Structure of Atom
Topic: Sub-Atomic Particles and Atomic Models
Q1. Given below are two statements: one is labelled as Assertion (A) and the other as
Reason (R).
Assertion (A): Loss of electron from hydrogen atom results in nucleus of ∼ 1.5 × 10−3 pm
size.
Reason (R): Proton (H+ ) always exists in combined form.
In the light of the above statements, choose the most appropriate answer from the options
given below:
(a) Both (A) and (R) are correct and (R) (b) (A) is correct but (R) is not correct
is the correct explanation of (A)
(c) (A) is not correct but (R) is correct (d) Both (A) and (R) are correct but (R)
is not the correct explanation of (A)
JEE Main 06 April, 2023 (Shift-I)
Solution
Step 1: Removing the electron from H gives a proton (size ∼ 1–1.5 × 10−15 m = 1–1.5 × 10−3 pm)
⇒ (A) true.
Step 2: In ordinary media, a free proton exists as combined species (e.g. H3 O+ ) ⇒ (R) true.
Last 7 Year PYQ Questions
Step 3: But (R) does not explain the size mentioned in (A). Hence option (d).
Q2. Electrons in a cathode ray tube have been emitted with a velocity of 1000 m s−1 .
RANSHO
The number of the following statements that are true about the emitted radiation is .
Given: h = 6 × 10−34 J s, me = 9 × 10−31 kg.
(1) The de Broglie wavelength is 666.67 nm.
(2) The characteristics of electrons depend on the material of the electrodes.
(3) Cathode rays start from cathode and move towards anode.
(4) The nature of emitted electrons depends on the nature of the gas present in the tube.
JEE Main 01 Feb, 2023 (Shift-I)
Solution
h 6 × 10−34
Step 1: λ = = = 6.67 × 10−7 m = 666.7 nm ⇒ (1) true.
mv (9 × 10−31 )(103 )
Step 2: Cathode ray properties are independent of electrode material and gas ⇒ (2) false, (4)
false.
Step 3: They originate at cathode and move to anode ⇒ (3) true.
Number true = 2 ⇒ 2 .
Q3. If the Thomson model of the atom were correct, then the result of Rutherford’s gold
foil experiment would have been:
1
Covalent 2.O Physical Chemistry © Copyright @ Ransho Revision
, Structure of Atom
(a) All α-particles get bounced back by 180◦ (b) α-particles are deflected over a wide
range of angles
(c) All α-particles pass through the gold (d) α-particles pass through the gold foil
foil without decrease in speed deflected by small angles and with reduced
speed
JEE Main 27 July, 2021 (Shift-II)
Solution
Step 1: In Thomson’s diffuse positive charge model, the field is weak ⇒ only small deflections;
no large-angle scattering.
Step 2: Energy loss is minimal (speed nearly unchanged), but the key prediction is small-angle
deflection for all. Closest option: (d).
Q4. Amongst the following statements, that which was not proposed by Dalton was:
(a) Matter consists of invisible atoms (b) When gases combine in a chemical
reaction they do so in a simple ratio by
volume at the same T and P
(c) Chemical reactions involve (d) All atoms of a given element have
Last 7 Year PYQ Questions
reorganisation of atoms; atoms are neither identical properties (including mass);
created nor destroyed atoms of different elements differ in mass
JEE Main 07 Jan, 2020 (Shift-I)
Solution
RANSHO
Step 1: (a), (c), (d) are Dalton’s postulates.
Step 2: (b) is Gay–Lussac’s law of combining volumes, not Dalton’s. Hence (b).
Q5. Consider an imaginary ion 48 3−
22 X . The nucleus contains ‘a%‘ more neutrons than the
number of electrons in the ion. The value of a is (nearest integer).
JEE Main 26 July, 2022 (Shift-II)
Solution
Step 1: For A
Z X : protons = Z, neutrons N = A − Z, electrons e = Z + q.
q−
Step 2: Here Z = 22, A = 48, q = 3. So N = 48 − 22 = 26; e = 22 + 3 = 25.
Step 3: Percentage more neutrons than electrons:
N −e 26 − 25
a= × 100 = × 100 = 4%.
e 25
Step 4: Nearest integer: 4 .
