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NEET Physics: Modern Physics and Electronics · Chapter 2

Atoms

What to remember

  • Rutherford's alpha-scattering experiment showed that an atom has a tiny, dense, positively charged nucleus about 10⁻¹⁵ m in size, with electrons around it; the atom is mostly empty space.
  • In Bohr's model of hydrogen, rₙ = n² × 0.53 Å, vₙ ≈ 2.2 × 10⁶/n m s⁻¹ and Eₙ = −13.6/n² eV, with angular momentum mvr = nh/2π.
  • Light is emitted or absorbed when the electron jumps between levels: hν = Eᵢ − E_f. The lines for transitions ending on n = 1, 2, 3, 4, 5 are the Lyman, Balmer, Paschen, Brackett and Pfund series.
  • Rydberg formula: 1/λ = R(1/n₁² − 1/n₂²) with R = 1.097 × 10⁷ m⁻¹; an atom excited to level n can show n(n − 1)/2 different lines.

Alpha-particle scattering and Rutherford's model

Geiger and Marsden, working under Rutherford, directed a narrow beam of alpha particles at a very thin gold foil, about 2 × 10⁻⁷ m thick. An alpha particle is a helium nucleus (charge +2e, mass about 4 u). The particles had a kinetic energy of about 5.5 MeV. A rotating detector counted the particles scattered at different angles.

ObservationConclusion
Most alpha particles passed straight throughMost of the atom is empty space
A small fraction was scattered through small anglesAlpha particles are repelled by a small positive centre
About 1 in 8000 was deflected by more than 90°The positive charge and nearly all the mass are packed in a very small nucleus
A few were turned back through nearly 180°Head-on approach to a heavy, positive nucleus

Rutherford's nuclear model. The atom has a tiny central nucleus. It carries all the positive charge and almost all the mass. Electrons move around it in orbits, held by the electrostatic attraction. The size of the nucleus is about 10⁻¹⁵ m to 10⁻¹⁴ m. The size of the atom is about 10⁻¹⁰ m. So the atom is about 10⁴ to 10⁵ times larger than the nucleus.

Impact parameter (qualitative). The impact parameter b is the perpendicular distance of the initial line of the alpha particle from the centre of the nucleus.

Impact parameterScattering
Large bSmall deflection
Small bLarge deflection
b ≈ 0 (head-on)Deflection near 180°; the particle retraces its path

Only a few alpha particles have a very small impact parameter. This is why only a few are scattered through large angles.

Distance of closest approach. In a head-on approach the alpha particle slows down until all its kinetic energy K has become electrostatic potential energy. At that point it stops momentarily at distance r₀ from the nucleus of charge Ze:

K = (1/4πε₀)(2e)(Ze)/r₀, so r₀ = 2Ze²/(4πε₀K) = 9 × 10⁹ × 2Ze²/K.

If the kinetic energy doubles, r₀ becomes half. The nuclear radius is smaller than r₀. A larger Z gives a larger r₀ for the same K.

Limitations of Rutherford's model.

  • An electron moving in a circle is accelerating. Classical electromagnetic theory says it should radiate energy continuously, lose energy and spiral into the nucleus. The atom would collapse in about 10⁻⁸ s. Real atoms are stable.
  • Because the radiated frequency would change continuously, the model predicts a continuous spectrum. Atoms actually emit line spectra with definite wavelengths.

Atomic spectra and the hydrogen series

A heated or excited gas emits light of only certain wavelengths. A spectrometer shows these as bright lines on a dark background. This is the emission line spectrum. Each element has its own characteristic set of lines. For hydrogen the wavelengths follow a simple pattern.

Rydberg formula:

1/λ = R (1/n₁² − 1/n₂²), where n₂ > n₁ and R = 1.097 × 10⁷ m⁻¹.

SeriesFinal level n₁Initial levels n₂Region
Lyman12, 3, 4, …Ultraviolet
Balmer23, 4, 5, …Visible (and near ultraviolet)
Paschen34, 5, 6, …Infrared
Brackett45, 6, 7, …Infrared
Pfund56, 7, 8, …Infrared

For each series, the first line (n₂ = n₁ + 1) has the longest wavelength. The series limit (n₂ = ∞) has the shortest wavelength.

