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Atoms: questions with answers

Class 12 Physics, Chapter 12 (NCERT). 15 questions by concept, 1 with pictures. Open “Show answer” for the answer and why.

How to teach this chapter · Common misconceptions & worked visual for this chapter

Evidence for the nuclear atom

Question 1

In thin-gold-foil scattering, most alpha particles pass nearly straight through and very few reverse direction. Which distribution of positive charge and most atomic mass explains both observations?

  1. (a)Concentrated in a shell at the outer atomic surface
  2. (b)Concentrated in a tiny nucleus inside a mostly empty atom
  3. (c)Spread approximately uniformly through the whole atomic volume
  4. (d)Distributed only among the atomic electrons
Show answer

(b) Concentrated in a tiny nucleus inside a mostly empty atom

Rare close encounters with a small, dense, positively charged nucleus give large repulsive deflections. Most incident paths miss that small region and are nearly undeviated.

Question 2

Which comparison between the Thomson and Rutherford atom models is correct?

  1. (a)Rutherford makes the whole atom as small as its nucleus
  2. (b)Both have comparable atomic size, but Rutherford concentrates positive charge in a much smaller nucleus
  3. (c)Thomson assigns all positive charge to electrons
  4. (d)Both distribute positive charge uniformly throughout the entire atomic volume
Show answer

(b) Both have comparable atomic size, but Rutherford concentrates positive charge in a much smaller nucleus

The nuclear model changes the internal distribution of positive charge and mass, not the order of magnitude of the atom size. A diffuse positive distribution is the Thomson picture.

Head-on distance of closest approach

Question 3

A proton and an alpha particle approach the same fixed heavy nucleus head-on with equal initial speeds. Take q_alpha=2e and m_alpha=4m_proton. What is d_proton:d_alpha?

  1. (a)2:1
  2. (b)4:1
  3. (c)1:1
  4. (d)1:2
Show answer

(a) 2:1

At equal speed, K=mv²/2 and d is proportional to q/m. The proton has twice the alpha value of q/m, so its turning distance is twice as large. Comparing charge alone reverses this result.

Impact parameter and scattering angle

Question 4

The diagram shows only the initial paths of two alpha particles P and Q approaching the same heavy positive nucleus. Their incident kinetic energies are equal. Which particle will undergo the larger scattering angle?

PQN
  1. (a)Both have the same scattering angle
  2. (b)Particle Q
  3. (c)Particle P
  4. (d)Neither is deflected because both initial paths are straight
Show answer

(b) Particle Q

At fixed energy and nuclear charge, the path with the smaller perpendicular offset from the nucleus has the larger deflection. The diagram supplies only the incoming offsets, not the scattered paths.

Question 5

An alpha particle undergoes a 180 degree reversal in a head-on Coulomb encounter with a fixed heavy nucleus. What was its impact parameter?

  1. (a)Equal to the nuclear radius
  2. (b)Zero
  3. (c)Equal to the nonzero distance of closest approach
  4. (d)The largest impact parameter possible
Show answer

(b) Zero

Head-on incidence means that the initial line of motion passes through the nuclear centre, so its perpendicular offset b is zero. The turning distance remains nonzero because electrostatic repulsion stops the particle before contact.

Classical instability and Bohr stationary states

Question 6

In the classical Rutherford picture, an electron travels at constant speed along a circular orbit. Which statement about its acceleration and radiation is correct?

  1. (a)It has centripetal acceleration and classical theory predicts radiation
  2. (b)Its acceleration is zero, so classical theory predicts no radiation
  3. (c)Its acceleration is tangential while its direction stays fixed
  4. (d)Its speed must be zero because the atom is neutral
Show answer

(a) It has centripetal acceleration and classical theory predicts radiation

Constant speed does not mean constant velocity: the direction changes continuously. A charged particle with centripetal acceleration radiates in classical electromagnetism, causing the stability problem.

Bohr radii and unequal radial spacing

Question 7

For hydrogen Bohr orbits, what is (r3-r2):(r2-r1)?

  1. (a)1:1
  2. (b)5:3
  3. (c)9:4
  4. (d)3:5
Show answer

(b) 5:3

Since r_n=a0 n², the two gaps are (9-4)a0=5a0 and(4-1)a0=3a0. Integer n does not make the radii equally spaced.

Quantised angular momentum

Question 8

In Bohr model of hydrogen, what is the electron angular momentum in orbit n=2?

  1. (a)h/(2π)
  2. (b)2h/π
  3. (c)h/π
  4. (d)2h
Show answer

(c) h/π

Bohr quantisation gives L=n h/(2π). Substitute the stated n; the unit of angular momentum is already carried by h.

