one electron:

E_{n} = - 13.6 \frac{Z^{2}}{n^{2}} e V

The helium nucleus contains two protons. Its atomic number is therefore two, even though the ion has a net charge of only one positive elementary charge:

Z = 2, q_{i o n} = + e

Net ionic charge does not determine the nuclear atomic number. Using one here would treat the ion as hydrogen. Substitute the nuclear atomic number and the given energy:

- 3.4 = - \frac{13.6 \times 2^{2}}{n^{2}} = - \frac{54.4}{n^{2}}

3.4 n^{2} = 54.4

n^{2} = \frac{54.4}{3.4} = 16

The principal quantum number must be positive: $n = 4$

Now use the subshell mapping:

l = 0, 1, 2, 3 ⟷ s, p, d, f

The supplied azimuthal quantum number selects the d subshell:

n = 4, l = 2 \Rightarrow 4 d

How many angular and radial nodes does this state have?

There are two angular nodes and one radial node, making B correct and C incorrect. Calculate these separately before checking their sum. The magnetic quantum number is allowed here, but it does not replace the azimuthal quantum number in either node calculation.

The angular-node count is: $N_{a n g u l a r} = l = 2$

The radial-node count is:

N_{r a d i a l} = n - l - 1 = 4 - 2 - 1 = 1

Check both against the total:

N_{t o t a l} = n - 1 = 4 - 1 = 3

N_{r a d i a l} + N_{a n g u l a r} = 1 + 2 = 3

The supplied quantum numbers must also form an allowed combination. For the fourth shell, the allowed azimuthal values include the given value:

n = 4 : l = 0, 1, 2, 3

For the d subshell, the magnetic quantum number obeys: $- l \leq m_{l} \leq l$

l = 2 : m_{l} = - 2, - 1, 0, 1, 2

Thus, zero is allowed. The magnetic quantum number specifies the orbital angular-momentum projection; it does not change these node counts. A zero magnetic quantum number does not mean zero angular nodes.

Why is the nuclear-charge claim in option D false?

D is false because there are no other electrons to provide shielding. The helium ion contains one electron and a nucleus with two protons. In the question’s one-electron model, the electron experiences the full nuclear charge: $q_{n u c l e u s} = + 2 e$

Do not confuse nuclear charge with attraction at a particular distance. Greater electron–nucleus separation can weaken the Coulomb interaction, but it does not reduce the nuclear charge.

The complete option audit is:

A, true: The energy and azimuthal quantum number identify a 4d state.
B, true: The state has two angular nodes.
C, false: It has one radial node, not three.
D, false: No other electrons are present to shield the nuclear charge.

The final multiple-correct answer is A and B only.

Why does using total nodes lead to the wrong answer C?

One way to reach option C is to calculate the total number of nodes and then label it as the radial-node count. The arithmetic is correct, but the label is wrong. This error happens after finding the principal quantum number, not during the energy calculation.

The incorrect pathway starts with:

n = 4 \Rightarrow n - 1 = 3

Three counts all nodes, not just radial nodes. Option C asks specifically about radial nodes, so subtract the angular nodes:

N_{r a d i a l} = (n - 1) - l = 3 - 2 = 1

Use this check whenever both node types appear in the options:

N_{r a d i a l} + N_{a n g u l a r} = n - 1

Write “total”, “angular” and “radial” beside your calculations. Keep each label attached to its count when checking the options.

Can you solve three related Structure of Atom questions?

Use the same sequence: find the shell from energy, check the allowed quantum numbers, then count nodes. These are original practice variations, not verified JEE questions or additional 2019 PYQs. Each changes one part of the original setup.

What subshell and nodes follow for He+ at −13.6 eV with an azimuthal quantum number of one?

The state is 2p, with one angular node and no radial nodes. Use the helium nuclear atomic number in the energy equation:

n^{2} = \frac{54.4}{13.6} = 4, n = 2

n = 2, l = 1 \Rightarrow 2 p

N_{a n g u l a r} = 1, N_{r a d i a l} = 2 - 1 - 1 = 0

Can hydrogen have energy −3.4 eV and an azimuthal quantum number of two?

No, that combination is forbidden. Hydrogen has a different nuclear atomic number from the original helium ion:

Z = 1, n^{2} = \frac{13.6}{3.4} = 4, n = 2

The largest allowed azimuthal quantum number is: $l_{max} = n - 1 = 1$

The proposed d state therefore cannot exist at this energy. Always check that the azimuthal quantum number is allowed for the shell found from energy.

What changes if the original magnetic quantum number becomes minus two?

Replacing zero with minus two leaves the energy, subshell and node counts unchanged within the model used in the question. The replacement is allowed:

l = 2, - 2 \leq m_{l} \leq 2

m_{l} = - 2 : E = - 3.4 e V, state = 4 d

N_{a n g u l a r} = 2, N_{r a d i a l} = 1

For your next attempt, write the nuclear atomic number first and find the shell from energy. Then check the allowed quantum numbers and label each node count before selecting options.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Frequently asked questions

What is the answer to the JEE Advanced 2019 He+ question?

A and B only are correct. The electron is in a 4d state with two angular nodes and one radial node. It experiences the full nuclear charge of +2e, so the reduced-charge claim in option D is false.

How do you find n for He+ at an energy of -3.4 eV?

Use the hydrogen-like energy formula E_n = -13.6 Z²/n² eV with Z = 2 for helium. Substituting -3.4 eV gives n² = 54.4/3.4 = 16, so n = 4. The supplied l = 2 then identifies the state as 4d.

How many radial and angular nodes does a 4d orbital have?

A 4d orbital has one radial node and two angular nodes. Angular nodes equal l = 2, while radial nodes equal n - l - 1 = 4 - 2 - 1 = 1. The total is three nodes, but that is not the radial-node count.

Why does the electron in He+ experience a nuclear charge of +2e?

The helium nucleus contains two protons, giving it a charge of +2e. He+ has only one electron, so no other electrons are present to provide shielding in the question's model. Its net ionic charge of +e is not its nuclear charge.

Does m_l = 0 mean there are no angular nodes?

No; the angular-node count is determined by l, not m_l. For l = 2, the state has two angular nodes even when m_l = 0. The magnetic quantum number specifies the orbital angular-momentum projection.