2
© Copyright @ Ransho Revision Covalent 2.O Physical Chemistry
, Structure of Atom
Topic: Developments Leading to Bohr’s Model of Atom
Q6. Heat treatment of muscular pain involves radiation of wavelength about 900 nm.
Which spectral line of H atom is suitable for this?
(a) Lyman series, ∞ → 1 (b) Balmer series, ∞ → 2
(c) Paschen series, 5 → 3 (d) Paschen series, ∞ → 3
JEE Main 23 Jan, 2025 (Shift-I)
Solution
1 1 1
Step 1: For H: = RH 2 − 2 .
λ nf ni
1
Step 2: For ∞ → 3: λ = RH
9 ⇒ λ = 9 × 10−5 cm = 900 nm.
Step 3: Hence correct option: (d) (Paschen limit).
Q7. Assertion (A): In the photoelectric effect, electrons are ejected as soon as light of
frequency greater than threshold strikes the metal.
Reason (R): When the photon of any energy strikes an electron in the atom, transfer of
Last 7 Year PYQ Questions
energy from photon to electron takes place.
Choose the most appropriate answer:
(a) Both (A) and (R) are correct but (R) (b) (A) is correct but (R) is not correct
is NOT the correct explanation of (A)
(c) Both (A) and (R) are correct and (R) (d) (A) is not correct but (R) is correct
RANSHO
is the correct explanation of (A)
JEE Main 11 April, 2023 (Shift-I)
Solution
Step 1: (A) True: emission is instantaneous when ν > ν0 .
Step 2: (R) False: energy transfer requires absorption; photons of any energy do not necessarily
transfer energy (and below ν0 no emission).
Step 3: Hence option (b).
Q8. The number of incorrect statement(s) about a black body is .
(1) Emits/absorbs energy as electromagnetic radiation.
(2) Frequency distribution of emitted radiation depends on temperature.
(3) At a given temperature, intensity–frequency curve has a maximum.
(4) The peak of the curve shifts to higher frequency at higher temperature.
JEE Main 10 April, 2023 (Shift-I)
Solution
Step 1: (1)–(4) are all standard properties (Planck law, Wien’s law).
Step 2: Number incorrect = 0 .
3
Covalent 2.O Physical Chemistry © Copyright @ Ransho Revision
, Structure of Atom
Q9. The number of the following statements which is/are incorrect is .
(1) Line emission spectra are used to study electronic structure.
(2) Emission spectra of atoms in the gas phase show a continuous spread of wavelength
from red to violet.
(3) An absorption spectrum is like the photographic negative of an emission spectrum.
(4) Helium was discovered in the Sun by spectroscopic method.
JEE Main 08 April, 2023 (Shift-I)
Solution
Step 1: (1) True; (2) False — atomic emission is line, not continuous; (3) True; (4) True.
Step 2: Number incorrect = 1 .
Q10. The ionization energy of sodium in kJ mol−1 , if electromagnetic radiation of
wavelength 242 nm is just sufficient to ionize sodium atom, is .
JEE Main 31 Jan, 2024 (Shift-I)
Solution
hc hc
Step 1: Ephoton = ; per mole Em = NA .
Last 7 Year PYQ Questions
λ λ
Step 2: Substitute: h = 6.626 × 10−34 J s, c = 3.0 × 108 m s−1 , λ = 242 × 10−9 m.
6.626 × 10−34 × 3.0 × 108
Step 3: Em = × 6.022 × 1023 ≈ 4.95 × 102 kJ mol−1 .
242 × 10−9
RANSHO
Step 4: Nearest integer: 495 .
Q11. The wavenumber for radiation of wavelength 5800 Å is x × 104 cm−1 . The value of
x is .
JEE Main 08 April, 2024 (Shift-II)
Solution
1
Step 1: ν̃ = ; 5800 Å = 5.8 × 10−5 cm.
λ(cm)
1
Step 2: ν̃ = = 1.724 × 104 cm−1 .
5.8 × 10−5
Step 3: Hence x = 1.724 ( 1.72).
Q12. The wavelength of an electron of kinetic energy 4.50 × 10−29 J is ×10−5 m (nearest
integer).
Given: me = 9 × 10−31 kg, h = 6.6 × 10−34 J s.
