CUET UG Physics Booster Test 3-Atomic Structure and Masses -

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CUET UG Physics Booster Test 3-Atomic Structure and Masses

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 3-Atomic Structure and Masses

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QUESTION 1 OF 20

Incorrect statement about nuclear composition and mass:

Answer Review

1 Incorrect statement about nuclear composition and mass:

A The nucleus of the lightest hydrogen isotope constitutes a basic building block called a proton.

B The mass of a proton is mathematically equivalent to the mass of a hydrogen atom plus the mass of an electron.

C A proton carries exactly one fundamental unit of positive charge.

D Protons and neutrons together contribute to the vast majority of an atom's mass.

💡 Short Explanation

�� Proton is the nucleus of protium. �� Proton carries +e charge. �� Proton mass equals hydrogen mass minus electron mass.

✅ Correct Answer Explanation

Statement B is incorrect because a hydrogen atom contains both a proton and an electron. mH = mp + me Therefore, mp = mH − me Statements A, C and D are correct descriptions of nuclear composition and mass. Hence, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement about the proton.
�� Option C → Correct statement; proton carries +1 elementary charge.
�� Option D → Correct statement; nucleons contribute almost all atomic mass.

🎯 Strategy

Used

Elimination

Application:

Identify the statement inconsistent with proton mass relations.

Final Logic:

Proton mass is obtained by subtracting electron mass from hydrogen atom mass.

🧠 Memory Trick

Hydrogen = Proton + Electron

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Proton

2 Statements regarding Chadwick's hypothesis of a new neutral particle:
1. Conservation of energy and momentum showed photon radiation could not possess the energy required to knock out protons.
2. Beryllium spontaneously decayed into protons under normal conditions without bombardment.
3. Electrons could not mathematically exist inside the nuclear volume.
4. Alpha-particles fused entirely with beryllium to create an unstable carbon isotope.

A (1)

B (2)

C (3)

D (4)

💡 Short Explanation

�� Chadwick analyzed collision data. �� Gamma-ray explanation failed. �� Discovery led to the neutron.

✅ Correct Answer Explanation

Statement 1 is correct because conservation laws showed that if the radiation were photons, unrealistically high energies would be required to explain the observed proton ejection. Statement 2 is incorrect because beryllium does not spontaneously emit protons under normal conditions. Statement 3 is unrelated to Chadwick's primary reasoning. Statement 4 does not explain why the neutron hypothesis was proposed. Therefore, only statement 1 is correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement 2 is incorrect.
�� Option C → Statement 3 is not the basis of Chadwick's conclusion.
�� Option D → Statement 4 is incorrect in this context.

🎯 Strategy

Used

Elimination

Application:

Check which statement directly explains Chadwick's reasoning.

Final Logic:

Only statement 1 justifies the neutron hypothesis.

🧠 Memory Trick

Photon Failed → Neutron Found

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Discovery of Neutron

3 Statements evaluating the utility of the atomic mass unit:
1. It resolves the inconvenience of measuring minuscule atomic masses in kilograms.
2. Elements' atomic masses in (u) are generally close to integral multiples of a hydrogen atom's mass.
3. 1 u = 1.660539 × 10⁻²⁷ kg.
4. The unit relies completely on the mass of a free, isolated proton.

A (1), (2), (3)

B (1), (3), (4)

C (2), (3), (4)

D (1), (2), (4)

💡 Short Explanation

�� Kilogram is inconvenient at atomic scale. �� Atomic masses are expressed in u. �� Carbon-12 defines the unit.

✅ Correct Answer Explanation

Statement 1 is correct because atomic masses are extremely small when expressed in kilograms. Statement 2 is correct because many atomic masses are close to integral multiples of hydrogen mass. Statement 3 is correct because 1 u = 1.660539 × 10⁻²⁷ kg. Statement 4 is incorrect because the atomic mass unit is based on carbon-12, not a free proton. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 4, which is incorrect.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Compare each statement with the definition of atomic mass unit.

