Class 9 Science Chapter 4 Structure of The Atom

Key Concept: Impact parameter & scattering angle, Nuclear size estimation

c) Scattering angle increases but less than doubles

[Solution Description]

Rutherford’s scattering formula states $b \propto \cot\left(\frac{\theta}{2}\right)$. If $b$ is halved, $\cot\left(\frac{\theta}{2}\right)$ also halves. To find the new angle, let original cotangent value = $2x$, then new value = $x$. Therefore, $\frac{\theta_{\text{new}}}{2}$ must decrease to increase $\tan\left(\frac{\theta_{\text{new}}}{2}\right)$ by a factor of 2. This implies $\theta_{\text{new}}$ increases. Using exact relationship: $\cot(\theta_1/2)/\cot(\theta_2/2) = b_1/b_2 = 2$. Solving shows $\theta_2 > \theta_1$.

Key Concept: Shell Capacity & Sequential Filling

b) 19

[Solution Description]

Step 1: Fill K-shell ($2 \times 1^2 = 2$).

Step 2: Fill L-shell ($2 \times 2^2 = 8$).

Step 3: Fill M-shell up to 8 electrons (as outermost shell follows valence limit). Total electrons = $2 + 8 + 8 = 18$.

Step 4: To occupy N-shell, add 1 electron. Total atomic number = $18 + 1 = 19$.

Key Concept: Average Atomic Mass Calculation

c) 63.6 u

[Solution Description]

Formula: Weighted average = (Fraction1 × Mass1) + (Fraction2 × Mass2)

First convert   to fractions: 70% = 0.7 and 30% = 0.3

Calculation: (63 × 0.7) + (65 × 0.3)

63 × 0.7 = 44.1

65 × 0.3 = 19.5

Total = 44.1 + 19.5 = 63.6 u

Key Concept: Nuclear Decay, Isobar Formation

b) ${}^{234}_{91}\text{Pa}$

[Solution Description]

$\beta^-$ decay converts a neutron to a proton, increasing atomic number by 1. Mass number remains unchanged. Original nucleus: ${}^{234}_{90}\text{Th}$. New atomic number = 91, mass number = 234. The product is ${}^{234}_{91}\text{Pa}$, which is an isobar of thorium-234.

Key Concept: Isotopes and Neutrons

c) Number of neutrons

[Solution Description]

Isotopes have the same atomic number ($Z$) but different mass numbers ($A$). Using $N = A – Z$, the difference in mass number implies a difference in neutron count. Carbon-12 has $12 – 6 = 6$ neutrons, while Carbon-14 has $14 – 6 = 8$ neutrons.

Key Concept: Atom Notation

c) 23

[Solution Description]

In atomic notation $^A_Z\text{X}$, the superscript $A$ is the mass number. Here, $A = 23$ (superscript value).

Key Concept: Rule 1 Valency Calculation

d) 4

[Solution Description]

For elements with 1–4 valence electrons, valency equals the number of valence electrons. Here, valence electrons = 4. Therefore, valency = 4.

Key Concept: Electron Distribution Calculation

c) 3

[Solution Description] The total electrons are 13. First shell (K) holds $2 \times 1^2 = 2$ electrons. Second shell (L) holds $2 \times 2^2 = 8$ electrons. Remaining electrons are $13 – 2 – 8 = 3$, which go to the M-shell.

Key Concept: Valency Calculation – Non-Metal

a) 3

[Solution Description] Phosphorus needs 8 electrons in its outermost shell for stability. Since it has 5 valence electrons, it gains $8 – 5 = 3$ electrons. Valency is the number of electrons gained, so the valency is 3.

Key Concept: Noble Gas Valency

c) They have a completely filled outermost shell

[Solution Description]

Noble gases have 8 electrons in their outermost shell (except helium, which has 2). Since their outermost shell is already complete, they neither gain nor lose electrons. Thus, their combining capacity (valency) is 0.

Key Concept: Valence electron significance

d) High reactivity to gain one electron

[Solution Description]

Valence electrons determine reactivity. A nearly full shell (needing one electron to complete 8) indicates high reactivity. Noble gases with 8 electrons are stable.

Key Concept: Rutherford’s α-scattering experiment observations vs conclusions

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Solution:

– Observation (i) states “Most α-particles passed straight through”.

– Conclusion (i) concludes “Most space inside atom is empty” because no obstruction caused deflection.

– Both Assertion and Reason match experimental data.

– The Reason directly explains why the Assertion happened (empty space allows passage).

– Thus, both are true with correct explanation.

Key Concept: Charge Distribution, Electrical Neutrality

c) The charge density remains unchanged

[Solution Description]

Original positive charge density = $\frac{Q}{\frac{4}{3}\pi R^{3}}$. Electrons have charge $-ne$ where $e$ is elementary charge. By Thomson’s model, $Q = ne$ for neutrality. Removing two electrons doesn’t alter the positive sphere since electrons are separate. Thus, charge density remains $\frac{ne}{\frac{4}{3}\pi R^{3}}$.