JEE Main 06 April, 2023 (Shift-I)
4
© Copyright @ Ransho Revision Covalent 2.O Physical Chemistry
Topic: Sub-Atomic Particles and Atomic Models
Q1. Given below are two statements: one is labelled as Assertion (A) and the other as
Reason (R).
Assertion (A): Loss of electron from hydrogen atom results in nucleus of ∼ 1.5 × 10−3 pm
size.
Reason (R): Proton (H+ ) always exists in combined form.
In the light of the above statements, choose the most appropriate answer from the options
given below:
(a) Both (A) and (R) are correct and (R) (b) (A) is correct but (R) is not correct
is the correct explanation of (A)
(c) (A) is not correct but (R) is correct (d) Both (A) and (R) are correct but (R)
is not the correct explanation of (A)
JEE Main 06 April, 2023 (Shift-I)
Solution
Step 1: Removing the electron from H gives a proton (size ∼ 1–1.5 × 10−15 m = 1–1.5 × 10−3 pm)
⇒ (A) true.
Step 2: In ordinary media, a free proton exists as combined species (e.g. H3 O+ ) ⇒ (R) true.
Last 7 Year PYQ Questions
Step 3: But (R) does not explain the size mentioned in (A). Hence option (d).
Q2. Electrons in a cathode ray tube have been emitted with a velocity of 1000 m s−1 .
RANSHO
The number of the following statements that are true about the emitted radiation is .
Given: h = 6 × 10−34 J s, me = 9 × 10−31 kg.
(1) The de Broglie wavelength is 666.67 nm.
(2) The characteristics of electrons depend on the material of the electrodes.
(3) Cathode rays start from cathode and move towards anode.
(4) The nature of emitted electrons depends on the nature of the gas present in the tube.
JEE Main 01 Feb, 2023 (Shift-I)
Solution
h 6 × 10−34
Step 1: λ = = = 6.67 × 10−7 m = 666.7 nm ⇒ (1) true.
mv (9 × 10−31 )(103 )
Step 2: Cathode ray properties are independent of electrode material and gas ⇒ (2) false, (4)
false.
Step 3: They originate at cathode and move to anode ⇒ (3) true.
Number true = 2 ⇒ 2 .
Q3. If the Thomson model of the atom were correct, then the result of Rutherford’s gold
foil experiment would have been:
1
Covalent 2.O Physical Chemistry © Copyright @ Ransho Revision
, Structure of Atom
(a) All α-particles get bounced back by 180◦ (b) α-particles are deflected over a wide
range of angles
(c) All α-particles pass through the gold (d) α-particles pass through the gold foil
foil without decrease in speed deflected by small angles and with reduced
speed
JEE Main 27 July, 2021 (Shift-II)
Solution
Step 1: In Thomson’s diffuse positive charge model, the field is weak ⇒ only small deflections;
no large-angle scattering.
Step 2: Energy loss is minimal (speed nearly unchanged), but the key prediction is small-angle
deflection for all. Closest option: (d).
Q4. Amongst the following statements, that which was not proposed by Dalton was:
(a) Matter consists of invisible atoms (b) When gases combine in a chemical
reaction they do so in a simple ratio by
volume at the same T and P
(c) Chemical reactions involve (d) All atoms of a given element have
Last 7 Year PYQ Questions
reorganisation of atoms; atoms are neither identical properties (including mass);
created nor destroyed atoms of different elements differ in mass
JEE Main 07 Jan, 2020 (Shift-I)
Solution
RANSHO
Step 1: (a), (c), (d) are Dalton’s postulates.
Step 2: (b) is Gay–Lussac’s law of combining volumes, not Dalton’s. Hence (b).
Q5. Consider an imaginary ion 48 3−
22 X . The nucleus contains ‘a%‘ more neutrons than the
number of electrons in the ion. The value of a is (nearest integer).
JEE Main 26 July, 2022 (Shift-II)
Solution
Step 1: For A
Z X : protons = Z, neutrons N = A − Z, electrons e = Z + q.
q−
Step 2: Here Z = 22, A = 48, q = 3. So N = 48 − 22 = 26; e = 22 + 3 = 25.
Step 3: Percentage more neutrons than electrons:
N −e 26 − 25
a= × 100 = × 100 = 4%.
e 25
Step 4: Nearest integer: 4 .