SeriesLongest wavelength (about)Series limit (about)
Lyman121.6 nm91.2 nm
Balmer656 nm364.6 nm
Paschen1875 nm820 nm
Brackett4051 nm1458 nm

Within a series, lines come closer together at shorter wavelengths.

Bohr model of the hydrogen atom

Bohr kept Rutherford's nuclear picture and added three postulates.

  • 1. Stationary orbits. An electron can revolve only in certain circular orbits without radiating energy. These are stable, stationary states.
  • 2. Quantised angular momentum. The allowed orbits are those in which the angular momentum of the electron is an integral multiple of h/2π:

mvr = nh/2π, n = 1, 2, 3, …

  • 3. Emission and absorption. When the electron jumps from an orbit of higher energy Eᵢ to one of lower energy E_f, a photon is emitted:

hν = Eᵢ − E_f.

The same energy must be absorbed for the reverse jump.

Results for hydrogen. The electrostatic force provides the centripetal force: mv²/r = e²/(4πε₀r²). Combining this with the second postulate gives:

QuantityFormulaValue
Radiusrₙ = n² × 0.53 År₁ = 0.53 Å; rₙ ∝ n²
Speedvₙ = (1/n)(e²/2ε₀h)v₁ ≈ 2.2 × 10⁶ m s⁻¹; vₙ ∝ 1/n
Kinetic energyKₙ = +13.6/n² eVKₙ ∝ 1/n²
Potential energyUₙ = −27.2/n² eVUₙ = −2Kₙ
Total energyEₙ = −13.6/n² eVEₙ = −Kₙ
Time periodTₙ = 2πrₙ/vₙTₙ ∝ n³

The negative sign of the total energy shows that the electron is bound to the nucleus. The energy is zero when the electron is at rest at infinite distance.

Energy level diagram (in words). The levels are numbered upward from the ground state, n = 1. The energies are −13.6 eV (n = 1), −3.4 eV (n = 2), −1.51 eV (n = 3), −0.85 eV (n = 4), −0.54 eV (n = 5), and so on, approaching 0 eV at n = ∞. The levels come closer together as n increases. Lyman transitions are the downward jumps to the lowest level. Balmer jumps end at −3.4 eV.

Level nEnergy (eV)Radius (Å)
1−13.600.53
2−3.402.12
3−1.514.77
4−0.858.48
5−0.5413.25

Ionisation energy and excitation energy.

  • Ionisation energy is the energy needed to remove the electron completely from the atom. For hydrogen in the ground state it is 13.6 eV. From n = 2 it is 3.4 eV, and from n = 3 it is 1.51 eV.
  • Excitation energy is the energy needed to raise the electron from a lower level to a higher one. From n = 1 to n = 2 it is 10.2 eV (this is the first excitation energy), from n = 1 to n = 3 it is 12.09 eV.
  • The longest-wavelength photon that can ionise hydrogen from the ground state has λ = 1240/13.6 ≈ 91 nm.

Spectral lines and the Bohr model. A photon of wavelength λ has energy 1240/λ eV when λ is in nm. The Rydberg formula follows from the energy differences, and the Bohr model gives R = 1.097 × 10⁷ m⁻¹. Absorption lines appear when electrons take photons of exactly the right energy from a continuous beam.

Number of spectral lines. A hydrogen atom excited to level n can return to lower levels by many routes. If a large number of atoms are present, the total number of different lines is

N = n(n − 1)/2.

Highest level n23456
Lines1361015

Limits of Bohr's model. It works for hydrogen, but it does not explain the relative intensities of lines and cannot be applied to atoms with many electrons.

De Broglie's explanation of the second postulate

De Broglie reasoned that an electron in a circular orbit behaves like a wave. For a stable orbit the wave must fit the circle as a standing wave, so the circumference must contain a whole number of wavelengths:

2πrₙ = nλ.

With λ = h/mv, this gives mvr = nh/2π, which is exactly Bohr's second postulate. In the n-th orbit, the circumference holds n de Broglie wavelengths. For the first orbit, λ = 2π × 0.53 Å ≈ 3.33 Å.

Worked examples

Example 1 (closest approach). An alpha particle of 5.5 MeV is fired head-on at a gold nucleus (Z = 79). Use k = 9 × 10⁹ N m² C⁻² and e = 1.6 × 10⁻¹⁹ C.