Speed and period of orbital revolution

Question 9

A hydrogen electron changes from Bohr orbit n=2 to n=1. By what percentage does its orbital period decrease?

  1. (a)12.5%
  2. (b)50%
  3. (c)75%
  4. (d)87.5%
Show answer

(d) 87.5%

T is proportional to n³, so final period is 1/8 of initial. The decrease is (1-1/8)×100%=87.5%. Using n² would give 75%.

Kinetic, potential and total atomic energy

Question 10

An electron-nucleus system has negative total energy when separated stationary particles at infinity define zero energy. What does that sign indicate?

  1. (a)The electron kinetic energy must be negative
  2. (b)The atom must radiate continuously even in a stationary state
  3. (c)The electron is already free because its total energy is negative
  4. (d)The electron is bound and energy must be supplied to free it
Show answer

(d) The electron is bound and energy must be supplied to free it

Negative energy lies below the separated-particle threshold. A positive amount of energy is needed to reach that threshold; the negative sign does not make kinetic energy negative.

Photon energy and transition comparisons

Question 11

A hydrogen atom emits a Balmer photon in a direct transition n=4 to n=2. Which interpretation is consistent with Bohr model?

  1. (a)Two photons must have exactly equal energies in every event
  2. (b)The event is impossible because only adjacent-orbit transitions are permitted
  3. (c)The electron must gain energy because the final quantum number is smaller
  4. (d)One photon carries E4-E2 without a required stop at n=3
Show answer

(d) One photon carries E4-E2 without a required stop at n=3

The discrete levels restrict allowed states, not transitions to neighbours only. The n=4 to n=2 energy difference is carried by a single photon in this specified direct transition.

Hydrogen spectral series and line counts

Question 12

At which principal quantum number do the downward transitions of the Balmer series terminate?

  1. (a)n=2
  2. (b)n=1
  3. (c)n=4
  4. (d)n=3
Show answer

(a) n=2

Balmer emission ends at n=2. Transitions ending at the ground state n=1 form the Lyman series.

Selective absorption and excitation thresholds

Question 13

Ground-state hydrogen is exposed separately to 12.5 eV photons and 12.5 eV incident electrons. The excitation energies to n = 2, 3, 4 are 10.2, 12.09, 12.75 eV. Ignore linewidth and multiphoton processes. Which comparison is correct?

  1. (a)An electron can excite n=3 and keep the extra energy; the photon cannot be absorbed into a bound level
  2. (b)Both can excite n=3 by giving up 12.09 eV and keeping the extra energy
  3. (c)The photon can excite n=4; the electron cannot excite any level
  4. (d)Both always ionise the atom
Show answer

(a) An electron can excite n=3 and keep the extra energy; the photon cannot be absorbed into a bound level

An incident electron can transfer only 12.09 eV and leave with remaining kinetic energy. A photon is absorbed as a whole, and 12.5 eV matches no bound-state gap and is below ionisation. Thus the two incident particles have different possibilities.

de Broglie standing-wave condition

Question 14

An electron moves in Bohr orbit n=4 of radius r. What is its de Broglie wavelength according to the standing-wave condition?

  1. (a)πr/2
  2. (b)r
  3. (c)4πr
  4. (d)r/4
Show answer

(a) πr/2

The full circumference fits n wavelengths: 2πr=n lambda. Therefore lambda=2πr/n; substituting the radius for the circumference loses the 2π factor.

Limits of the Bohr model

Question 15

Why does the unmodified single-electron Bohr calculation not correctly describe neutral helium?

  1. (a)It omits the interaction between the two electrons of helium
  2. (b)Helium has no positive nucleus
  3. (c)The Planck constant becomes zero in helium
  4. (d)It does describe helium exactly because only the nuclear charge changes
Show answer

(a) It omits the interaction between the two electrons of helium

Neutral helium contains two electrons that interact with one another as well as the nucleus. Those additional forces are absent from the single-electron calculation, so simply changing nuclear charge is insufficient.

Common misconception in Atoms

Open full misconception guide →

Is an emitted photon set by the upper energy level alone?

Wrong

The photon energy equals the atom's energy at the upper level.

Why it feels right. The light comes from an atom leaving that level.

Right

The photon carries the difference between the two levels.

Worked example

Both transitions begin at −2 eV. The lower destination gives the larger energy gap.
  1. In a model atom, a transition from −2 eV to −5 eV emits a 3 eV photon.
  2. A transition from the same upper level to −8 eV emits a 6 eV photon.