Atomic Structure Questions and Answers

Q: Assertion (S): The total angular momentum for light atoms (low atomic number) is obtained by Russell-Saunders coupling, whereas jj-coupling is used for heavy atoms (high atomic number).Reasoning (R): The spin-orbit interactions are weak in light atoms (low atomic number) and strong in heavy atoms (high atomic number).The correct option is

The correct option is S and R are true; and R is the correct reason for S.In light atoms the electrostatic interactions between electrons are much stronger than spin-orbit coupling, so the individual orbital momenta first combine to a total L and the spins to a total S, which then couple weakly to give J: Russell-Saunders (LS) coupling. Spin-orbit coupling grows steeply with nuclear charge (roughly as Z4 for a given orbital), so in heavy atoms each electron's own l and s couple first to a j, and the j values then combine: jj-coupling.The strength of the spin-orbit interaction is precisely what decides which scheme applies, so R both is true and explains S, ruling out the other three options.

Q: The correct statement(s) about 4D5/2 state of an atom is (are):

The correct statement is that it can show spectral transition to the 4P3/2 state.In the term symbol 4D5/2, the superscript 4 = 2S + 1 gives S = 3/2, D gives L = 2, and J = 5/2. A transition to 4P3/2 has \(\Delta S = 0\), \(\Delta L = -1\) and \(\Delta J = -1\), all allowed by the selection rules.The other statements fail. S is 3/2, not 1/2. The s1p2 configuration gives the terms 4P, 2P, 2D and 2S, with no quartet D. And a level with J = 5/2 splits into 2J + 1 = 6 levels in a magnetic field, not five. The paper allows several answers; the key gives this one only.

Q: The Vividh Bharati station of All India Radio, Kozhikode, broadcasts on a frequency of 1500 kHz. What is the wavelength of the electromagnetic radiation emitted by transmitter?( c = 3 × 108ms-1 )

The wavelength is 200 m.\[ \begin{aligned} \lambda &= \frac{c}{\nu} = \frac{3 \times 10^{8}}{1500 \times 10^{3}} \\ &= \frac{3 \times 10^{8}}{1.5 \times 10^{6}} = 200\ \text{m} \end{aligned} \]The common slip is in the prefix: kHz is \(10^{3}\) Hz. 300 m would need a frequency of 1000 kHz, and 100 m one of 3000 kHz.

Q: Which of the following experimental phenomenon is explained by the wave nature of electromagnetic radiation?

The phenomenon is Diffraction.Diffraction, the bending of radiation round obstacles and the spreading through narrow openings, along with interference, is explained by the wave nature of electromagnetic radiation.The other four are the classic phenomena the wave theory could not explain and which needed quantisation: black-body radiation (Planck), the photoelectric effect (Einstein's photons), the fall of heat capacity of solids at low temperature, and the line spectra of atoms such as hydrogen (Bohr).

Q: Which of the following statements are correct about canal rays?(i) They carry positively charged particles.(ii) The mass of the particles of these rays does not depend upon the gas present in the cathode ray tube.(iii) The particles behave in a different manner in electric field as those of cathode rays.(iv) The charge to mass ratio of the particles does not depends on the gas present in the cathode ray tube.

The correct statements are (i) and (iii).Canal rays (anode rays) are streams of positive ions formed from the gas in the discharge tube, so (i) is right, and being positively charged they are deflected in the opposite direction to cathode rays in an electric field, so (iii) is right.(ii) and (iv) are wrong: the particles are ions of whatever gas fills the tube, so both their mass and their charge-to-mass ratio depend on that gas (hydrogen gives the lightest, the proton). This is the key difference from cathode rays, whose e/m is the same for every gas.

Q: In which of the following spectral region Balmer series lines are observed for atomic hydrogen?

Balmer lines lie in the visible region.The Balmer series is produced when the electron in a hydrogen atom falls from a higher level (n = 3, 4, 5 ...) to n = 2. These transitions release photons of wavelength roughly 365 nm to 656 nm, which is visible light; the red H-alpha line at 656 nm is the best known.Transitions ending on n = 1 (the Lyman series) carry more energy and lie in the ultraviolet. Those ending on n = 3, 4 and 5 (Paschen, Brackett, Pfund) lie in the infrared. No hydrogen series of this kind falls in the microwave or radiowave regions.