K = 5.5 × 10⁶ × 1.6 × 10⁻¹⁹ = 8.8 × 10⁻¹³ J.

r₀ = k × 2 × 79 × e²/K = 9 × 10⁹ × 158 × 2.56 × 10⁻³⁸ / 8.8 × 10⁻¹³ ≈ 4.1 × 10⁻¹⁴ m.

Example 2 (wavelength of a transition). Find the wavelength of the photon emitted when an electron falls from n = 3 to n = 2. Use hc = 1240 eV nm.

ΔE = 3.4 − 1.51 = 1.89 eV. λ = 1240/1.89 ≈ 656 nm.

Check by the Rydberg formula: 1/λ = 1.097 × 10⁷ × (1/4 − 1/9) = 1.097 × 10⁷ × 5/36, so λ = 6.56 × 10⁻⁷ m.

Example 3 (number of lines). Ground-state hydrogen atoms absorb photons of energy 12.09 eV. The electron goes to level n where 13.6 − 13.6/n² = 12.09, so n = 3. The number of lines is 3 × 2/2 = 3. They are 3 → 2, 3 → 1 and 2 → 1.

Example 4 (ratios). For n = 2 compared with n = 1: radius is 4 times, speed is half, time period is 8 times, kinetic energy is one-fourth, and the magnitude of total energy is one-fourth.

Formula summary

QuantityFormulaUnits
Closest approachr₀ = 2Ze²/(4πε₀K)m
Rydberg formula1/λ = R(1/n₁² − 1/n₂²)m⁻¹
Bohr second postulatemvr = nh/2πkg m² s⁻¹
Frequency conditionhν = Eᵢ − E_fJ or eV
Radiusrₙ = 0.53 n² ÅÅ
Speedvₙ ≈ 2.2 × 10⁶/n m s⁻¹m s⁻¹
EnergyEₙ = −13.6/n² eVeV
Linesn(n − 1)/2none
Orbit condition2πrₙ = nλm

NEET traps

  • Lyman is ultraviolet, Balmer is visible, Paschen, Brackett and Pfund are infrared. Paschen is not visible.
  • The series are named by the final level: Lyman 1, Balmer 2, Paschen 3, Brackett 4, Pfund 5.
  • The first line of a series has the longest wavelength; the series limit has the shortest.
  • Energy of the electron is negative, but the kinetic energy is positive and equals −Eₙ.
  • Orbit radius goes as n², speed goes as 1/n, and energy goes as 1/n². Do not mix these.
  • Ionisation energy from n = 2 is 3.4 eV, not 13.6 eV.
  • First excitation energy of hydrogen is 10.2 eV, not 13.6 eV.
  • Lines from level n are n(n − 1)/2 for many atoms. A single atom shows one photon at a time.
  • A small impact parameter means a large angle of scattering, not a small one.
  • Closest approach is inversely proportional to kinetic energy and is not the size of the nucleus.

One-liners

  • Geiger and Marsden carried out the alpha scattering experiment on thin gold foil.
  • Most alpha particles pass undeviated; about 1 in 8000 turns by more than 90°.
  • The nucleus is about 10⁻¹⁵ m across; the atom is about 10⁻¹⁰ m.
  • Rutherford's model predicts that atoms collapse and give a continuous spectrum.
  • Bohr's electron does not radiate in a stationary orbit.
  • Angular momentum in the n-th orbit is nh/2π.
  • r₁ = 0.53 Å.
  • E₁ = −13.6 eV.
  • Ionisation energy of ground-state hydrogen is 13.6 eV.
  • Lyman series is ultraviolet.
  • Balmer series is in the visible region.
  • Paschen, Brackett and Pfund series are infrared.
  • Excited level n gives n(n − 1)/2 lines.
  • Time period of the electron in the n-th orbit goes as n³.
  • De Broglie's condition for a stable orbit is 2πr = nλ.

Practice questions

  1. The experiment on scattering of alpha particles by a thin gold foil was carried out by

    1. Bohr and de Broglie
    2. Geiger and Marsden
    3. Thomson and Millikan
    4. Hertz and Lenard
    Answer

    B. Geiger and Marsden

    Geiger and Marsden, working under Rutherford, counted alpha particles scattered by a thin gold foil.