Final Logic:

1 u is based on carbon-12, not a proton.

🧠 Memory Trick

u → Carbon-12 Standard

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Atomic Mass Unit

4 Statements concerning mass spectrometry applications and findings:
1. It practically proves the existence of isotopes.
2. It accurately measures individual atomic masses.
3. It demonstrates elements are mixtures of atomic species with identical mass but different chemical properties.
4. It shows that practically every element consists of a mixture of several isotopes.

A (1), (2), (4)

B (1), (3)

C (2), (4)

D (1), (2), (3), (4)

💡 Short Explanation

�� Mass spectrometry measures atomic masses. �� It identifies isotopes. �� Most elements occur as isotopic mixtures.

✅ Correct Answer Explanation

Statement 1 is correct because mass spectrometry provides evidence for isotopes. Statement 2 is correct because it accurately measures atomic masses. Statement 3 is incorrect because isotopes have similar chemical properties but different masses, not identical masses with different chemical properties. Statement 4 is correct because most elements occur as mixtures of isotopes. Therefore, statements 1, 2 and 4 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 3, which is incorrect.
�� Option C → Omits statement 1, which is correct.
�� Option D → Includes statement 3, which is incorrect.

🎯 Strategy

Used

Elimination

Application:

Evaluate each statement using isotope concepts.

Final Logic:

Only statements 1, 2 and 4 are correct.

🧠 Memory Trick

Mass Spectrometer → Isotopes

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Mass Spectrometry

5 Match List I with List II.

List I	List II
1. ⁴⁰₁₈Ar and ⁴⁰₂₀Ca	a. Isobars
2. ¹⁴₆C and ¹⁵₇N	b. Isotones
3. ¹H and ²H	c. Isotopes
4. ¹H and ³H	d. Isotopes

A (1)-a, (2)-b, (3)-c, (4)-d

B (1)-b, (2)-c, (3)-a, (4)-d

C (1)-a, (2)-d, (3)-b, (4)-c

D (1)-c, (2)-a, (3)-d, (4)-b

💡 Short Explanation

�� Isobars have the same mass number. �� Isotones have the same neutron number. �� Isotopes have the same atomic number.

✅ Correct Answer Explanation

1 → a : ⁴⁰₁₈Ar and ⁴⁰₂₀Ca have the same mass number (A = 40) but different atomic numbers, so they are isobars. 2 → b : ¹⁴₆C and ¹⁵₇N both contain 8 neutrons, so they are isotones. 3 → c : ¹H and ²H have the same atomic number (Z = 1) but different mass numbers, so they are isotopes. 4 → d : ¹H and ³H also have the same atomic number (Z = 1) but different mass numbers, so they are isotopes. Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Isobars and isotones are incorrectly interchanged.
�� Option C → ¹⁴₆C and ¹⁵₇N are isotones, not isotopes.
�� Option D → Multiple incorrect pairings of nuclear species.

🎯 Strategy

Used

Option Grouping

Application:

Match each nuclide pair using the definitions of isotopes, isobars and isotones.

Final Logic:

Same A → Isobars; Same N → Isotones; Same Z → Isotopes.

🧠 Memory Trick

A-Bar, N-Tone, Z-Tope

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Isotopes, Isobars and Isotones

6 Given a hypothetical element X with exactly two isotopes of masses 10 u and 11 u, if the weighted average mass is 10.8 u, what is the fractional abundance (where 1 = 100%) of the 11 u isotope?

A 0.8

B 0.2

C 0.5

D 0.1

💡 Short Explanation

�� Use weighted average formula. �� Let abundance of 11 u isotope be x. �� Solve linear equation.

✅ Correct Answer Explanation

Let abundance of the 11 u isotope = x. Then abundance of the 10 u isotope = (1 − x). 10(1 − x) + 11x = 10.8 10 + x = 10.8 x = 0.8 Therefore, the fractional abundance of the 11 u isotope is 0.8. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Gives average mass 10.2 u.
�� Option C → Gives average mass 10.5 u.
�� Option D → Gives average mass 10.1 u.

