Atoms NEET PYQReal NEET & AIPMT 2005–2020 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Atoms question from the NEET & AIPMT papers on this site (2005–2020), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Q1NEET 2020 · Q136
For which one of the following, Bohr model is not valid ?
Not quite — the answer is D.
Bohr's model applies only to one-electron (hydrogen-like) systems. Hydrogen, deuterium and He⁺ each have a single electron, but Ne⁺ still has 9 electrons, so electron–electron interactions make the model invalid.
Q2NEET 2019 · Q27
The total energy of an electron in an atom in an orbit is −3.4 eV. Its kinetic and potential energies are, respectively:
Not quite — the answer is C.
In the Bohr model K = −E and U = 2E. With E = −3.4 eV, K = +3.4 eV and U = −6.8 eV. Kinetic energy can never be negative, which rules out the choices with negative KE.
Q3NEET 2018 · Q11
The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is
Not quite — the answer is B.
In a Bohr orbit, K = ke²/2r and the total energy E = −ke²/2r, so E = −K and K : E = 1 : −1. The ratio 1 : −2 is kinetic to potential energy, not to total energy.
Q4NEET 2017 · Q143
The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is:
Not quite — the answer is C.
The last (series limit) line has n₁ → ∞. Balmer: 1/λ = R(1/4), so λ = 4/R; Lyman: 1/λ = R(1/1), so λ = 1/R. Their ratio is 4. Taking the ratio the other way round gives 0.25, not an option.
Q5NEET 2017 (Gujarati paper) · Q172
If the longest wavelength in the ultraviolet region of hydrogen spectrum is λ₀ then the shortest wavelength in its infrared region is:
Not quite — the answer is C.
The longest Lyman (UV) line is n = 2 → 1: 1/λ₀ = R(1 − 1/4) = 3R/4. The shortest infrared wavelength is the Paschen series limit, ∞ → 3: 1/λ = R/9, so λ = 9/R = 9 × (3λ₀/4) = 27λ₀/4. 36/5 λ₀ is the longest Balmer line (3 → 2), which is in the visible region.
Q6NEET 2016 Phase 2 · Q41
If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength λ. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be
Not quite — the answer is C.
1/λ ∝ (1/2² − 1/3²) = 5/36 and 1/λ′ ∝ (1/3² − 1/4²) = 7/144. So λ′/λ = (5/36)/(7/144) = 20/7, giving λ′ = 20λ/7. The 4→3 photon has less energy, so its wavelength must be longer than λ, which rules out the fractions smaller than 1.
Q7NEET 2016 Phase 1 · Q23
When an α-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze', its distance of closest approach from the nucleus depends on m as:
Not quite — the answer is A.
At closest approach all kinetic energy becomes potential energy: ½mv² = k(Ze)(2e)/r₀, so r₀ = 4kZe²/(mv²). With v fixed, r₀ ∝ 1/m: a heavier particle carries more kinetic energy and gets closer. A 1/√m or 1/m² dependence does not follow from this energy balance.
Q8NEET 2016 Phase 1 · Q38
Given the value of Rydberg constant is 10⁷ m⁻¹, the wave number of the last line of the Balmer series in hydrogen spectrum will be:
Not quite — the answer is C.
The last (series-limit) line of Balmer corresponds to n = ∞ → 2: wave number = R(1/2² − 0) = 10⁷/4 = 0.25 × 10⁷ m⁻¹. The first Balmer line would use 1/4 − 1/9 instead.
Q9AIPMT 2015
Consider 3rd orbit of He⁺ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = 9 × 10⁹ constant, Z = 2 and h (Planck's Constant) = 6.6 × 10⁻³⁴ J s]
Not quite — the answer is C.
Bohr speed v = 2πKZe²/(nh) = 2π × 9 × 10⁹ × 2 × (1.6 × 10⁻¹⁹)²/(3 × 6.6 × 10⁻³⁴) ≈ 1.46 × 10⁶ m/s (equivalently 2.19 × 10⁶ × Z/n). 2.92 × 10⁶ m/s would be the n = 1.5 value, i.e. a mistake in n.
