NEET Previous Year Questions (2014-2024): Atoms | Physics Class 12 PDF Download

2024

Q1: Given below are two statements:
Statement I: Atoms are electrically neutral as they contain equal number of positive and negative charges.
Statement II: Atoms of each element are stable and emit their characteristic spectrum.
In the light of the above statements, choose the most appropriate answer from the options given below.
(a) Both Statement I and Statement II are correct
(b) Both Statement I and Statement II are incorrect
(c) Statement I is correct but Statement II is incorrect
(d) Statement I is incorrect but Statement II is correct     [2024]
Ans: (c)

Let's analyze each statement one by one to determine the correctness:

• Statement I: "Atoms are electrically neutral as they contain equal number of positive and negative charges."
  This statement is correct. Atoms are composed of protons, neutrons, and electrons. Protons have a positive charge, electrons have a negative charge, and neutrons have no charge. In a neutral atom, the number of protons (positive charges) equals the number of electrons (negative charges), thus making the atom electrically neutral. The positive and negative charges balance each other, resulting in no overall charge.
• Statement II: "Atoms of each element are stable and emit their characteristic spectrum."
  This statement requires modification for accuracy. Atoms emit their characteristic spectrum when they are excited and then return to a lower energy state. The emission of a spectrum is not linked directly to the stability of an atom; it is linked to the electronic transitions within the atom. The term "stable" in relation to atoms typically refers to the overall energy state of an atom being at its lowest or in a ground state. However, atoms can emit radiation whether they are in a stable ground state or in an excited state. Moreover, not all atoms are inherently stable; some may be radioactive and unstable. The part that atoms emit their characteristic spectrum under certain conditions (like excitation) is correct, but linking this property strictly to stability is misleading.

Based on the analysis:
Statement I is correct.
Statement II is partially correct, as it is accurate that each element can emit a characteristic spectrum when excited, but incorrect in generalizing that all such atoms are stable.
Therefore, the correct option is:
Option C: Statement I is correct but Statement II is incorrect.

Q2: Match List I with List II:Choose the correct answer from the options given below:
(a) A-II, B-I, C-IV, D-III
(b) A-III, B-IV, C-II, D-I
(c) A-IV, B-III, C-I, D-II
(d) A-I, B-II, C-III, D-IV      [2024]
Ans: (b)

2023

Q.2. In hydrogen spectrum, the shortest wavelength in the Balmer series is λ. The shortest wavelength in the Bracket series is         (2023)  
(a) 2λ
(b) 4λ
(c) 9λ
(d) 16λ
Ans: b

λ = 4/R ...(1)
For Bracket, (n₂ = 4, n₁ = ∞)

λ' = 16/R ...(2)

Q.3. The radius of inner most orbit of hydrogen atom is 5.3 × 10 ^–11 m. What is the radius of third allowed orbit of hydrogen atom?         (2023)  
(a) 0.53 Å
(b) 1.06 Å
(c) 1.59 Å
(d) 4.77 Å
Ans: d

r₂ = 9r₁ = 5.3 x 10^-11 x 9
= 47.7 × 10^–11
= 4.77 Å

2022

Q.4. Let T1 and T2 be the energy of an electron in the first and second excited states of hydrogen atoms, respectively. According to Bohr’s model of an atom, the ratio T1 : T2 is         (2022)  
(a) 4 : 1 
(b) 4 : 9 
(c) 9 : 4 
(d) 1 : 4
Ans: c

En = E_o / n² 
For first excited state: n = 2
For second excited state  n = 3  

2020

Q.5. For which one of the following, Bohr model is not valid?     (2020)
(a) Deuteron atom
(b) Singly ionised neon atom (Ne⁺)
(c) Hydrogen atom
(d) Singly ionised helium atom (He⁺)
Ans: b

Bohr Model is valid only for those atoms which have one electron in orbit.