  2. Most alpha particles pass nearly undeviated through a thin gold foil. This shows that

    1. the atom is a uniform sphere of positive charge
    2. the nucleus is negatively charged
    3. electrons are heavier than alpha particles
    4. most of the atom is empty space
    Answer

    D. most of the atom is empty space

    Only a tiny nucleus deflects alpha particles; the rest of the atom is empty space.

  3. The Balmer series of hydrogen lies mainly in the

    1. X-ray region
    2. far infrared region
    3. visible region
    4. ultraviolet region
    Answer

    C. visible region

    Balmer lines (n → 2) lie in the visible region and just beyond into the near ultraviolet.

  4. The total energy of the electron in the n = 4 orbit of a hydrogen atom is

    1. −0.85 eV
    2. −3.40 eV
    3. −1.70 eV
    4. +0.85 eV
    Answer

    A. −0.85 eV

    E₄ = −13.6/16 = −0.85 eV.

  5. Taking the radius of the first Bohr orbit as 0.53 Å, the radius of the third orbit is about

    1. 3.18 Å
    2. 14.3 Å
    3. 1.59 Å
    4. 4.77 Å
    Answer

    D. 4.77 Å

    r₃ = 3² × 0.53 = 9 × 0.53 = 4.77 Å.

  6. Hydrogen atoms are excited to the n = 5 level. The maximum number of different spectral lines that can be emitted is

    1. 5
    2. 4
    3. 20
    4. 10
    Answer

    D. 10

    N = n(n − 1)/2 = 5 × 4/2 = 10.

  7. The energy needed to remove the electron from a hydrogen atom in the n = 2 state is

    1. 10.2 eV
    2. 3.4 eV
    3. 13.6 eV
    4. 6.8 eV
    Answer

    B. 3.4 eV

    Ionisation energy from n = 2 = 13.6/4 = 3.4 eV.

  8. Take hc = 1240 eV nm. The wavelength of the photon emitted when a hydrogen electron falls from n = 3 to n = 2 is about

    1. 656 nm
    2. 122 nm
    3. 1875 nm
    4. 486 nm
    Answer

    A. 656 nm

    ΔE = 3.4 − 1.51 = 1.89 eV; λ = 1240/1.89 ≈ 656 nm. Rydberg check: 1/λ = R(1/4 − 1/9).

  9. Take hc = 1240 eV nm. The shortest wavelength in the Lyman series of hydrogen is about

    1. 122 nm
    2. 656 nm
    3. 91 nm
    4. 365 nm
    Answer

    C. 91 nm

    The series limit has ΔE = 13.6 eV, so λ = 1240/13.6 ≈ 91 nm. Check: 1/R = 91 nm.

  10. Transitions of the electron in hydrogen that end on the level n = 3 form the

    1. Lyman series
    2. Paschen series
    3. Balmer series
    4. Brackett series
    Answer

    B. Paschen series

    Paschen: final level n = 3 (infrared).

  11. The ratio of the kinetic energy of the electron in the first Bohr orbit to that in the third Bohr orbit of hydrogen is

    1. 27 : 1
    2. 1 : 9
    3. 9 : 1
    4. 3 : 1
    Answer

    C. 9 : 1

    K ∝ 1/n², so K₁/K₃ = 3²/1² = 9.

  12. The angular momentum of the electron in the second Bohr orbit of hydrogen is

    1. h/π
    2. h/2π
    3. h
    4. 2h/π
    Answer

    A. h/π

    L = nh/2π = 2h/2π = h/π.

  13. The minimum energy needed to excite a ground-state hydrogen atom to its first excited state is

    1. 10.2 eV
    2. 1.89 eV
    3. 13.6 eV
    4. 3.4 eV
    Answer

    A. 10.2 eV

    E₂ − E₁ = −3.4 − (−13.6) = 10.2 eV.

  14. Take R = 1.1 × 10⁷ m⁻¹. The longest wavelength in the Lyman series of hydrogen is about

    1. 364 nm
    2. 121 nm
    3. 91 nm
    4. 102 nm
    Answer

    B. 121 nm

    1/λ = R(1 − 1/4) = 3R/4, λ = 4/(3 × 1.1 × 10⁷) = 1.21 × 10⁻⁷ m. Check: 1240/10.2 ≈ 122 nm.