Q10AIPMT 2014
Hydrogen atom in ground state is excited by a monochromatic radiation of λ = 975 Å. Number of spectral lines in the resulting spectrum emitted will be:
Not quite — the answer is B.
Photon energy = 12400/975 ≈ 12.7 eV, which matches E₄ − E₁ = 13.6(1 − 1/16) = 12.75 eV, so the atom reaches n = 4. Lines emitted = n(n − 1)/2 = 6.
Q11NEET 2013
Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is
Not quite — the answer is A.
Longest wavelengths: Lyman (2 → 1): 1/λ = R(1 − 1/4) = 3R/4, λ = 4/(3R). Balmer (3 → 2): 1/λ = R(1/4 − 1/9) = 5R/36, λ = 36/(5R). Ratio = (4/3)/(36/5) = 20/108 = 5/27.
Q12AIPMT 2012 Mains
The transition from the state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition from :
Not quite — the answer is D.
Infrared needs a much smaller energy gap than the ultraviolet 3 → 1 line. The energy factors (1/n₁² − 1/n₂²) are 3/4 for 2 → 1, 5/36 for 3 → 2, 3/16 for 4 → 2 and only 7/144 for 4 → 3, so 4 → 3 (Paschen-type) is the infrared one; 3 → 2 and 4 → 2 are visible for hydrogen.
Q13AIPMT 2012 Prelims
An electron of a stationary hydrogen atom passes from the fifth energy level to the ground level. The velocity that the atom acquired as a result of photon emission will be (m is the mass of the electron, R the Rydberg constant and h Planck's constant)
Not quite — the answer is B.
The photon energy for n = 5 → 1 is E = hcR(1 − 1/25) = (24/25)hcR. Its momentum p = E/c = (24/25)hR, and the atom recoils with equal momentum, so v = p/m = 24hR/(25m). The inverted forms have wrong dimensions for velocity.
Q14AIPMT 2012 Prelims
Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths λ₁ : λ₂ emitted in the two cases is
Not quite — the answer is A.
Third excited state n = 4 → second excited n = 3: 1/λ₁ = R(1/9 − 1/16) = 7R/144. Second excited n = 3 → first excited n = 2: 1/λ₂ = R(1/4 − 1/9) = 5R/36 = 20R/144. Ratio λ₁/λ₂ = 20/7. Taking excited state numbers as n values directly gives a wrong ratio.
Q15AIPMT 2011 Mains
An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, then the value of n is
Not quite — the answer is D.
Photon energy = work function + eV₀ = 2.75 + 10 = 12.75 eV. For hydrogen, 13.6(1 − 1/n²) = 12.75 gives 1/n² = 0.0625, so n = 4. The n = 3 → 1 transition gives only 12.09 eV, too little.
Q16AIPMT 2011 Prelims
The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is
Not quite — the answer is A.
Lyman first line: 1/λ = R(1 − 1/4) = 3R/4. Balmer second line (n = 4 → 2) for the ion: 1/λ = Z²R(1/4 − 1/16) = 3Z²R/16. Equating gives Z² = 4, Z = 2 (He⁺). Using n = 3 → 2 as the second Balmer line gives a non-integer Z.
Q17AIPMT 2010 Prelims · Q43
The energy of a hydrogen atom in the ground state is −13.6 eV. The energy of a He⁺ ion in the first excited state will be
Not quite — the answer is B.
Eₙ = −13.6 Z²/n² eV. For He⁺, Z = 2 and first excited state n = 2: E = −13.6 × 4/4 = −13.6 eV. −54.4 eV is the He⁺ ground state.
Q18AIPMT 2010 Prelims · Q45
A alpha nucleus of energy (1/2)mv² bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
Not quite — the answer is D.
At closest approach all kinetic energy becomes potential energy: ½mv² = (1/4πε₀)(Ze)(2e)/r₀, so r₀ = 4Ze²/(4πε₀mv²). Thus r₀ ∝ 1/m, ∝ Ze and ∝ 1/v², so 1/v⁴ and v² are wrong powers and 1/(Ze) has the wrong dependence.
