NEET Structure of Atom Previous Year Questions with Solutions

A radial node is a spherical surface around the nucleus at a finite distance r > 0 where the radial wavefunction ψ_r (and consequently the radial probability density ψ_r²) passes through zero.

• • Graphically, each radial node corresponds to the wavefunction curve crossing the r-axis (zero-line) at r > 0.

1. Plot A: ψ_r has a non-zero maximum at r = 0 and decays asymptotically to zero without ever cutting the zero line. Number of radial nodes = 0 (Represents 1s orbital). 2. Plot B: ψ_r starts positive at r = 0, crosses the zero axis once, dips into a negative minimum, and approaches zero asymptotically. Number of radial nodes = 1 (Represents 2s orbital). 3. Plot C: ψ_r starts positive at r = 0, crosses the zero axis first into negative values, turns around and crosses the zero axis a second time into positive values, and then approaches zero asymptotically. Number of radial nodes = 2 (Represents 3s orbital). 4. Plot D: ψ_r starts at 0 at r = 0 (since l > 0), rises to a positive peak, crosses the zero axis once into negative, and approaches zero. Number of radial nodes = 1 (Represents 3p or 4d orbital).

• • Plot C represents 2 radial nodes.

The figure representing two radial nodes is C.

• • Matches Option (3).

According to the Rydberg formula, the reciprocal of the absorbed wavelength for hydrogen is:

• • Inverting gives: λ₁ = 36 / (5 R_H).

For the transition between n = 4 and n = 6:

• • Inverting gives: λ₂ = 144 / (5 R_H).

Divide the two wavelengths:

• • Matches Option (4) exactly.

The energy of a stationary state in a hydrogen-like species is given by Bohr’s formula:

• • For He⁺ (Z = 2): E₁(He⁺) = -2.18 × 10⁻¹⁸ × 4 = -8.72 × 10⁻¹⁸ J.
• • For Li²⁺ (Z = 3): E₁(Li²⁺) = -2.18 × 10⁻¹⁸ × 9 = -19.62 × 10⁻¹⁸ J.

The radius of a Bohr orbit is inversely proportional to the atomic number Z:

• • For He⁺ (Z = 2): r₁(He⁺) = 52.9 / 2 = 26.45 pm ≈ 26.4 pm.
• • For Li²⁺ (Z = 3): r₁(Li²⁺) = 52.9 / 3 = 17.63 pm ≈ 17.6 pm.

Compare with the multiple choice options:

• • Option (1) has the exact values and proper order of magnitude (10⁻¹⁸ J).

The principal quantum number n can be any positive integer (1, 2, 3, ...). It identifies the principal electron shell, governs the effective size of the electron cloud (radius r ∝ n²/Z), and predominantly determines the orbital energy:

For a given n, l takes integral values from 0 to (n - 1). It dictates the orbital angular momentum and defines the three-dimensional geometric shape of the orbital:

For a given l, m_l ranges from -l to +l through zero (total 2l + 1 values). It specifies the spatial orientation of the orbital with respect to an external magnetic field:

The electron spin quantum number accounts for the intrinsic angular momentum of the electron, with two allowable spin orientations: +1/2 (spin-up) and -1/2 (spin-down):

For any one-electron hydrogen-like ion of atomic number Z, the energy of the stationary electron state with principal quantum number n is given by:

For helium ion He+, Z = 2. In the ground state, n = 1:

For beryllium ion Be3+, Z = 4. For the n = 2 quantum level:

Substituting -4 RH = -x directly into the expression:

For an electron in an atom, the azimuthal (orbital angular momentum) quantum number l defines the subshell shape. For any given value of l, the magnetic orbital quantum number m_l designates the spatial orientation of the orbital and can take all integer values from -l to +l, including zero:

The total number of permissible orientations n_m in a subshell of azimuthal quantum number l is given by counting all values in this range:

The question asks for the relation expressing l in terms of n_m. Subtract 1 from both sides and divide by 2:

Check for s (l=0), p (l=1), and d (l=2) subshells:

Atoms of all elements (with the sole exception of the hydrogen isotope protium, ¹H, which lacks a neutron) are composed of three subatomic fundamental particles: electrons, protons, and neutrons. Stating that all elements are composed of "two" fundamental particles is false.

The rest mass of an electron measured to six significant figures in SI units is 9.10939 × 10⁻³¹ kg (or 9.10939 × 10⁻²⁸ g). This matches the official NCERT physical constant table verbatim.

Chemical properties of an atom are determined by the number of valence electrons and electronic configuration, which depends strictly on the atomic number Z. Because all isotopes of a given element have identical atomic numbers Z, they exhibit identical chemical properties (differing only in nuclear reactions and slight physical mass-dependent rate isotope effects).

Protons and neutrons reside within the atomic nucleus and are collectively referred to as nucleons. Electrons orbit outside the nucleus and are not nucleons.

Dalton’s atomic theory (1808) postulated that all matter is composed of extremely small, indivisible particles called atoms, regarding the atom as the ultimate indivisible building block of matter.

For any electromagnetic radiation propagating in vacuum or air, the speed of light $c$ is related to frequency $\nu$ and wavelength $\lambda$ by:

The given frequency is in kilohertz ($1\text{ kHz} = 10^3\text{ Hz} = 10^3\text{ s}^{-1}$):

Substitute $c = 3.0 \times 10^8\text{ m/s}$ and $\nu = 1.368 \times 10^6\text{ s}^{-1}$:

For any nuclide represented as $$^{A}_{Z}\text{X}$$: - $$Z$$ is the atomic number (subscript), which represents the number of protons in the nucleus. - $$A$$ is the mass number (superscript), which represents the total number of nucleons (protons + neutrons).

For $$^{175}_{71}\text{Lu}$$: - Atomic number $$Z = 71$$. - Therefore, $$\text{Number of protons} = 71$$. - Because the atom is neutral (net charge = 0), the number of electrons equals the number of protons: $$\text{Number of electrons} = 71$$

$$\text{Number of neutrons} = A - Z = 175 - 71 = 104$$

The question explicitly demands: "protons, neutrons and electrons, respectively": $$\text{Protons} = 71, \quad \text{Neutrons} = 104, \quad \text{Electrons} = 71$$ This sequence corresponds exactly to Option (1).