NEET Atoms Previous Year Questions with Solutions

The problem defines that the magnetic flux passing through the circular orbit of the electron is quantized as Φ = n(h/e). For the ground state (lowest energy state, n = 1):

• • A = πr² is the orbital area enclosed by the electron path.

Under the magnetic Lorentz force F = e v B = m v² / r, the angular frequency of the electron is the cyclotron frequency ω = eB / m. The equivalent steady electric current loop formed by the circulating electron is:

• • T is the cyclotron orbital period.

The magnetic dipole moment μ of a planar current loop is the product of current and enclosed area:

• • Magnetic field B cancels cleanly from numerator and denominator.
• • Notice this fundamental result is directly proportional to the Bohr magneton μ_B = eh / (4πm) with factor of 2.

The particle of mass m moves in a circular orbit of radius r under the action of a constant inward radial force F. Equating centripetal force to F:

• • Because F and m are constant, orbital speed is proportional to the square root of radius: v ∝ r^(1/2).

According to Bohr’s postulate, the orbital angular momentum is an integral multiple of h / 2π:

• • Substitute v ∝ r^(1/2) into the angular momentum equation:
• • m (C r^(1/2)) r = n ħ ⟹ r^(3/2) ∝ n.

Solve for r and v as functions of the principal quantum number n:

• • Now substitute r ∝ n^(2/3) back into the speed relation:
• • v ∝ r^(1/2) ∝ (n^(2/3))^(1/2) = n^(1/3).
• • Thus: r ∝ n^(2/3) and v ∝ n^(1/3), corresponding to Option (3).

For transitions in the hydrogen atom ending at n1 = 2 (the Balmer series in the visible spectrum), the wavelength λ is given by the Rydberg formula:

The transition from n2 = 3 to n1 = 2 has the smallest energy difference:

The transition from n2 = 4 to n1 = 2 has intermediate energy:

The transition from n2 = 5 to n1 = 2:

The highest energy transition in the set from n2 = 6 to n1 = 2 produces the shortest wavelength:

An isolated neutral atom of atomic number Z contains exactly Z positively charged protons in its nucleus (charge +Ze) and Z negatively charged extranuclear electrons (charge -Ze). The algebraic sum of charges is zero:

In the opening section of NCERT Physics, the textbook introduces atomic models by noting that ordinary matter is stable and that atoms of each chemical element emit a unique, characteristic line emission spectrum when energized. Under this textbook definition, Statement II is treated as correct, leading to Option (3): "Both Statement I and Statement II are correct".

Following the examination, subject matter experts and candidates formally challenged Statement II on two solid physical grounds: 1. Not all atoms are stable: elements with Z > 82 (e.g., Uranium, Polonium, Radium) as well as radioactive isotopes of lighter elements (e.g., Carbon-14, Tritium) undergo spontaneous nuclear decay and are inherently unstable. 2. Atoms in their ground state are stable but do NOT emit radiation; emission occurs only when atoms transition downward from excited states. Under this scientific rigor, Statement II is incorrect, leading to Option (1): "Statement I is correct but Statement II is incorrect".

After expert committee review, the National Testing Agency (NTA) officially upheld both perspectives and awarded full credit for BOTH Option (1) and Option (3) in the NEET 2024 Final Answer Key.

The wavelength λ of radiation emitted during an electronic transition from state n₂ to state n₁ in hydrogen is given by the Rydberg formula:

The shortest wavelength corresponds to the maximum energy transition, which occurs from n₂ = ∞ to the ground level of that series (n₁):

For the Balmer series, n₁ = 2. For the Brackett series, n₁ = 4:

Substitute 1/R into the expression for Brackett series limit:

According to Bohr’s model of the hydrogen atom, by equating the electrostatic Coulomb attraction to the centripetal force and applying angular momentum quantization L = mvr = nh / 2π, the radius of the n-th stationary orbit is derived as:

For hydrogen (Z = 1), the radius of the innermost (ground state, n = 1) orbit is given as:

Setting principal quantum number n = 3:

Multiplying the values:

Niels Bohr formulated his quantized atomic model based on Coulomb's inverse-square law between the nuclear positive charge ($$+Ze$$) and a single orbiting electron ($$-e$$): $$\frac{m v^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{Z e^2}{r^2}$$ Because it considers only two-body central force interactions, it cannot account for multi-electron repulsions.

- Hydrogen atom ($$H$$): Atomic number $$Z = 1$$, contains 1 electron. - Singly ionized helium ($$He^+$$): Helium has $$Z = 2$$. After removing one electron, $$He^+$$ retains $$2 - 1 = 1$$ electron. - Deuteron atom ($$^2_1H$$): Isotope of hydrogen with 1 proton, 1 neutron, and 1 orbiting electron. - Singly ionized neon ($$Ne^+$$): Neon has atomic number $$Z = 10$$. Removal of one electron leaves $$10 - 1 = 9$$ electrons.

Because $$Ne^+$$ is a 9-electron system, electron-electron repulsions and screening effects dominate, rendering the simple hydrogenic Bohr quantization condition invalid.

Therefore, the Bohr model is not valid for singly ionized neon ($$Ne^+$$).