68 free MCQs on Atoms with worked answers and explanations. Thomson's plum-pudding model, Rutherford's nuclear model from alpha-scattering, Bohr's postulates and hydrogen spectrum, spectral series (Lyman to Pfund), Rydberg formula, and limitations of each model.
Take the timed Atoms chapterwise test →Below are 68 practice questions on Atoms, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Atoms notes.
The Geiger–Marsden alpha-scattering result: a few alpha particles bounce back at large angles, which rules out Thomson’s diffuse model and demands a tiny, massive, positively charged nucleus. Image: Kurzon, CC BY 3.0, via Wikimedia Commons.
Q1.
Energy of hydrogen electron in nth orbit:
Answer: C. -13.6/n<sup>2</sup> eV
Why: En = -13.6/n<sup>2</sup> eV. Ground state (n=1): -13.6 eV. Negative means the electron is bound to the atom.
Q2.
Visible light emission from hydrogen atom (Balmer series) involves transitions to:
Answer: B. n=2
Why: Balmer series: electrons fall to n=2. These transitions produce visible light (H-alpha, H-beta, etc.).
Q3.
Rutherford's gold foil experiment showed that:
Answer: A. Atoms are mostly empty with a tiny dense nucleus
Why: Most alpha particles passed through; a few deflected greatly. This proved the atom is mostly empty space with a tiny, dense, positive nucleus.
Q4.
Heisenberg's uncertainty principle states that:
Answer: B. Position and momentum cannot both be known precisely
Why: delta_x x delta_p >= h/(4pi). Position and momentum cannot simultaneously be known with arbitrary precision.
Q5.
Ionization energy of hydrogen atom is:
Answer: A. 13.6 eV
Why: Ionization energy = energy needed to remove electron from ground state = 13.6 eV (positive, as it takes this much energy).
Q6.
Lyman series of hydrogen spectrum lies in the:
Answer: C. Ultraviolet region
Why: Lyman series: transitions to n=1. High energy, short wavelength: ultraviolet region.
Q7.
X-rays are produced when:
Answer: B. Fast electrons suddenly decelerate hitting a metal target
Why: X-rays (Bremsstrahlung) are produced when fast electrons abruptly decelerate in a metal target, converting KE to X-ray photons.
Q8.
The tiny, dense, positively charged core of an atom is the:
Answer: A. nucleus
Why: Almost all the mass and the positive charge of an atom are concentrated in the nucleus.
Q9.
In an atom, electrons revolve around the nucleus in:
Answer: A. orbits
Why: Electrons occupy definite orbits (energy levels) around the nucleus.
Q10.
The model in which electrons move in fixed circular orbits was proposed by:
Answer: A. Niels Bohr
Why: Bohr proposed the model of quantised electron orbits in 1913.
Q11.
Besides protons, the nucleus contains:
Answer: A. neutrons
Why: The nucleus is made of protons and neutrons (collectively nucleons).
Q12.
The atomic number of an element equals its number of:
Answer: A. protons
Why: Atomic number Z equals the number of protons in the nucleus.
Q13.
Rutherford’s alpha-particle scattering experiment led to the discovery of the:
Answer: A. nucleus
Why: The large-angle scattering revealed a tiny, dense, positive nucleus.
Q14.
Most of the volume of an atom is:
Answer: A. empty space
Why: The nucleus is minute compared with the atom, so an atom is mostly empty space.
Q15.
The energy of an electron in an atom is:
Answer: A. quantised
Why: Electrons can only have certain discrete (quantised) energy values.
Q16.
When an electron jumps to a lower orbit, it ___ energy:
Answer: A. emits
Why: A downward jump releases energy as a photon.
Q17.
When an electron jumps to a higher orbit, it ___ energy:
Answer: A. absorbs
Why: Moving to a higher level requires the electron to absorb energy.
Q18.
The lowest energy state of an atom is called the ___ state:
Answer: A. ground
Why: The ground state is the most stable, lowest-energy configuration.
Q19.
The emission spectrum of hydrogen consists of discrete:
Answer: A. lines
Why: Hydrogen emits light only at specific wavelengths, giving a line spectrum.
Q20.
The alpha particle used in Rutherford’s experiment is a ___ nucleus:
Answer: A. helium
Why: An alpha particle is a helium nucleus (2 protons + 2 neutrons).
Q21.
Photon energy for the transition n=3 to n=1 in hydrogen:
Answer: A. 12.09 eV
Why: E = 13.6(1/1² - 1/3²) = 13.6(1 - 1/9) = 13.6 × 8/9 = 12.09 eV. This is the Lyman-alpha series.
Q22.
The energy levels of hydrogen-like atoms (atomic number Z):
Answer: B. En = -13.6 Z²/n² eV
Why: For hydrogen-like ions: En = -13.6 Z²/n² eV. For helium ion (Z=2): E<sub>1</sub> = -54.4 eV.
Q23.
The series of hydrogen spectrum in ultraviolet region is:
Answer: C. Lyman
Why: Lyman series: transitions to n=1. These photons are in the ultraviolet range. Balmer series (visible) transitions to n=2.
Q24.