Key Concept: Electric Force Inside Sphere

c) $\frac{kQe}{2R^2}$

[Solution Description]

Step 1: For a uniformly charged sphere, the electric field inside at distance $r$ is $E = \frac{kQr}{R^3}$.

Step 2: At $r = R/2$, $E = \frac{kQ(R/2)}{R^3} = \frac{kQ}{2R^2}$.

Step 3: Force on electron $F = eE = \frac{kQe}{2R^2}$.

Key Concept: Element Identity Determination

a) Number of protons

[Solution Description]

The atomic number ($Z$) determines an element’s identity because every element has a unique number of protons. Even if neutron count varies (isotopes), proton count remains fixed.

Key Concept: Proton mass comparison

b) 2000 times

[Solution Description]

Syllabus states: “Mass: $\sim 2000 \times$ mass of electron”. Proton’s mass is nearly 1836 times electron’s mass, but approximate value given in syllabus ($\sim$2000) takes priority for correct answer.

Key Concept: Neutron-Proton Stability

c) Yes, because N > Z

[Solution Description]

For heavy elements (atomic number >20), stable nuclei require more neutrons than protons. Here, protons (26) and neutrons (30). Neutron-to-proton ratio = 30/26 ≈ 1.15. Heavy elements (like Iron, Z=26) need $N > Z$ for stability. Since 30 >26, the nucleus is stable.

Key Concept: Rutherford’s α-particle scattering experiment

c) Atoms contain a small, dense nucleus

[Solution Description]

Rutherford concluded that atoms have mostly empty space since most α-particles passed through undeflected. The dense nucleus occupies a tiny volume, causing only occasional deflections. This ruled out Thomson’s uniform sphere model.

Key Concept: Prediction vs Reality Conflict

c) Long-term stability of matter structures

[Solution Description] Rutherford’s model predicted atomic collapse due to energy loss via radiation. However in reality, atoms remain stable without electron-nucleus collisions. This contradiction exposes model’s drawback.

Key Concept: Properties of Protons and Electrons

b) Electron mass is 2000 times the mass of a hydrogen atom

[Solution Description]

Electrons have a mass of approximately $\frac{1}{2000}$ times that of a hydrogen atom, while protons have nearly the same mass as a hydrogen atom. Charge of proton is +1 and electron is -1. Atoms contain protons internally, not electrons. So, the incorrect statement is “Electron mass is 2000 times the mass of a hydrogen atom”.

Key Concept: Energy Transition Between Orbits

c) Photon emission occurs with specific wavelength

[Solution Description] When an electron transitions from higher energy level ($n=3$) to lower energy level ($n=2$), the excess energy is emitted as electromagnetic radiation (photon). The energy difference between these levels determines the photon’s wavelength using:

$\Delta E = E_3 – E_2 \propto \frac{1}{\lambda}$

Key Concept: Average atomic mass calculation

b) 28.6 u

[Solution Description]

Average mass = (Mass₁ × fraction₁) + (Mass₂ × fraction₂)

= $(28 × \frac{70}{100}) + (30 × \frac{30}{100})$

= $(28 × 0.7) + (30 × 0.3)$

= 19.6 + 9 = 28.6 u

Key Concept: Average Atomic Mass

b) 80 u

[Solution Description]

Use the formula:

$\text{Average mass} = \left(79 \times \frac{50.69}{100}\right) + \left(81 \times \frac{49.31}{100}\right)$

Calculate each term:

– $79 \times 0.5069 = 40.0451$

– $81 \times 0.4931 = 39.9411$

Add them: $40.0451 + 39.9411 = 79.9862 \approx 80 \, \text{u}$.

Key Concept: Filling Order and Outer Shell Limit

b) 2, 8, 8, 2

[Solution Description]

Fill shells sequentially with $2n^2$ values and check outer shell limit:

– K-shell ($n=1$): 2 electrons. Remaining: $20 – 2 = 18$.

– L-shell ($n=2$): 8 electrons. Remaining: $18 – 8 = 10$.

– M-shell ($n=3$): Max capacity $18$, but since it becomes the outermost after filling K and L, it can only hold 8 electrons. Assign 8 to M-shell. Remaining: $10 – 8 = 2$.

– N-shell ($n=4$): Assign remaining 2 electrons. Final distribution: 2, 8, 8, 2.

Key Concept: Electron Configuration Application

a) K=2, L=8, M=3

[Solution Description]

Total electrons = 13.

First fill K-shell (max $2 \times 1^2 = 2$). Remaining electrons: $13 – 2 = 11$.

Next fill L-shell (max $2 \times 2^2 = 8$). Remaining electrons: $11 – 8 = 3$.

Remaining 3 electrons go to M-shell. Final distribution: K=2, L=8, M=3.