🎯 Strategy

Used

Substitution

Application:

Apply weighted average mass formula.

Final Logic:

10(1−x) + 11x = 10.8 gives x = 0.8.

🧠 Memory Trick

Average Near 11 → High 11 Abundance

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Average Atomic Mass

7 Constituents of the Deuterium nucleus, Constituents of the Tritium nucleus:

A 1 proton + 1 neutron, 1 proton + 2 neutrons

B 1 proton + 0 neutrons, 1 proton + 1 neutron

C 2 protons + 1 neutron, 3 protons + 1 neutron

D 1 proton + 2 neutrons, 1 proton + 1 neutron

💡 Short Explanation

�� Deuterium contains one neutron. �� Tritium contains two neutrons. �� Both contain one proton.

✅ Correct Answer Explanation

Deuterium (²H) contains: 1 proton + 1 neutron Tritium (³H) contains: 1 proton + 2 neutrons Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents protium and deuterium.
�� Option C → Incorrect proton counts.
�� Option D → Deuterium and tritium are interchanged.

🎯 Strategy

Used

Definition Recall

Application:

Recall the composition of hydrogen isotopes.

Final Logic:

²H → 1p + 1n; ³H → 1p + 2n.

🧠 Memory Trick

D = One N, T = Two N

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Isotopes of Hydrogen

8 If mp is proton mass and mn is neutron mass, the expected mass of an unbound tritium nucleus configuration (before accounting for any mass defect) is:

A mp + mn

B 2mp + mn

C mp + 2mn

D 3mp

💡 Short Explanation

�� Tritium contains one proton. �� Tritium contains two neutrons. �� Mass defect ignored.

✅ Correct Answer Explanation

A tritium nucleus contains: 1 proton + 2 neutrons Ignoring mass defect, Mass = mp + 2mn Therefore, option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents deuterium.
�� Option B → Represents helium-3 composition.
�� Option D → Tritium does not contain three protons.

🎯 Strategy

Used

Substitution

Application:

Insert proton and neutron counts into the mass expression.

Final Logic:

Tritium = 1p + 2n.

🧠 Memory Trick

Tritium = Triple Mass Hydrogen

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Tritium

9 The mass of a carbon-12 atom is used as the universal standard for atomic masses primarily because

A it was discovered first by James Chadwick during alpha-particle bombardment.

B it provides a convenient scale (1 u = 1/12th of its mass) resulting in most isotopic masses being close to integral multiples of hydrogen's mass.

C it contains exactly equal numbers of electrons, protons, and antineutrinos in its core.

D carbon-12 has the highest theoretical binding energy per nucleon of all known elements.

💡 Short Explanation

�� Carbon-12 defines the atomic mass scale. �� 1 u is based on carbon-12. �� Provides a convenient reference standard.

✅ Correct Answer Explanation

Carbon-12 was chosen as the reference standard because: 1 u = (1/12) × mass of a carbon-12 atom This creates a convenient mass scale for atomic and isotopic masses. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Carbon-12 was not discovered by Chadwick.
�� Option C → Antineutrinos are not constituents of carbon nuclei.
�� Option D → Iron-group nuclei have higher binding energies.

🎯 Strategy

Used

Elimination

Application:

Identify the scientific basis of the carbon-12 standard.

Final Logic:

Carbon-12 provides a convenient and universal mass scale.

🧠 Memory Trick

Carbon-12 ÷ 12 = 1 u

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Carbon-12 Standard

10 If an alpha particle is simply a helium nucleus (⁴₂He), its charge in terms of the fundamental charge is:

A +2e

B +e

C +4e

D 0

💡 Short Explanation

�� Alpha particle is a helium nucleus. �� Helium has atomic number 2. �� Nuclear charge equals +Ze.