Q19AIPMT 2009 Prelims · Q29
In a Rutherford scattering experiment when a projectile of charge z₁ and mass M₁ approaches a target nucleus of charge z₂ and mass M₂, the distance of closest approach is r₀. The energy of the projectile is:
Not quite — the answer is A.
At the distance of closest approach all kinetic energy becomes electrostatic potential energy: E = kz₁z₂e²/r₀. So, for a given r₀, the energy is directly proportional to z₁z₂ and does not depend on the masses.
Q20AIPMT 2009 Prelims · Q47
The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between:
Not quite — the answer is C.
Number of lines from level n = n(n − 1)/2 = 6, so n = 4. The longest wavelength needs the smallest energy gap, which is between n = 4 and n = 3. The 3 → 1 or 2 → 1 lines have larger gaps and shorter wavelengths.
Q21AIPMT 2008 Prelims · Q43
The ground state energy of hydrogen atom is −13.6 eV. When its electron is in the first excited state, its excitation energy is
Not quite — the answer is D.
The first excited state is n = 2 with energy −13.6/4 = −3.4 eV. Excitation energy = −3.4 − (−13.6) = 10.2 eV. 3.4 eV is the energy needed to ionise from n = 2, not the excitation energy from the ground state.
Q22AIPMT 2007 Prelims · Q35
The total energy of electron in the ground state of hydrogen atom is – 13.6 eV. The kinetic energy of an electron in the first excited state is
Not quite — the answer is D.
En = −13.6/n² eV, so in the first excited state (n = 2) E = −3.4 eV. For a Bohr orbit kinetic energy = −E = 3.4 eV. 13.6 eV is the ground-state value.
Q23AIPMT 2006 Prelims · Q12
Ionization potential of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy 12.1 eV. According to Bohr's theory, the spectral lines emitted by hydrogen will be :
Not quite — the answer is B.
After absorbing 12.1 eV the electron's energy is −13.6 + 12.1 = −1.51 eV, which is the n = 3 level (−13.6/9). From n = 3 the possible downward transitions are 3→2, 3→1 and 2→1, i.e. n(n − 1)/2 = 3 lines. Counting only the direct jump to the ground state would give one line.
Q24AIPMT 2005 Prelims · Q11
Energy levels A, B and C of a certain atom correspond to increasing values of energy i.e., E_A < E_B < E_C. If λ₁, λ₂ and λ₃ are wavelengths of radiations corresponding to transitions C to B, B to A and C to A respectively, which of the following relations is correct ?
Not quite — the answer is B.
Photon energies add: hc/λ₃ = hc/λ₁ + hc/λ₂, so 1/λ₃ = 1/λ₁ + 1/λ₂ and λ₃ = λ₁λ₂/(λ₁ + λ₂). Adding the wavelengths directly is wrong because energy, not wavelength, is additive.
Q25AIPMT 2005 Prelims · Q48
The total energy of an electron in the first excited state of hydrogen is about −3.4 eV. Its kinetic energy in this state is :
Not quite — the answer is D.
In a Bohr orbit KE = −E_total, so KE = +3.4 eV. Kinetic energy can never be negative, which rules out −3.4 eV and −1.7 eV.
How often Atoms came up
Questions from this chapter in each paper on this site.
- NEET 2020: 1
- NEET 2019: 1
- NEET 2018: 1
- NEET 2017: 1
- NEET 2017 (Gujarati paper): 1
- NEET 2016 Phase 2: 1
- NEET 2016 Phase 1: 2
- AIPMT 2015: 1
- AIPMT 2014: 1
- NEET 2013: 1
- AIPMT 2012 Mains: 1
- AIPMT 2012 Prelims: 2
- AIPMT 2011 Mains: 1
- AIPMT 2011 Prelims: 1
- AIPMT 2010 Prelims: 2
- AIPMT 2009 Prelims: 2
- AIPMT 2008 Prelims: 1
- AIPMT 2007 Prelims: 1
- AIPMT 2006 Prelims: 1
- AIPMT 2005 Prelims: 2
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