Chapter 8: Journey Inside the Atom

Given questions

1. Are atoms the smallest indivisible particles?
2. Why do electrons not fall into the nucleus even though they are attracted to the protons in it?
3. Why did scientists keep modifying atomic models?

(1) Are atoms the smallest indivisible particles?

• Dalton’s Atomic Theory (1808) stated that matter is made of tiny, indivisible particles called atoms.
• About a century later, three kinds of sub-atomic particles were discovered : electrons (J. J. Thomson, 1897), protons (E. Goldstein, 1886; Rutherford, 1919) and neutrons (J. Chadwick, 1932).
• Hence an atom itself can be split into still smaller constituents, so it is not the smallest indivisible particle of matter.

Conclusion : An atom is divisible into electrons, protons and neutrons; therefore atoms are not the ultimate, smallest or indivisible particles.

(2) Why do electrons not fall into the nucleus even though they are attracted to protons in it?

The behaviour of electrons can be understood with the help of Bohr’s postulates.

• Bohr proposed that an electron moves round the nucleus only in certain permitted circular paths called orbits or energy levels.
• While it remains in any one orbit, the centrifugal force of its motion exactly balances the electrostatic force of attraction between the electron and the nucleus, so its total energy stays constant.
• Mathematically, for an electron of mass $$m$$ and speed $$v$$ moving in an orbit of radius $$r$$ around a nucleus of charge $$+Ze$$ (for hydrogen, $$Z = 1$$):
  $$\frac{mv^{2}}{r} = \frac{1}{4\pi\varepsilon_{0}}\;\frac{Ze^{2}}{r^{2}}$$
• Because the energy is constant in an orbit, the electron does not spiral into the nucleus.
• An electron can move to a lower orbit only by emitting exactly the energy difference $$\Delta E$$ as radiation. The lowest orbit (ground state) has minimum but still finite radius, so the electron never reaches the nucleus.

Hence, the quantised angular-momentum condition $$mvr = n\hbar$$ (where $$n = 1,2,3\ldots$$ and $$\hbar = \frac{h}{2\pi}$$) prevents a continuous loss of energy, so electrons remain outside the nucleus.

(3) Why did scientists keep modifying atomic models?

Each atomic model was proposed to explain the experimental facts known at that time. As new experiments revealed additional facts that the earlier model could not explain, scientists refined or replaced the model.

Thus, atomic theory evolves because science is evidence-based; when new evidence conflicts with an existing theory, the theory must be modified so that predictions agree with all observations.

Step 1 – Recall Thomson’s “plum–pudding” model

According to J. J. Thomson (1904):

• The entire atom is a single positively-charged sphere. We may picture this uniform positive charge as a soft, continuous “pudding”. In symbols we write the net positive charge as $$+q$$ spread over the whole volume.
• Negatively-charged electrons (each carrying charge $$-e$$) are embedded everywhere inside that sphere, like tiny “plums”.
• The sum of all negative charges equals the total positive charge, so the atom as a whole is electrically neutral:

\[ \sum (-e)= -q, \qquad +q + (-q) = 0. \]

Step 2 – Describe an orange / lemon

• The fruit is roughly spherical and filled with soft juicy pulp.
• Several hard seeds are scattered inside the pulp.
• From outside the fruit looks neutral; only by cutting it open do we see the seeds.

Step 3 – Where the analogy matches Thomson’s idea

1. Mixed components: Both objects contain two distinctly different parts – soft, continuous material (pulp / positive charge) and small, discrete inclusions (seeds / electrons).
2. Enclosure: The smaller parts are completely surrounded by the softer medium, just as electrons are embedded inside the positive sphere.
3. Overall neutrality: Looking from outside we cannot tell that an orange holds seeds, just as an atom shows no net charge from outside.

Step 4 – Where the analogy fails

1. Uniform distribution: In Thomson’s model electrons are sprinkled uniformly throughout the positive sphere. Orange seeds are few and cluster in specific locules, not uniformly throughout every bit of pulp.
2. Relative mass: Seeds are far heavier and larger than the pulp surrounding them, whereas electrons are about $$1/1836$$ the mass of a proton; the positive matter actually carries almost the entire mass of the atom.
3. Amount of positive “pulp”: In the fruit the pulp is a genuine physical substance, not an electric charge. In the atom Thomson had to imagine a continuous positive charge, something that does not have a macroscopic counterpart.
4. Charge vs. matter: Seeds are electrically neutral; they do not possess the charge $$-e$$ required by the analogy. Therefore the fruit is not an electrical model at all.
5. Removability: Electrons can be knocked out only with very high energy (ionisation). Seeds can be removed easily with fingers – so the binding force in the fruit is nothing like the electrostatic attraction inside an atom.
6. Inner empty space: Later experiments (Rutherford 1911) showed that an atom is mostly empty space with a tiny nucleus; an orange is almost completely filled with matter. Hence the orange model could not predict Rutherford’s results.