✅ Correct Answer Explanation

For a helium nucleus: Z = 2 Nuclear charge = +Ze = +2e Therefore, an alpha particle carries a charge of +2e. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Corresponds to one proton only.
�� Option C → Charge depends on proton count, not mass number.
�� Option D → Alpha particle is positively charged.

🎯 Strategy

Used

Substitution

Application:

Use Charge = +Ze with Z = 2.

Final Logic:

Helium nucleus contains two protons, giving charge +2e.

🧠 Memory Trick

Alpha = Helium = +2e

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Nuclear Charge and Alpha Particle

11 The mass number of a heavy nucleus is 235 and its atomic number is 92. How many neutrons does it possess?

A 92

B 143

C 235

D 327

💡 Short Explanation

�� Mass number = Protons + Neutrons. �� Atomic number = Protons. �� Neutrons = A − Z.

✅ Correct Answer Explanation

For a nucleus: A = Z + N Therefore, N = A − Z = 235 − 92 = 143 Hence, the nucleus contains 143 neutrons. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents the atomic number (protons).
�� Option C → Represents the mass number.
�� Option D → Obtained by adding A and Z, which is incorrect.

🎯 Strategy

Used

Substitution

Application:

Apply N = A − Z directly.

Final Logic:

235 − 92 = 143.

🧠 Memory Trick

Neutrons = Mass − Protons

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Mass Number and Neutron Number

12 Incorrect statement about nucleons and their behavior:

A The number of nucleons in an atom is strictly equal to its mass number A.

B Both protons and neutrons are exclusively classified as nucleons.

C Nucleons are the primary constituents of the atomic nucleus.

D The specific number of nucleons exactly determines the chemical properties of an element.

💡 Short Explanation

�� Nucleons are protons and neutrons. �� Mass number equals nucleon count. �� Chemical properties depend on electrons.

✅ Correct Answer Explanation

Chemical properties are determined primarily by the electronic configuration of an atom, which depends on atomic number and electron arrangement. Statements A, B and C are correct because: • Mass number equals total nucleons. • Nucleons are protons and neutrons. • Nucleons form the nucleus. Statement D is incorrect because nucleon count does not directly determine chemical properties. Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Correct definition of nucleons.
�� Option C → Correct statement regarding nuclear composition.

🎯 Strategy

Used

Elimination

Application:

Identify which statement contradicts atomic structure principles.

Final Logic:

Chemistry depends on electrons, not nucleon count.

🧠 Memory Trick

Chemistry → Electrons, Not Nucleons

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Nucleons

13 Statements regarding the concept of a tiny dense nucleus surrounded by electrons:
1. The volume of a nucleus is about 10⁻¹² times the volume of the atom, making the atom mostly empty space.
2. Electrons are scattered evenly throughout the nuclear volume in light atoms.
3. Nuclear mass is less than 0.1% of the total atomic mass due to heavy outer electrons.
4. The atomic radius is smaller than the nuclear radius by a massive factor of 10⁴.

A (1)

B (2)

C (3)

D (4)

💡 Short Explanation

�� Nucleus occupies extremely small volume. �� Most atomic volume is empty space. �� Electrons exist outside the nucleus.

✅ Correct Answer Explanation

Statement 1 is correct because the nuclear volume is approximately 10⁻¹² times the atomic volume, implying that atoms are largely empty space. Statement 2 is incorrect because electrons are outside the nucleus. Statement 3 is incorrect because almost all atomic mass resides in the nucleus. Statement 4 is incorrect because the atomic radius is much larger than the nuclear radius. Therefore, only statement 1 is correct.

❓ Why Other Options Are Incorrect

�� Option B → Statement 2 is incorrect.
�� Option C → Statement 3 is incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

Elimination

Application:

Compare each statement with the nuclear model of the atom.

Final Logic:

Only statement 1 correctly describes atomic structure.