Q: A 150 watt bulb emits monochromatic light of wavelength 662 nm. How many number of photons emitted per second by the bulb?(Planck's constant, h = \(6.62 \times 10^{-34}\) J s, c = \(3 \times 10^{8}\) m \(s^{-1}\))

The bulb emits \(5 \times 10^{20}\) photons per second.Energy of one photon:\[ \begin{aligned} E &= \frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{662 \times 10^{-9}} \\ &= 3 \times 10^{-19}\ \text{J} \\ n &= \frac{150}{3 \times 10^{-19}} = 5 \times 10^{20}\ \text{s}^{-1} \end{aligned} \]The numbers are chosen so that 6.62 and 662 cancel neatly. The other options come from slips in the photon energy.

Q: The orbital described by the quantum numbers n=4 and \(l\) = 2 is

The orbital is 4d.The principal quantum number gives the shell (4), and \(l = 2\) is a d subshell (\(l = 0, 1, 2, 3\) are s, p, d, f). So the orbital is 4d.4s has \(l = 0\) and 4f has \(l = 3\). 3d has \(n = 3\). 2d does not exist, because \(l\) must be less than \(n\).A general rule worth carrying: for a shell \(n\), the allowed \(l\) values run from 0 to \(n - 1\), so the first d subshell is 3d and the first f subshell is 4f.

Q: A sub-atomic particle of mass 6.63 × 10-31 kg is moving with a velocity of 1 × 106 ms-1. What is the de Broglie wave length (in nm) associated with it (h = 6.63 × 10-34 Js)?

The de Broglie wavelength is 1.0 nm.\[ \begin{aligned} \lambda &= \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{6.63 \times 10^{-31} \times 10^{6}} \\ &= 10^{-9}\ \text{m} \end{aligned} \]Since 1 nm = \(10^{-9}\) m, the wavelength is 1.0 nm. The other options are off by powers of ten (a slip in the exponents) or by a factor of 2 or 5.

Q: For hydrogen atom, the orbitals with the lowest energy among the given orbitals are(i) 4s (ii) 2px (iii) 3dz2 (iv) 2py

The lowest-energy orbitals are (ii) & (iv), the 2px and 2py orbitals.In the hydrogen atom, with a single electron, the energy of an orbital depends only on the principal quantum number n: \(E_{n} = -\frac{13.6}{n^{2}}\) eV. Orbitals with the same n are degenerate.Here 2px and 2py have n = 2, while 3d (n = 3) and 4s (n = 4) are higher. The (n + l) rule that places 4s below 3d applies to many-electron atoms, not to hydrogen. Naming only (ii) misses that 2py has exactly the same energy.

Assertion (S): The total angular momentum for light atoms (low atomic number) is obtained by Russell-Saunders coupling, whereas jj-coupling is used for heavy atoms (high atomic number).

Reasoning (R): The spin-orbit interactions are weak in light atoms (low atomic number) and strong in heavy atoms (high atomic number).

The correct option is

A.S and R are true; and R is the correct reason for S
B.S and R are true; but R is NOT the correct reason for S
C.S is true but R is false
D.S is false but R is true

Exam Year(s)

GATE Chemistry - 2023,

Answer: A

The correct option is S and R are true; and R is the correct reason for S.

In light atoms the electrostatic interactions between electrons are much stronger than spin-orbit coupling, so the individual orbital momenta first combine to a total L and the spins to a total S, which then couple weakly to give J: Russell-Saunders (LS) coupling. Spin-orbit coupling grows steeply with nuclear charge (roughly as Z⁴ for a given orbital), so in heavy atoms each electron's own l and s couple first to a j, and the j values then combine: jj-coupling.

The strength of the spin-orbit interaction is precisely what decides which scheme applies, so R both is true and explains S, ruling out the other three options.

Discuss in Forum

The correct statement(s) about ⁴D_5/2 state of an atom is (are):

A.it corresponds to \(L = 2, S = 1/2\), and \(J = 5/2\).
B.it can originate from s\(^{1}\)p\(^{2}\) electronic configuration.
C.it splits into five levels in the presence of magnetic field.
D.it can show spectral transition to \(^{4}P_{3/2}\) state.

