Structure of Atom — JEE Main Questions

1. What is the spin quantum number of an electron in a 2p orbital with m_l = 0?

1. 1
2. 0
3. +½
4. -½

Answer: +½

The spin quantum number m_s is an intrinsic property of the electron, independent of n, l, or m_l. It can only be +½ or -½. For any orbital, an electron can have either spin; the question does not specify which, so both are possible. However, the options include +½ and -½; since the question asks for 'the' spin quantum number, either is correct, but +½ is a valid answer.

2. In a multi-electron atom, which of the following correctly represents the order of energy of orbitals with n = 3?

1. 3d < 3p < 3s
2. 3s = 3p = 3d
3. 3s < 3p < 3d
4. 3p < 3s < 3d

Answer: 3s < 3p < 3d

Due to shielding and penetration, for the same principal quantum number n, s orbitals have the highest penetration (lowest energy), followed by p, then d. Thus in multi-electron atoms, the energy order is 3s < 3p < 3d. In hydrogen, they are degenerate, but not in multi-electron atoms.

3. UV light of wavelength 200 nm falls on a metal with work function 2.5 eV. What is the de Broglie wavelength (in nm) of the ejected photoelectron?

1. 0.089
2. 0.486
3. 0.637
4. 0.091

Answer: 0.637

First, photon energy E = hc/λ = (1240 eV·nm)/(200 nm) = 6.20 eV. KE_max = E − W = 6.20 − 2.5 = 3.70 eV = 5.93×10⁻¹⁹ J. Momentum p = √(2m KE) = √(2 × 9.109×10⁻³¹ × 5.93×10⁻¹⁹) = 1.04×10⁻²⁴ kg m/s. de Broglie wavelength λ = h/p = 6.626×10⁻³⁴ / 1.04×10⁻²⁴ = 6.37×10⁻¹⁰ m = 0.637 nm.

4. A hydrogen atom is excited from n=1 to n=5. How many distinct emission lines are possible when it de-excites?

Answer: 10.0

The number of distinct spectral lines when an electron de-excites from level n to the ground state is given by n(n−1)/2. Here n=5, so lines = 5×4/2 = 10. This counts all possible downward transitions from n=5 to lower levels (5→4, 5→3, 5→2, 5→1, 4→3, 4→2, 4→1, 3→2, 3→1, 2→1).

5. For hydrogen 1s orbital, the radial probability density P(r) = (4/a₀³) r² e^(-2r/a₀). At what distance (in a₀) is P(r) maximum?

1. 1.0
2. 0.5
3. 2.0
4. 0.0

Answer: 1.0

To find the most probable distance, differentiate P(r) with respect to r and set dP/dr = 0. Using the product rule, dP/dr = (4/a₀³)[2r e^(-2r/a₀) + r²(-2/a₀) e^(-2r/a₀)] = (8r/a₀³) e^(-2r/a₀)[1 - r/a₀] = 0. Since r ≠ 0, we get r = a₀. Thus the maximum occurs at r = 1.0 a₀.

6. What is the maximum number of electrons that can be accommodated in the M shell (n = 3) according to the Pauli exclusion principle?

Answer: 18.0

The M shell corresponds to n = 3. The maximum number of electrons in a shell is given by 2n² = 2(3²) = 18. Alternatively, subshells: 3s (2), 3p (6), 3d (10) sum to 18. Pauli's principle allows each orbital to hold at most two electrons with opposite spins.

7. According to de Broglie, the condition for a stable Bohr orbit is that the circumference equals an integer multiple of the electron's de Broglie wavelength. Which expression correctly represents this condition?

1. 2πr = nλ
2. πr = nλ
3. 2πr = nλ/2
4. πr = nλ/2

Answer: 2πr = nλ

For a stable orbit, the electron wave must be a standing wave around the circle. This requires the circumference 2πr to be an integer multiple of the wavelength: 2πr = nλ. Substituting λ = h/mv gives mvr = nh/2π, which is Bohr's quantisation condition.

8. The electronic configuration of Fe (Z=26) is [Ar] 3d⁶ 4s². What is the correct configuration of Fe³⁺?

1. [Ar] 3d⁷
2. [Ar] 3d⁶ 4s¹
3. [Ar] 3d⁴ 4s²
4. [Ar] 3d⁵

Answer: [Ar] 3d⁵

When forming cations, electrons are removed from the orbital with highest principal quantum number first. For Fe, 4s (n=4) is higher than 3d (n=3). So remove two 4s electrons to get Fe²⁺ ([Ar] 3d⁶), then remove one 3d electron to get Fe³⁺ ([Ar] 3d⁵).

9. How many unpaired electrons are present in the ground state of a sulfur atom (Z=16)?

Answer: 2.0

Sulfur has electronic configuration [Ne] 3s² 3p⁴. According to Hund's rule, the 3p⁴ configuration has two unpaired electrons: three p orbitals each get one electron first, then the fourth electron pairs in one orbital, leaving two unpaired.

10. According to classical electrodynamics, why would an electron in Rutherford's model spiral into the nucleus?

1. Because the electron is attracted to the nucleus by electrostatic force
2. Because the electron gains mass as it moves faster
3. Because the electron loses energy by radiating electromagnetic waves
4. Because the electron collides with other electrons

Answer: Because the electron loses energy by radiating electromagnetic waves

According to Maxwell's theory, an accelerated charged particle emits electromagnetic radiation. The electron in a circular orbit is accelerating (centripetal acceleration), so it continuously loses energy and spirals into the nucleus.

11. Which of the following is an experimental observation of the photoelectric effect?

1. Electrons are emitted for any light intensity regardless of frequency.
2. Electrons are emitted only if the light frequency is above a threshold.
3. The kinetic energy of ejected electrons depends on light intensity.
4. Emission of electrons occurs after a time delay of a few seconds.

Answer: Electrons are emitted only if the light frequency is above a threshold.

The photoelectric effect shows that electrons are emitted only when the incident light frequency exceeds a threshold frequency ν₀, regardless of intensity. This is a key experimental observation explained by Einstein's quantum theory.

12. Which of the following is NOT a limitation of Bohr's model of the atom?

1. It fails to explain the spectrum of multi-electron atoms.
2. It cannot explain the fine structure of spectral lines.
3. It cannot explain the hydrogen spectrum.
4. It cannot explain the Zeeman effect.

Answer: It cannot explain the hydrogen spectrum.

Bohr's model successfully explains the hydrogen spectrum, so that is not a limitation. Its limitations include failure for multi-electron atoms, inability to explain fine structure, and inability to explain Zeeman and Stark effects.