🧠 Memory Trick

Atom = Mostly Empty Space

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Nuclear Model of Atom

14 Match List I with List II regarding charges in a neutral atom.

List I	List II
1. Nucleus total charge	a. Zero
2. Total electron charge	b. Zero
3. Atom net overall charge	c. +Ze
4. Single neutron charge	d. −Ze

A (1)-c, (2)-d, (3)-a, (4)-b

B (1)-d, (2)-c, (3)-b, (4)-a

C (1)-c, (2)-d, (3)-b, (4)-b

D (1)-a, (2)-b, (3)-c, (4)-d

💡 Short Explanation

�� Nucleus is positively charged. �� Electrons contribute negative charge. �� Neutron is neutral.

✅ Correct Answer Explanation

1 → c : Nucleus carries charge +Ze. 2 → d : Total electronic charge is −Ze. 3 → a : Neutral atom has net charge zero. 4 → b : Neutron has zero charge. Thus, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Nucleus and electron charges are interchanged.
�� Option C → Matching is incomplete and incorrect.
�� Option D → Multiple mismatches.

🎯 Strategy

Used

Option Grouping

Application:

Match each particle/system with its electrical charge.

Final Logic:

+Ze, −Ze, 0, 0 is the correct sequence.

🧠 Memory Trick

Nucleus Plus, Electrons Minus, Neutron Zero

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Electrical Neutrality

15 Statements relating to chlorine isotopes:
1. The atomic mass of the lighter isotope is 34.98 u.
2. The lighter isotope contributes roughly 75.4% to the weighted average mass.
3. Both the 34.98 u and 36.98 u isotopes behave identically in chemical reactions.
4. The heavier isotope has a precise mass of 36.98 u.

A (1), (2), (3), (4)

B (1), (2), (4)

C (2), (3), (4)

D (1), (3), (4)

💡 Short Explanation

�� Chlorine has two major isotopes. �� Lighter isotope is more abundant. �� Isotopes have similar chemical properties.

✅ Correct Answer Explanation

Statement 1 is correct because the lighter chlorine isotope has a mass of 34.98 u. Statement 2 is correct because its abundance is about 75.4%. Statement 3 is correct because isotopes have identical electronic configurations and hence nearly identical chemical behavior. Statement 4 is correct because the heavier isotope has a mass of 36.98 u. Therefore, all four statements are correct.

❓ Why Other Options Are Incorrect

�� Option B → Omits statement 3, which is correct.
�� Option C → Omits statement 1, which is correct.
�� Option D → Omits statement 2, which is correct.

🎯 Strategy

Used

Elimination

Application:

Verify each statement using chlorine isotope data.

Final Logic:

All four statements are correct.

🧠 Memory Trick

35 → 75%, 37 → 25%

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Chlorine Isotopes

16 If m₁ and m₂ are masses of two distinct isotopes with percentage abundances x and y (where x + y = 100), the average atomic mass is calculated by:

A (m₁ + m₂)/(x + y)

B (m₁ + m₂)/2

C (m₁m₂)/(x + y)

D (m₁x + m₂y)/100

💡 Short Explanation

�� Average atomic mass is weighted. �� Abundance must be considered. �� Simple mean is generally incorrect.

✅ Correct Answer Explanation

Average atomic mass is calculated by multiplying each isotopic mass by its percentage abundance and dividing by 100: Average Atomic Mass = (m₁x + m₂y)/100 Therefore, option D is correct.

❓ Why Other Options Are Incorrect

�� Option A → Not the weighted average formula.
�� Option B → Ignores isotopic abundances.
�� Option C → Incorrect mathematical expression.

🎯 Strategy

Used

Definition Recall

Application:

Recall the weighted-average formula.

Final Logic:

Mass × Abundance is the key principle.

🧠 Memory Trick

Weighted = Mass × Percent

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Average Atomic Mass

17 Statements comparing nucleons in free states:
1. A free proton is a fundamentally stable particle.
2. A free neutron is fundamentally unstable outside a nucleus.
3. A free neutron decays with a mean life of about 1000 s.
4. A free proton decays rapidly into an electron and an antineutrino.

A (1), (2), (3)

B (1), (3), (4)

C (2), (3), (4)

D (1), (4)

💡 Short Explanation

�� Proton is stable. �� Free neutron is unstable. �� Mean life is approximately 1000 s.