Key Concept: Valency Determination

b) 3

[Solution Description]

The outermost shell is M with 5 electrons. Valency for elements close to 8 electrons in the outermost shell is calculated as $8 – \text{electrons in outermost shell}$. Here, $8 – 5 = 3$. Thus, the valency is 3.

Key Concept: Electron Charge

d) $-1$

[Solution Description]

The syllabus explicitly states that electrons carry a charge of $-1$ unit. Protons have a charge of $+1$, but the question specifically asks about electrons. Since there is an option matching $-1$, it is correct.

Key Concept: Bohr’s Energy Radiation Postulate

b) revolve in discrete orbits

[Solution Description]

Bohr’s first postulate states that electrons revolving in discrete orbits (like K, L, M shells) do not emit energy. Radiation occurs only when electrons jump between orbits.

Key Concept: Drawbacks of Rutherford’s Model

c) Electron stability issue due to energy loss

[Solution Description] According to classical physics, accelerating electrons (as they revolve) should emit radiation, lose energy, and eventually collapse into the nucleus. This stability issue is a key drawback mentioned in the syllabus.

Key Concept: Electron distribution limitation, Valency prediction error

c) 12

[Solution Description]

Step 1: Atomic number 14 = Silicon. Actual electronic configuration: K(2), L(8), M(4). Valency = 4 (or 4⁻).

Step 2: Rutherford’s model lacks concept of electron shells. Total electrons = 14. Scientists might assume all electrons participate equally since there are no defined energy levels.

Step 3: Incorrect prediction: Valency = 14 (total electrons) or 14 – 2 = 12 (incorrectly subtracting inner electrons). Actual options show mismatch between naive electron count vs shell-based valency.

Key Concept: Bohr’s Energy Levels

d) Assertion is false, but Reason is true.

[Solution Description]

According to Bohr’s model, energy levels are quantized and designated by $n$. The energy of an electron {increases} as the orbit number ($n$) increases. $K$-shell corresponds to $n = 1$ (lowest energy) and $L$-shell to $n = 2$. Thus, Assertion (A) is false because $K$-shell electrons have {lower} energy. Reason (R) is true as energy indeed increases with $n$.

Key Concept: Isotope Notation Interpretation

c) Mass number

[Solution Description]

The standard notation $^A_Z\text{X}$ shows mass number ($A$) as the left superscript. Here, $^{23}$ before $_{11}$ corresponds to mass number ($A$).

Key Concept: Valency Calculation for 5 Valence Electrons

c) 3

[Solution Description] Valency for elements with 5–7 valence electrons is calculated as $8 – \text{(number of valence electrons)}$. Here, valence electrons = 5. Substituting values: $8 – 5 = 3$.

Key Concept: Isobar Exclusion, Chemical Specificity

c) Chemical behavior varies with atomic number

[Solution Description]

Isobars like $^{131}\text{Te}$ share the same mass number but differ in atomic number (protons). Chemical properties depend on electron configuration tied to protons. The thyroid only absorbs iodine-based compounds, rejecting tellurium despite identical mass.

Key Concept: Isotope vs. Isobar

b) $^{59}\text{Co}$ and $^{60}\text{Co}$

[Solution Description]

Isotopes have same atomic number but different mass numbers. For cobalt (atomic number 27), $^{58}\text{Co}$ and $^{60}\text{Co}$ both have 27 protons but different neutrons.

Key Concept: Discovery Recognition

c) J.J. Thomson

[Solution Description] J.J. Thomson discovered electrons with $\frac{1}{2000}$ hydrogen atom mass through cathode ray deflection. Goldstein worked on protons, Rutherford on nucleus, and Bohr on orbits.

Key Concept: Charge Neutrality

a) $+2$

[Solution Description]

Each proton has a charge of $+1$ and each electron $-1$. Total charge = (Number of protons $\times$ charge per proton) + (Number of electrons $\times$ charge per electron). Calculation: $(12 \times +1) + (10 \times -1) = +12 – 10 = +2$.

Key Concept: Thomson’s Model

c) Thomson’s model

[Solution Description] Thomson’s model describes atoms as a positively charged sphere with electrons embedded like currants in a Christmas pudding. Rutherford focused on the nucleus, Bohr on discrete orbits, and Chadwick discovered neutrons.

Key Concept: Exception Case for Neutrons

a) Hydrogen

[Solution Description]

As per the syllabus, all atoms except hydrogen have neutrons. Hydrogen’s nucleus contains only protons (atomic mass ≈ $1$ u = $1$ proton, neutrons = $1 – 1 = 0$). Other options (helium, lithium, carbon) have neutrons.

Key Concept: Discovery of Electrons

c) J.J. Thomson

[Solution Description]

J.J. Thomson used cathode-ray tube experiments (1897). Rays were deflected towards positive plate, proving they carry negative charge. These particles were named electrons.