Exam Year(s)

GATE Chemistry - 2024,

Answer: D

The correct statement is that it can show spectral transition to the ⁴P_3/2 state.

In the term symbol ⁴D_5/2, the superscript 4 = 2S + 1 gives S = 3/2, D gives L = 2, and J = 5/2. A transition to ⁴P_3/2 has \(\Delta S = 0\), \(\Delta L = -1\) and \(\Delta J = -1\), all allowed by the selection rules.

The other statements fail. S is 3/2, not 1/2. The s¹p² configuration gives the terms ⁴P, ²P, ²D and ²S, with no quartet D. And a level with J = 5/2 splits into 2J + 1 = 6 levels in a magnetic field, not five. The paper allows several answers; the key gives this one only.

Discuss in Forum

The Vividh Bharati station of All India Radio, Kozhikode, broadcasts on a frequency of 1500 kHz. What is the wavelength of the electromagnetic radiation emitted by transmitter?( c = 3 × 10⁸ms^-1 )

A.200 m
B.300 m
C.100 m
D.250 m
E.150 m

Exam Year(s)

KEAM - 2025,

Answer: A

The wavelength is 200 m.

\[ \begin{aligned} \lambda &= \frac{c}{\nu} = \frac{3 \times 10^{8}}{1500 \times 10^{3}} \\ &= \frac{3 \times 10^{8}}{1.5 \times 10^{6}} = 200\ \text{m} \end{aligned} \]

The common slip is in the prefix: kHz is \(10^{3}\) Hz. 300 m would need a frequency of 1000 kHz, and 100 m one of 3000 kHz.

Discuss in Forum

Which of the following experimental phenomenon is explained by the wave nature of electromagnetic radiation?

A.Black-body radiation
B.Photoelectric effect
C.Diffraction
D.Variation of heat capacity of solids as a function of temperature
E.Line spectra of atoms with reference to hydrogen

Exam Year(s)

KEAM - 2025,

Answer: C

The phenomenon is Diffraction.

Diffraction, the bending of radiation round obstacles and the spreading through narrow openings, along with interference, is explained by the wave nature of electromagnetic radiation.

The other four are the classic phenomena the wave theory could not explain and which needed quantisation: black-body radiation (Planck), the photoelectric effect (Einstein's photons), the fall of heat capacity of solids at low temperature, and the line spectra of atoms such as hydrogen (Bohr).

Discuss in Forum

Which of the following statements are correct about canal rays?

(i) They carry positively charged particles.
(ii) The mass of the particles of these rays does not depend upon the gas present in the cathode ray tube.
(iii) The particles behave in a different manner in electric field as those of cathode rays.
(iv) The charge to mass ratio of the particles does not depends on the gas present in the cathode ray tube.

A.(ii) and (iv)
B.(i) and (iv)
C.(i) and (ii)
D.(ii) and (iii)
E.(i) and (iii)

Exam Year(s)

KEAM - 2025,

Answer: E

The correct statements are (i) and (iii).

Canal rays (anode rays) are streams of positive ions formed from the gas in the discharge tube, so (i) is right, and being positively charged they are deflected in the opposite direction to cathode rays in an electric field, so (iii) is right.

(ii) and (iv) are wrong: the particles are ions of whatever gas fills the tube, so both their mass and their charge-to-mass ratio depend on that gas (hydrogen gives the lightest, the proton). This is the key difference from cathode rays, whose e/m is the same for every gas.

Discuss in Forum

In which of the following spectral region Balmer series lines are observed for atomic hydrogen?

A.Visible
B.Ultraviolet
C.Microwave
D.Infrared
E.Radiowave

Exam Year(s)

KEAM - 2025,

Answer: A

Balmer lines lie in the visible region.

The Balmer series is produced when the electron in a hydrogen atom falls from a higher level (n = 3, 4, 5 ...) to n = 2. These transitions release photons of wavelength roughly 365 nm to 656 nm, which is visible light; the red H-alpha line at 656 nm is the best known.