Wave nature of electrons was confirmed by:
Answer: B. Davisson-Germer electron diffraction experiment
Why: Davisson-Germer (1927): electrons diffracted by nickel crystal, confirming de Broglie wave hypothesis.
Q25.
The velocity of electron in nth orbit of hydrogen (v₁ = 2.18 × 10⁶ m/s):
Answer: A. v₁/n
Why: vn = v₁/n. Speed decreases as orbit number increases. Ground state has highest speed.
Q26.
Electron in orbit has:
Answer: B. Angular momentum = nh/2π
Why: Bohr postulate: angular momentum L = nh/(2π) = n × hbar. Quantized in integer multiples of hbar.
Q27.
In Bohr’s model, the angular momentum of an orbiting electron is quantised in units of:
Answer: A. h/2π
Why: Bohr’s condition: angular momentum = nh/2π.
Q28.
The radius of the nth Bohr orbit is proportional to:
Answer: A. n²
Why: r<sub>n</sub> ∝ n², so higher orbits are much larger.
Q29.
Using E<sub>n</sub> = −13.6/n² eV, the ground-state (n = 1) energy of hydrogen is:
Answer: A. −13.6 eV
Why: E<sub>1</sub> = −13.6/1² = −13.6 eV.
Q30.
The energy of the n = 2 level of hydrogen is:
Answer: A. −3.4 eV
Why: E<sub>2</sub> = −13.6/2² = −3.4 eV.
Q31.
As n increases, the energy levels of hydrogen become:
Answer: A. closer together
Why: The levels crowd together and converge toward 0 eV as n grows.
Q32.
The Balmer series of the hydrogen spectrum lies in the ___ region:
Answer: A. visible
Why: Balmer lines (transitions to n = 2) fall in the visible region.
Q33.
The Lyman series of the hydrogen spectrum lies in the ___ region:
Answer: A. ultraviolet
Why: Lyman lines (transitions to n = 1) fall in the ultraviolet.
Q34.
The Paschen series of the hydrogen spectrum lies in the ___ region:
Answer: A. infrared
Why: Paschen lines (transitions to n = 3) fall in the infrared.
Q35.
The energy needed to remove the electron from the ground state of hydrogen (its ionisation energy) is:
Answer: A. 13.6 eV
Why: Raising the electron from −13.6 eV to 0 eV needs 13.6 eV.
Q36.
The radius of the first Bohr orbit of hydrogen is about:
Answer: A. 0.53 Å
Why: The Bohr radius is approximately 0.53 ångström (5.3 × 10⁻¹¹ m).
Q37.
According to Bohr, an electron in a stationary orbit does not ___ energy:
Answer: A. radiate
Why: In a stationary state the electron does not radiate, despite accelerating - a Bohr postulate.
Q38.
A spectral line’s wavelength relates to the energy difference by ΔE equals:
Answer: A. hc/λ
Why: ΔE = hν = hc/λ.
Q39.
In hydrogen, the transition from n = 3 to n = 2 produces a line in the ___ series:
Answer: A. Balmer
Why: Any transition ending at n = 2 belongs to the Balmer series.
Q40.
The number of neutrons in a nucleus equals the mass number minus the:
Answer: A. atomic number
Why: Number of neutrons = mass number A − atomic number Z.
Q41.
De Broglie wavelength of a thermal neutron at temperature T:
Answer: C. h/sqrt(3mkT)
Why: Thermal energy = (3/2)kT = p²/2m. p = sqrt(3mkT). lambda = h/sqrt(3mkT).
Q42.
Rydberg formula for hydrogen: 1/lambda = R(1/n<sub>1</sub>² - 1/n<sub>2</sub>²). Rydberg constant R =
Answer: A. 1.097 × 10⁷ m⁻¹
Why: Rydberg constant RH = 1.097 × 10⁷ m⁻¹. It appears in the Rydberg formula for spectral lines of hydrogen.
Q43.
Bohr radius a₀ in terms of fundamental constants:
Answer: B. 4pi eps<sub>0</sub> hbar²/(m<sub>e</sub> e²)
Why: a₀ = (4pi eps<sub>0</sub> hbar²)/(m<sub>e</sub> e²) = 0.529 Angstrom. It sets the scale of atomic orbitals.
Q44.
Quantization of angular momentum in Bohr model arises from:
Answer: B. De Broglie standing wave condition: n lambda = 2 pi r
Why: Standing wave condition: for stable orbit, circumference = n wavelengths. 2pi r = n lambda = n h/mv. This gives L = mvr = nh/2pi.
Q45.
Zeeman effect is the splitting of spectral lines in:
Answer: B. Magnetic field
Why: Zeeman effect: spectral lines split in magnetic field due to interaction of orbital magnetic moment with field.
Q46.
The number of spectral lines when electron jumps from nth orbit to ground state:
Answer: C. n(n-1)/2
Why: Transitions possible from n levels: any pair can transition. Total lines = n(n-1)/2 (combinations of 2 from n levels). Wait, for transitions from nth orbit to 1: possible intermediate stops give n(n-1)/2 total lines.
Q47.