  15. In a head-on approach of an alpha particle to a nucleus, if the kinetic energy of the alpha particle is doubled, the distance of closest approach

    1. doubles
    2. remains the same
    3. becomes half
    4. becomes one-fourth
    Answer

    C. becomes half

    r₀ = 2Ze²/(4πε₀K) ∝ 1/K.

  16. The order of the size of an atomic nucleus is

    1. 10⁻⁵ m
    2. 10⁻¹⁰ m
    3. 10⁻¹⁸ m
    4. 10⁻¹⁵ m
    Answer

    D. 10⁻¹⁵ m

    The nucleus is about 10⁻¹⁵ m to 10⁻¹⁴ m across; the atom is about 10⁻¹⁰ m.

  17. Given below are two statements: Statement I: Rutherford's model cannot explain the line spectra of atoms. Statement II: In Rutherford's model an accelerating orbiting electron should radiate energy and spiral into the nucleus. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both are correct limitations of the nuclear model.

  18. Given below are two statements: Statement I: In Bohr's model an electron radiates energy continuously while it stays in a stationary orbit. Statement II: In a stationary orbit the angular momentum of the electron is an integral multiple of h/2π. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    D. Statement I is incorrect but Statement II is correct

    No radiation occurs in a stationary orbit, so I is incorrect. II is Bohr's second postulate.

  19. Which of the following statements are correct? 1. The Lyman series lies in the ultraviolet region. 2. The Balmer series lies partly in the visible region. 3. The Paschen series lies in the visible region.

    1. 2 and 3 only
    2. 1 and 3 only
    3. 1, 2 and 3
    4. 1 and 2 only
    Answer

    D. 1 and 2 only

    Paschen lines are in the infrared.

  20. Given below are two statements: Statement I: The energy of the electron in a hydrogen atom is given by Eₙ = −13.6/n² eV. Statement II: The energy of the n = 2 level is −6.8 eV. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    E₂ = −13.6/4 = −3.4 eV, so II is incorrect.

  21. Given below are two statements: Statement I: In Bohr's model the radius of the n-th orbit is proportional to n². Statement II: In Bohr's model the speed of the electron increases as n increases. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    vₙ ∝ 1/n, so the speed decreases with n; II is incorrect.

  22. Which of the following statements are correct for the Bohr model of hydrogen? 1. The orbit radius is proportional to n². 2. The orbital speed is proportional to 1/n. 3. The total energy is proportional to 1/n.

    1. 1 and 2 only
    2. 1 and 3 only
    3. 2 and 3 only
    4. 1, 2 and 3
    Answer

    A. 1 and 2 only

    Eₙ ∝ 1/n², so 3 is incorrect.

  23. Given below are two statements: Statement I: A hydrogen atom gives a continuous spectrum. Statement II: A hydrogen atom excited to the n = 4 level can show only 4 different spectral lines. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    B. Both Statement I and Statement II are incorrect

    Hydrogen gives a line spectrum, and n = 4 gives 4 × 3/2 = 6 lines; both are incorrect.

  24. Given below are two statements: Statement I: In Rutherford scattering, a smaller impact parameter gives a larger angle of scattering. Statement II: For an impact parameter close to zero the alpha particle retraces its path, a scattering angle of about 180°. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    A. Both Statement I and Statement II are correct

    Both are correct.

  25. Given below are two statements: Statement I: The ionisation energy of a ground-state hydrogen atom is 13.6 eV. Statement II: The excitation energy of the first excited state of hydrogen is 13.6 eV. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    The first excitation energy is 10.2 eV, so II is incorrect.

  26. Given below are two statements: Statement I: De Broglie explained Bohr's second postulate by treating the orbiting electron as a standing wave. Statement II: In the n-th orbit the circumference holds (n + 1) de Broglie wavelengths. Choose the correct option.

    1. Both Statement I and Statement II are correct
    2. Both Statement I and Statement II are incorrect
    3. Statement I is correct but Statement II is incorrect
    4. Statement I is incorrect but Statement II is correct
    Answer

    C. Statement I is correct but Statement II is incorrect

    The circumference is 2πr = nλ, so n wavelengths; II is incorrect.

  27. Assertion (A): Rutherford's atomic model is unstable. Reason (R): An electron revolving round the nucleus is accelerated and should emit radiation, losing energy continuously.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    The radiating electron would spiral into the nucleus.

