JEE Challenger
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ChemMedAdv 2017 • Paper 1

Correct Combination of Quantum Wavefunction Properties for Hydrogen Atom

Comprehension Passage

Answer Q.31, Q.32 and Q.33 by appropriately matching the information provided in the three columns of the table below.

The wave function, ψn,l,ml\psi_{n,l,m_l}, is a mathematical expression depending upon spherical polar coordinates (r,θ,ϕ)(r, \theta, \phi) of the electron, characterized by quantum numbers nn, ll, and mlm_l. Here rr denotes the radial distance from the nucleus, θ\theta represents colatitude, and ϕ\phi is the azimuth. In the given formulas, ZZ represents the atomic number and aoa_o is the Bohr radius.

Column 1Column 2Column 3(I) 1s orbital(i) ψn,l,ml∝(Zao)32e−(Zrao)(P) Wavefunction plot vs r/ao(II) 2s orbital(ii) One radial node(Q) Probability density at nucleus ∝1ao3(III) 2pz orbital(iii) ψn,l,ml∝(Zao)52re−(Zr2ao)cos⁡θ(R) Probability density is maximum at nucleus(IV) 3dz2 orbital(iv) xy-plane is a nodal plane(S) Energy needed to excite electron from n=2 to n=4 state is 2732 times that from n=2 to n=6\begin{array}{|l|l|l|} \hline \text{Column 1} & \text{Column 2} & \text{Column 3} \\ \hline \text{(I) 1s orbital} & (i)\ \psi_{n,l,m_l} \propto \left(\frac{Z}{a_o}\right)^{\frac{3}{2}} e^{-\left(\frac{Zr}{a_o}\right)} & \text{(P) Wavefunction plot vs } r/a_o \\ \hline \text{(II) 2s orbital} & \text{(ii) One radial node} & \text{(Q) Probability density at nucleus } \propto \frac{1}{a_o^3} \\ \hline \text{(III) } 2p_z \text{ orbital} & (iii)\ \psi_{n,l,m_l} \propto \left(\frac{Z}{a_o}\right)^{\frac{5}{2}} r e^{-\left(\frac{Zr}{2a_o}\right)} \cos\theta & \text{(R) Probability density is maximum at nucleus} \\ \hline \text{(IV) } 3d_{z^2} \text{ orbital} & \text{(iv) } xy\text{-plane is a nodal plane} & \text{(S) Energy needed to excite electron from } n = 2 \text{ to } n = 4 \text{ state is } \frac{27}{32} \text{ times that from } n = 2 \text{ to } n = 6 \\ \hline \end{array}

For a hydrogen atom, select the only CORRECT combination:

Question Diagram 1

Options

A

(I) (i) (S)

Correct
B

(II) (i) (Q)

C

(I) (i) (P)

D

(I) (iv) (R)

Topics & Concepts

Structure of AtomQuantum NumbersBohr Model

Step-by-Step Solution

To determine the correct combination for a hydrogen atom, let us analyze the statements in each column for the combination (I) (i) (S):

  1. Matching (I) with (i): The wavefunction for a 1s1s orbital (n=1,l=0,ml=0n = 1, l = 0, m_l = 0) of a hydrogen-like species is given by: ψ1s=1π(Za0)32e−(Zra0)\psi_{1s} = \frac{1}{\sqrt{\pi}} \left(\frac{Z}{a_0}\right)^{\frac{3}{2}} e^{-\left(\frac{Zr}{a_0}\right)} This directly matches statement (i): ψn,l,ml∝(Za0)32e−(Zra0)\psi_{n,l,m_l} \propto \left(\frac{Z}{a_0}\right)^{\frac{3}{2}} e^{-\left(\frac{Zr}{a_0}\right)} Thus, (I) correctly corresponds to (i).

  2. Evaluating statement (S): The excitation energy for a transition from state n1n_1 to n2n_2 in a hydrogen atom (Z=1Z = 1) is given by the formula: ΔEn1→n2=13.6(1n12−1n22) eV\Delta E_{n_1 \to n_2} = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}

    • Energy required to excite an electron from n=2n = 2 to n=4n = 4: ΔE2→4=13.6(122−142)=13.6(14−116)=13.6(316) eV\Delta E_{2 \to 4} = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \left( \frac{3}{16} \right) \text{ eV}

    • Energy required to excite an electron from n=2n = 2 to n=6n = 6: ΔE2→6=13.6(122−162)=13.6(14−136)=13.6(836)=13.6(29) eV\Delta E_{2 \to 6} = 13.6 \left( \frac{1}{2^2} - \frac{1}{6^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{36} \right) = 13.6 \left( \frac{8}{36} \right) = 13.6 \left( \frac{2}{9} \right) \text{ eV}

    • Ratio of the two excitation energies: ΔE2→4ΔE2→6=31629=3×916×2=2732\frac{\Delta E_{2 \to 4}}{\Delta E_{2 \to 6}} = \frac{\frac{3}{16}}{\frac{2}{9}} = \frac{3 \times 9}{16 \times 2} = \frac{27}{32}

    Therefore, ΔE2→4=2732ΔE2→6\Delta E_{2 \to 4} = \frac{27}{32} \Delta E_{2 \to 6}, which makes statement (S) correct.

  3. Verification of incorrectness of other options:

    • (B) (II) (i) (Q): Statement (i) corresponds to the 1s1s orbital, not the 2s2s orbital (II).
    • (C) (I) (i) (P): Plot (P) shows a radial node where the wavefunction ψ\psi changes sign. The 1s1s orbital has zero radial nodes, so it cannot be matched with (P).
    • (D) (I) (iv) (R): The 1s1s orbital (I) is spherically symmetric and does not have a nodal plane, whereas (iv) describes an orbital with xyxy-plane as a nodal plane (e.g., 2pz2p_z).

Hence, the only correct combination is (I) (i) (S).

Correct Answer: A

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