In Rutherford scattering, the impact parameter b for deflection by angle theta:
Answer: A. b = Z e²/(4pi eps<sub>0</sub> × 2E) × cot(theta/2)
Why: Rutherford scattering: b = (Z e²/4pi eps<sub>0</sub>) × cot(theta/2) / (2E<sub>kin</sub>). Larger b gives smaller deflection angle.
Q48.
Fine structure of hydrogen spectral lines arises from:
Answer: B. Spin-orbit coupling (interaction of electron spin with orbital magnetic field)
Why: Fine structure: spin-orbit interaction. Electron spin magnetic moment interacts with magnetic field seen in electron rest frame. Splits energy levels.
Q49.
The de Broglie wavelength of a 1 kg ball moving at 1 m/s is negligible because:
Answer: B. Planck constant h = 6.63×10⁻³⁴ J·s is extremely small making lambda = h/mv = 6.63×10⁻³⁴ m
Why: lambda = h/mv = 6.63×10⁻³⁴/(1×1) = 6.63×10⁻³⁴ m. Far smaller than any measurable scale. Quantum effects negligible for macroscopic objects.
Q50.
The speed of an electron in the nth Bohr orbit is proportional to:
Answer: A. 1/n
Why: v<sub>n</sub> ∝ Z/n, so for hydrogen it falls off as 1/n.
Q51.
The line emitted in the n = 2 to n = 1 transition of hydrogen lies in the ___ region:
Answer: A. ultraviolet
Why: This first Lyman line is in the ultraviolet.
Q52.
For a hydrogen-like ion of nuclear charge Z, the energy levels scale as:
Answer: A. Z²
Why: E<sub>n</sub> = −13.6 Z²/n² eV, so the levels scale as Z².
Q53.
In the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²), the constant R is the ___ constant:
Answer: A. Rydberg
Why: R is the Rydberg constant, about 1.097 × 10⁷ m⁻¹.
Q54.
The kinetic energy of the electron in a hydrogen atom is ___ the magnitude of its total energy:
Answer: A. equal to
Why: KE = −E<sub>total</sub>, so its magnitude equals that of the total energy.
Q55.
The potential energy of the electron in a hydrogen atom is ___ its total energy:
Answer: A. twice its value, negative
Why: PE = 2 × E<sub>total</sub> (both negative), while KE = −E<sub>total</sub>.
Q56.
The series limit (shortest wavelength) of the Lyman series is the transition from n = ∞ to n =:
Answer: A. 1
Why: The Lyman series always ends on n = 1, so its limit is ∞ → 1.
Q57.
The frequency of the radiation emitted by an atom is proportional to the ___ between the two levels:
Answer: A. energy difference
Why: hν = ΔE, so frequency is proportional to the energy difference.
Q58.
A key limitation of Bohr’s model is that it fails to explain the spectra of ___ atoms:
Answer: A. multi-electron
Why: Bohr’s model works for one-electron systems but not for multi-electron atoms.
Q59.
The angular momentum of the electron in the ground state of hydrogen (n = 1) is:
Answer: A. h/2π
Why: Angular momentum = nh/2π, which for n = 1 is h/2π.
Q60.
The de Broglie explanation of Bohr’s quantisation requires the orbit circumference to hold a whole number of ___ wavelengths:
Answer: A. electron (de Broglie)
Why: A stable orbit fits an integer number of the electron’s de Broglie wavelengths.
Q61.
In the Bohr model of hydrogen, the radius of the n = 2 orbit (Bohr radius 0.53 Å) is:
Answer: C. 2.12 Å
Why: r<sub>n</sub> = n²·a₀ = 4·0.53 = 2.12 Å.
Q62.
The energy of the electron in the n = 2 level of hydrogen (E₁ = −13.6 eV) is:
Answer: C. −3.4 eV
Why: E<sub>n</sub> = −13.6/n² = −13.6/4 = −3.4 eV.
Q63.
The number of distinct spectral lines emitted when hydrogen atoms de-excite from the n = 4 level is:
Answer: C. 6
Why: Number of lines = n(n − 1)/2 = 4·3/2 = 6.
Q64.
The angular momentum of an electron in the n = 3 orbit of hydrogen is:
Answer: C. 3h/2π
Why: L = nh/2π = 3h/2π for n = 3.
Q65.
In the Bohr model, the speed of the electron is inversely proportional to n. In the n = 2 orbit the speed is:
Answer: A. half that of the ground state
Why: v ∝ 1/n, so at n = 2 the speed is half the ground-state value.
Q66.
The Balmer series of the hydrogen spectrum lies mainly in the:
Answer: A. visible region
Why: Transitions to n = 2 (Balmer series) produce lines in the visible region.
Q67.
The ionization energy of a hydrogen atom in its ground state is:
Answer: C. 13.6 eV
Why: The energy needed to remove the electron from n = 1 is 13.6 eV.
Q68.
The shortest wavelength (series limit) of the Lyman series of hydrogen (1/R ≈ 91.2 nm) is:
Answer: A. 91.2 nm
Why: The Lyman series limit corresponds to n = ∞ to n = 1, giving λ_min = 1/R ≈ 91.2 nm.