  28. Assertion (A): Atomic hydrogen emits a line spectrum. Reason (R): Photons are emitted when the electron jumps between discrete energy levels, each photon having energy equal to the level difference.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Fixed energy gaps give fixed wavelengths, hence lines.

  29. Assertion (A): Most alpha particles pass nearly straight through a thin gold foil. Reason (R): The nucleus occupies a very small part of the volume of the atom.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    A. Both A and R are true and R is the correct explanation of A

    Few alpha particles pass close to a nucleus, so most are not deviated.

  30. Assertion (A): The radius of the n = 2 orbit of hydrogen is four times that of the n = 1 orbit. Reason (R): The speed of the electron in the n = 2 orbit is half that in the n = 1 orbit.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    B. Both A and R are true but R is not the correct explanation of A

    rₙ ∝ n² and vₙ ∝ 1/n are both true, but the speed ratio does not explain the radius ratio.

  31. Assertion (A): The energy needed to ionise a hydrogen atom from the n = 2 level is 3.4 eV. Reason (R): The total energy of the electron in the n = 2 level is +3.4 eV.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    C. A is true but R is false

    E₂ = −3.4 eV, so R is false.

  32. Assertion (A): The lines of the Paschen series of hydrogen lie in the ultraviolet region. Reason (R): Paschen lines arise from transitions ending at n = 3.

    1. Both A and R are true and R is the correct explanation of A
    2. Both A and R are true but R is not the correct explanation of A
    3. A is true but R is false
    4. A is false but R is true
    Answer

    D. A is false but R is true

    Paschen lines are infrared, so A is false. R is true.

  33. Match List I with List II. List I: a. Lyman series b. Balmer series c. Paschen series d. Brackett series List II: i. final level n = 4 ii. final level n = 1 iii. final level n = 3 iv. final level n = 2

    1. a-ii, b-iii, c-iv, d-i
    2. a-ii, b-iv, c-iii, d-i
    3. a-iv, b-ii, c-iii, d-i
    4. a-i, b-iv, c-iii, d-ii
    Answer

    B. a-ii, b-iv, c-iii, d-i

    Lyman 1, Balmer 2, Paschen 3, Brackett 4.

  34. Match List I with List II. List I: a. n = 1 b. n = 2 c. n = 3 d. n = 4 List II: i. 4.77 Å ii. 0.53 Å iii. 8.48 Å iv. 2.12 Å

    1. a-iv, b-ii, c-i, d-iii
    2. a-ii, b-iv, c-i, d-iii
    3. a-ii, b-iv, c-iii, d-i
    4. a-ii, b-i, c-iv, d-iii
    Answer

    B. a-ii, b-iv, c-i, d-iii

    rₙ = 0.53n² Å gives 0.53, 2.12, 4.77 and 8.48 Å.

  35. Match List I with List II. List I: a. Energy of n = 1 b. Energy of n = 3 c. Energy of n = 4 d. Energy of n = 2 List II: i. −3.40 eV ii. −0.85 eV iii. −13.6 eV iv. −1.51 eV

    1. a-i, b-iv, c-ii, d-iii
    2. a-iii, b-i, c-ii, d-iv
    3. a-iii, b-iv, c-i, d-ii
    4. a-iii, b-iv, c-ii, d-i
    Answer

    D. a-iii, b-iv, c-ii, d-i

    Eₙ = −13.6/n²: −13.6, −3.4, −1.51, −0.85 eV for n = 1, 2, 3, 4.

  36. Match List I with List II. List I: a. Highest level n = 2 b. Highest level n = 3 c. Highest level n = 4 d. Highest level n = 5 List II: i. 10 lines ii. 3 lines iii. 1 line iv. 6 lines

    1. a-iii, b-iv, c-ii, d-i
    2. a-ii, b-iii, c-iv, d-i
    3. a-iii, b-ii, c-i, d-iv
    4. a-iii, b-ii, c-iv, d-i
    Answer

    D. a-iii, b-ii, c-iv, d-i

    n(n − 1)/2 gives 1, 3, 6 and 10 lines.