Transitions ending on n = 1 (the Lyman series) carry more energy and lie in the ultraviolet. Those ending on n = 3, 4 and 5 (Paschen, Brackett, Pfund) lie in the infrared. No hydrogen series of this kind falls in the microwave or radiowave regions.

Discuss in Forum

A 150 watt bulb emits monochromatic light of wavelength 662 nm. How many number of photons emitted per second by the bulb?
(Planck's constant, h = \(6.62 \times 10^{-34}\) J s, c = \(3 \times 10^{8}\) m \(s^{-1}\))

A.\(4 \times 10^{20}s^{-1}\)
B.\(3 \times 10^{20}s^{-1}\)
C.\(6 \times 10^{20}s^{-1}\)
D.\(5 \times 10^{20}s^{-1}\)
E.\(1 \times 10^{20}s^{-1}\)

Exam Year(s)

KEAM - 2025,

Answer: D

The bulb emits \(5 \times 10^{20}\) photons per second.

Energy of one photon:

\[ \begin{aligned} E &= \frac{hc}{\lambda} = \frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{662 \times 10^{-9}} \\ &= 3 \times 10^{-19}\ \text{J} \\ n &= \frac{150}{3 \times 10^{-19}} = 5 \times 10^{20}\ \text{s}^{-1} \end{aligned} \]

The numbers are chosen so that 6.62 and 662 cancel neatly. The other options come from slips in the photon energy.

Discuss in Forum

The orbital described by the quantum numbers n=4 and \(l\) = 2 is

A.4s
B.3d
C.2d
D.4f
E.4d

Exam Year(s)

KEAM - 2025,

Answer: E

The orbital is 4d.

The principal quantum number gives the shell (4), and \(l = 2\) is a d subshell (\(l = 0, 1, 2, 3\) are s, p, d, f). So the orbital is 4d.

4s has \(l = 0\) and 4f has \(l = 3\). 3d has \(n = 3\). 2d does not exist, because \(l\) must be less than \(n\).

A general rule worth carrying: for a shell \(n\), the allowed \(l\) values run from 0 to \(n - 1\), so the first d subshell is 3d and the first f subshell is 4f.

Discuss in Forum

A sub-atomic particle of mass 6.63 × 10^-31 kg is moving with a velocity of 1 × 10⁶ ms^-1. What is the de Broglie wave length (in nm) associated with it (h = 6.63 × 10^-34 Js)?

A.10.0
B.1.0
C.0.10
D.5.0
E.0.50

Exam Year(s)

KEAM - 2025,

Answer: B

The de Broglie wavelength is 1.0 nm.

\[ \begin{aligned} \lambda &= \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{6.63 \times 10^{-31} \times 10^{6}} \\ &= 10^{-9}\ \text{m} \end{aligned} \]

Since 1 nm = \(10^{-9}\) m, the wavelength is 1.0 nm. The other options are off by powers of ten (a slip in the exponents) or by a factor of 2 or 5.

Discuss in Forum

10.

For hydrogen atom, the orbitals with the lowest energy among the given orbitals are
(i) 4s (ii) 2p_x (iii) 3d_z2 (iv) 2p_y

A.(i) & (iii)
B.(ii) & (iv)
C.(ii) & (iii)
D.(ii) only
E.(i) only

Exam Year(s)

KEAM - 2025,

Answer: B

The lowest-energy orbitals are (ii) & (iv), the 2p_x and 2p_y orbitals.

In the hydrogen atom, with a single electron, the energy of an orbital depends only on the principal quantum number n: \(E_{n} = -\frac{13.6}{n^{2}}\) eV. Orbitals with the same n are degenerate.

Here 2p_x and 2p_y have n = 2, while 3d (n = 3) and 4s (n = 4) are higher. The (n + l) rule that places 4s below 3d applies to many-electron atoms, not to hydrogen. Naming only (ii) misses that 2p_y has exactly the same energy.

Atomic Structure Questions and Answers

Topics

Other Subjects

Tags: Atomic Structure Questions and Answers

About Us

Atomic Structure Questions and Answers

Topics

Other Subjects

Tags: Atomic Structure Questions and Answers

About Us

Stay informed about all important Dates & Notifications