Bohr Model

2019

Q.6. The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are, respectively:    (2019)
(a) –3.4 eV, –3.4 eV
(b) –3.4 eV, –6.8 eV
(c) 3.4 eV, –6.8 eV
(d) 3.4 eV, 3.4 eV
Ans: c

In Bohr's model of H atom

∴ K.E. = 3.4 eV
U = –6.8 eV

Q.7. α-particle consists of :    (2019)
(a) 2 protons and 2 neutrons only
(b) 2 electrons, 2 protons and 2 neutrons
(c) 2 electrons and 4 protons only
(d) 2 protons only
Ans: a

α-particle is nucleus of Helium which has two protons and two neutrons.

2018

Q.8. The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom,is :-    (2018)
(a) 1 : 1
(b) 1 : –1
(c) 2 : –1
(d) 1 : –2
Ans: b

2017

Q.9. Suppose the charge of a proton and an electron differ slightly. One of them is – e, the other is(e + De). If the net of electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then De is of the order of [Given mass of hydrogen mh = 1.67 × 10^–27 kg]     (2017)
(a) 10^–23 C
(b) 10^–37 C
(c) 10^–47 C
(d) 10^–20 C
Ans: b

We know, a hydrogen atom has one electron and one proton
so, net charge on each hydrogen atom = (e + ∆e - e) = ∆e
so, electrostatic force ,
gravitational force,
where M_h denotes mass of hydrogen
a/c to question,
gravitational force = electrostatic force


we know, K = 9 × 10⁹ Nm^²/C^², G = 6.67 × 10^-11 Nm^²/Kg^² , M_h = 1.67 × 10^-27 Kg
so, 9 × 10⁹ × ∆e^² = 6.67 × 10^-11 × (1.67 × 10^-27)^²
or, ∆e^² = 6.67 × 10^-11 × 1.67 × 1.67 × 10^-54/9 × 10⁹
or, ∆e = 1.43767 × 10^-37 C

Q.10. The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is :-     (2017)
(a) 1
(b) 4
(c) 0.5
(d) 2
Ans: b

For last line of Balmer : n₁ = 2 & n₂ = ∞

For last line Lyman series : n₁ = 1 & n₂ = ∞

2016

Q.11. When an α-particle of mass 'm' moving with velocity 'v' bombards on a heavy nucleus of charge 'Ze' its distance of closet approach from the nucleus depends on m as :    (2016)
(a) m
(b) 1/m
(c) 1/√m
(d) 1/m²
Ans: b

When an alpha particle moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy.
Initial Kinetic energy of the alpha particle = Potential energy of alpha particle at closest approach.
That is,


This is the required distance of closest approach to alpha particle from the nucleus.
When an alpha particle moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy.
Initial Kinetic energy of the alpha particle = Potential energy of alpha particle at closest approach.
That is,


This is the required distance of closest approach to alpha particle from the nucleus.

Q.12. 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 :     (2016)
(a) 2.5 x 10⁷m^-1
(b) 0.025 x 10⁴m^-1
(c) 0.5 x 10⁷m^-1
(d) 0.25 x 10⁷m^-1
Ans: d

Rydberg constant, r = 10⁷ m^-1
For last line in Balmer series,
n₂ = ∞; n₁ = 2
We know,

2015

Q.13. Consider 3^rd orbit of He⁺ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K = 9 x 10⁹ constant, Z = 2 and h(Planck's Constant) = 6.6 x 10^-34 J s]    (2015)
(a) 3.0 x 10⁸ m/s
(b) 2.92 x 10⁶ m/s
(c) 1.46 x 10⁶ m/s
(d) 0.73 x 10⁶ m/s
Ans: c

Q.14. If radius of thenucleus is taken to be R_Al, then the radius ofnucleus is nearly :    (2015)

(a)

Radius of the nucleus goes as R ∝ A^1/3, where A is the atomic mass. If R_Te is the radius of the nucleus of telurium atom and R_Al is the radius of the nucleus of aluminium atom we have

There is no Question for NEET 2021