  37. Match List I with List II. List I: a. Distance of closest approach b. Small impact parameter c. Rutherford's nucleus d. Bohr's second postulate List II: i. Large-angle scattering ii. mvr = nh/2π iii. 2Ze²/(4πε₀K) iv. Positive charge and mass concentrated in a small volume

    1. a-iii, b-i, c-ii, d-iv
    2. a-iii, b-i, c-iv, d-ii
    3. a-iii, b-iv, c-i, d-ii
    4. a-i, b-iii, c-iv, d-ii
    Answer

    B. a-iii, b-i, c-iv, d-ii

    r₀ = 2Ze²/(4πε₀K); small b gives large angles; nucleus is a small dense positive core; mvr = nh/2π.

  38. Match List I with List II. List I: a. Lyman first line b. Balmer first line c. Paschen first line d. Brackett first line List II: i. 656 nm ii. 4051 nm iii. 121.6 nm iv. 1875 nm

    1. a-i, b-iii, c-iv, d-ii
    2. a-iii, b-i, c-iv, d-ii
    3. a-iii, b-i, c-ii, d-iv
    4. a-iii, b-iv, c-i, d-ii
    Answer

    B. a-iii, b-i, c-iv, d-ii

    Using R = 1.097 × 10⁷ m⁻¹: 121.6 nm, 656 nm, 1875 nm and 4051 nm.

  39. An alpha particle of kinetic energy 5.5 MeV approaches a gold nucleus (Z = 79) head-on. Take k = 9 × 10⁹ N m² C⁻² and e = 1.6 × 10⁻¹⁹ C. The distance of closest approach is about

    1. 4.1 × 10⁻¹⁵ m
    2. 2.1 × 10⁻¹⁴ m
    3. 8.3 × 10⁻¹⁴ m
    4. 4.1 × 10⁻¹⁴ m
    Answer

    D. 4.1 × 10⁻¹⁴ m

    r₀ = k(2e)(79e)/K = 9 × 10⁹ × 158 × (1.6 × 10⁻¹⁹)² / (5.5 × 10⁶ × 1.6 × 10⁻¹⁹) ≈ 4.1 × 10⁻¹⁴ m.

  40. Take hc = 1240 eV nm. A hydrogen electron falls from n = 4 to n = 2. The wavelength of the emitted photon is about

    1. 122 nm
    2. 1875 nm
    3. 486 nm
    4. 656 nm
    Answer

    C. 486 nm

    ΔE = 3.4 − 0.85 = 2.55 eV; λ = 1240/2.55 ≈ 486 nm.

  41. Hydrogen atoms are excited to the n = 4 level. The largest photon energy that can be emitted is

    1. 12.75 eV
    2. 13.6 eV
    3. 12.09 eV
    4. 10.2 eV
    Answer

    A. 12.75 eV

    The largest jump is 4 → 1: −0.85 − (−13.6) = 12.75 eV.

  42. Ground-state hydrogen atoms absorb photons of energy 12.09 eV each. The number of different spectral lines the atoms can then emit is

    1. 6
    2. 3
    3. 1
    4. 2
    Answer

    B. 3

    13.6 − 13.6/n² = 12.09 gives n = 3. Lines: 3 → 2, 3 → 1, 2 → 1, that is 3 × 2/2 = 3.

  43. Take π = 3.14 and r₁ = 0.53 Å. The de Broglie wavelength of the electron in the first Bohr orbit of hydrogen is about

    1. 1.67 Å
    2. 0.53 Å
    3. 6.66 Å
    4. 3.33 Å
    Answer

    D. 3.33 Å

    2πr₁ = λ for n = 1, so λ = 2 × 3.14 × 0.53 = 3.33 Å.

  44. An alpha particle is fired directly at the centre of a gold nucleus (impact parameter almost zero). After reaching its distance of closest approach, it

    1. is absorbed by the nucleus
    2. is deflected through about 90°
    3. moves back along the same path, scattered through about 180°
    4. passes through the nucleus undeviated
    Answer

    C. moves back along the same path, scattered through about 180°

    Coulomb repulsion stops it momentarily and sends it straight back.

  45. Take hc = 1240 eV nm. The longest-wavelength photon that can ionise a hydrogen atom in the n = 2 state has wavelength about

    1. 656 nm
    2. 122 nm
    3. 365 nm
    4. 91 nm
    Answer

    C. 365 nm

    Ionisation energy from n = 2 is 3.4 eV; λ = 1240/3.4 ≈ 365 nm.

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