✅ Correct Answer Explanation

Statement 1 is correct because the free proton is stable. Statement 2 is correct because the free neutron is unstable. Statement 3 is correct because its mean life is approximately 1000 s. Statement 4 is incorrect because a free proton does not decay rapidly into an electron and an antineutrino. Therefore, statements 1, 2 and 3 are correct.

❓ Why Other Options Are Incorrect

�� Option B → Includes statement 4, which is incorrect.
�� Option C → Includes statement 4, which is incorrect.
�� Option D → Includes statement 4 and omits statements 2 and 3.

🎯 Strategy

Used

Elimination

Application:

Compare known stability properties of free nucleons.

Final Logic:

Only statements 1, 2 and 3 are correct.

🧠 Memory Trick

Proton Stable, Neutron Unstable

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Free Proton and Free Neutron

18 State of a free neutron, State of a free proton:

A Unstable, Stable

B Stable, Unstable

C Stable, Stable

D Unstable, Unstable

💡 Short Explanation

�� Free neutron undergoes decay. �� Free proton is stable. �� Neutron lifetime is finite.

✅ Correct Answer Explanation

A free neutron decays into a proton, electron and antineutrino and is therefore unstable. A free proton is considered stable. Hence, the correct combination is "Unstable, Stable". Therefore, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Stability states are reversed.
�� Option C → Free neutron is not stable.
�� Option D → Free proton is stable.

🎯 Strategy

Used

Definition Recall

Application:

Recall stability of free nucleons.

Final Logic:

Neutron decays, proton survives.

🧠 Memory Trick

Neutron Falls, Proton Persists

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Stability of Nucleons

19 In the symbolic representation of nuclides, the symbol X represents the chemical species, whereas

A the superscript A designates the number of atomic electrons.

B the subscript Z dictates its precise position within the periodic table by denoting atomic number.

C the symbol itself ensures the element contains zero isotopes.

D the mass number precisely limits the total nuclear binding energy directly.

💡 Short Explanation

�� X is the chemical symbol. �� Z is the atomic number. �� Atomic number determines periodic position.

✅ Correct Answer Explanation

In nuclide notation ᴬZX: • X = chemical symbol • Z = atomic number • A = mass number Atomic number determines the electronic configuration and periodic table position. Therefore, option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → A represents mass number, not electron count.
�� Option C → Elements may have multiple isotopes.
�� Option D → Binding energy is not determined directly by mass number alone.

🎯 Strategy

Used

Definition Recall

Application:

Use the standard nuclide notation.

Final Logic:

Z determines identity and periodic-table position.

🧠 Memory Trick

Z = Atomic Number = Position

📚 Class XI – NCERT Chemistry Part I – Chapter: Structure of Atom – Topic: Nuclide Notation

20 Consider the nuclide ²³⁵₉₂U. If this specific nucleus undergoes alpha decay (emitting a single alpha particle nucleus), what will be the new atomic number Z of the resulting nuclide fragment?

A 90

B 94

C 231

D 88

💡 Short Explanation

�� Alpha particle is ⁴₂He. �� Alpha decay reduces Z by 2. �� Alpha decay reduces A by 4.

✅ Correct Answer Explanation

During alpha decay: ²³⁵₉₂U → ⁴₂He + Daughter Nucleus Atomic number changes as: Z = 92 − 2 = 90 Therefore, the daughter nucleus has atomic number 90. Hence, option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Would require gain of two protons.
�� Option C → Represents neither atomic number nor daughter nucleus charge.
�� Option D → Reduction by four is incorrect.

🎯 Strategy

Used

Substitution

Application:

Apply alpha decay rule: Z decreases by 2.

Final Logic:

92 − 2 = 90.

🧠 Memory Trick

Alpha = Minus 2 Protons

📚 Class XII – NCERT Physics Part II – Chapter: Nuclei – Topic: Alpha Decay

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