2
© Copyright @ Ransho Revision Covalent 2.O Physical Chemistry
, Structure of Atom
Topic: Developments Leading to Bohr’s Model of Atom
Q6. Heat treatment of muscular pain involves radiation of wavelength about 900 nm.
Which spectral line of H atom is suitable for this?
(a) Lyman series, ∞ → 1 (b) Balmer series, ∞ → 2
(c) Paschen series, 5 → 3 (d) Paschen series, ∞ → 3
JEE Main 23 Jan, 2025 (Shift-I)
Solution
1 1 1
Step 1: For H: = RH 2 − 2 .
λ nf ni
1
Step 2: For ∞ → 3: λ = RH
9 ⇒ λ = 9 × 10−5 cm = 900 nm.
Step 3: Hence correct option: (d) (Paschen limit).
Q7. Assertion (A): In the photoelectric effect, electrons are ejected as soon as light of
frequency greater than threshold strikes the metal.
Reason (R): When the photon of any energy strikes an electron in the atom, transfer of
Last 7 Year PYQ Questions
energy from photon to electron takes place.
Choose the most appropriate answer:
(a) Both (A) and (R) are correct but (R) (b) (A) is correct but (R) is not correct
is NOT the correct explanation of (A)
(c) Both (A) and (R) are correct and (R) (d) (A) is not correct but (R) is correct
RANSHO
is the correct explanation of (A)
JEE Main 11 April, 2023 (Shift-I)
Solution
Step 1: (A) True: emission is instantaneous when ν > ν0 .
Step 2: (R) False: energy transfer requires absorption; photons of any energy do not necessarily
transfer energy (and below ν0 no emission).
Step 3: Hence option (b).
Q8. The number of incorrect statement(s) about a black body is .
(1) Emits/absorbs energy as electromagnetic radiation.
(2) Frequency distribution of emitted radiation depends on temperature.
(3) At a given temperature, intensity–frequency curve has a maximum.
(4) The peak of the curve shifts to higher frequency at higher temperature.
JEE Main 10 April, 2023 (Shift-I)
Solution
Step 1: (1)–(4) are all standard properties (Planck law, Wien’s law).
Step 2: Number incorrect = 0 .
3
Covalent 2.O Physical Chemistry © Copyright @ Ransho Revision
, Structure of Atom
Q9. The number of the following statements which is/are incorrect is .
(1) Line emission spectra are used to study electronic structure.
(2) Emission spectra of atoms in the gas phase show a continuous spread of wavelength
from red to violet.
(3) An absorption spectrum is like the photographic negative of an emission spectrum.
(4) Helium was discovered in the Sun by spectroscopic method.
JEE Main 08 April, 2023 (Shift-I)
Solution
Step 1: (1) True; (2) False — atomic emission is line, not continuous; (3) True; (4) True.
Step 2: Number incorrect = 1 .
Q10. The ionization energy of sodium in kJ mol−1 , if electromagnetic radiation of
wavelength 242 nm is just sufficient to ionize sodium atom, is .
JEE Main 31 Jan, 2024 (Shift-I)
Solution
hc hc
Step 1: Ephoton = ; per mole Em = NA .
Last 7 Year PYQ Questions
λ λ
Step 2: Substitute: h = 6.626 × 10−34 J s, c = 3.0 × 108 m s−1 , λ = 242 × 10−9 m.
6.626 × 10−34 × 3.0 × 108
Step 3: Em = × 6.022 × 1023 ≈ 4.95 × 102 kJ mol−1 .
242 × 10−9
RANSHO
Step 4: Nearest integer: 495 .
Q11. The wavenumber for radiation of wavelength 5800 Å is x × 104 cm−1 . The value of
x is .
JEE Main 08 April, 2024 (Shift-II)
Solution
1
Step 1: ν̃ = ; 5800 Å = 5.8 × 10−5 cm.
λ(cm)
1
Step 2: ν̃ = = 1.724 × 104 cm−1 .
5.8 × 10−5
Step 3: Hence x = 1.724 ( 1.72).
Q12. The wavelength of an electron of kinetic energy 4.50 × 10−29 J is ×10−5 m (nearest
integer).
Given: me = 9 × 10−31 kg, h = 6.6 × 10−34 J s.
JEE Main 06 April, 2023 (Shift-I)
4
© Copyright @ Ransho Revision Covalent 2.O Physical Chemistry