Step 5 – Conclusion

An orange or lemon is a partial but imperfect everyday illustration of Thomson’s “plum-pudding” atom. It helps beginners visualise small particles lodged in a continuous medium; however, it fails to capture the uniformity of charge distribution, the correct mass ratios, and—most importantly—the electrical nature of the atom. Therefore it should be used only as a loose analogy, not as a faithful physical model.

Step 1 : Setting up the experiment
Thomson passed very high voltage (of the order of 10 kV) through a discharge tube that had:

• a highly evacuated glass tube (pressure about $$10^{-4} ext{–}10^{-6} ext{ atm}$$),
• two metal electrodes sealed inside — the cathode (–) and the anode (+),
• a small amount of any chosen gas (for example $$\mathrm{H_2}$$, $$\mathrm{N_2}$$, $$\mathrm{CO_2}$$, neon, etc.).

Step 2 : What he observed
Whatever gas or metal electrodes he used, a straight beam started from the cathode and produced a greenish fluorescence on the glass wall. These were called cathode rays.

Step 3 : Proving that the rays are made of negatively-charged particles
(a) The beam was deflected toward the positive plate of an external electric field.
(b) The beam was deflected in a magnetic field exactly as a stream of negative charges (like current moving from – to +).
(c) Thomson measured the charge-to-mass ratio and found the value $$\displaystyle \frac e m = 1.758\times10^{11}\;\mathrm{C\,kg^{-1}}$$, the same no matter which gas filled the tube.

Step 4 : Key logical inference
Because

1. the nature of the cathode rays (charge, $${e/m}$$ ratio, velocity, ability to cast shadow, etc.) never changed with the kind of gas inside the tube, and
2. the same invariance was seen even when the cathode was made of different metals,

Thomson concluded that the particles making up the cathode rays are identical, universal constituents of matter. He named these particles electrons.

Step 5 : Final statement
Since cathode rays arose from the gas inside the tube and from the cathode itself, and since their properties were independent of what substance was present, the only possible explanation is that every atom of every element contains the same negatively-charged particles. Hence, electrons must be present in all atoms.

Step 1 · Recall the original (thin-foil) observations

• More than 99 % of the $$\alpha$$-particles passed straight through the foil without any deviation.
• About 1 in 10 000 was scattered through an angle >$$90^{\circ}$$ (large-angle scattering).
• About 1 in 20 000 was reflected almost backwards (rebound).

This proved that (i) most of the atom is empty space and (ii) every atom contains a very small, massive, positively charged nucleus.

Step 2 · What changes when the foil is made thicker?

Let the original foil have thickness $$t$$ and contain $$n$$ atomic layers. If the same metal is beaten to a thickness of, say, $$2t$$ the number of atomic layers becomes $$2n$$. Hence an incoming $$\alpha$$-particle now meets roughly twice as many gold atoms on its way through the foil.

Therefore the probability $$P$$ that an $$\alpha$$-particle will pass close to at least one nucleus is roughly proportional to the number of layers:

$$P \;\propto\; \text{number of layers}\;\propto\; \text{foil thickness}.$$

Step 3 · Predicted observations for a thicker foil

1. Fewer straight-through tracks
   Because more particles now experience at least one strong nuclear repulsion, the fraction that goes undeviated (<$$1^{\circ}$$) will decrease from the original >99 %.
2. More small- and medium-angle deflections
   Paths bent by a few degrees to, say, $$30^{\circ}$$ will become noticeably more frequent.
3. More large-angle (>$$90^{\circ}$$) scattering and rebounds
   Instead of 1 in 10 000 or 20 000, several in every 10 000 particles may now be observed returning towards the source.
4. Possible complete absorption
   After suffering many successive small deflections an $$\alpha$$-particle may lose all its kinetic energy inside the thick foil and fail to emerge at all; a few will therefore be stopped in the metal.

Step 4 Reasoning linked to the nuclear model

The nucleus still occupies an extremely small part of each atom, so the chance per layer of a head-on collision remains tiny. Making the foil thicker merely multiplies that tiny chance by the larger number of layers, hence the changes listed above. The basic conclusion of Rutherford’s experiment — a small, dense, positively charged nucleus — remains valid.

Step 5 Suggested diagram to draw

• Draw a wide grey bar (to represent the thicker gold foil) instead of a thin line.
• Mark a beam of $$\alpha$$-particles entering from the left.
• Inside the bar show several circles labelled “nucleus”.
• Draw many curved arrows:
  • A few arrows going straight through but now fewer in number.
  • More arrows emerging at small and medium angles.
  • Several arrows turning back towards the source to indicate rebounds.
  • Optionally, one or two arrows ending inside the bar to show absorbed $$\alpha$$-particles.
• Label the percentages qualitatively, e.g. “straight <99 %”, “deflected more”, “rebound larger fraction”.

This single sketch contrasts clearly with the textbook’s thin-foil diagram, highlighting every new feature expected when the foil is thicker.

Step 1 · Why Rutherford chose $$\alpha$$-particles

• Charge: each $$\alpha$$-particle carries two positive charges $$\left(+2e\right)$$.
• Mass: $$4\,\mathrm{u}$$, nearly 8000 times the mass of an electron, so their path is almost a straight line unless a very strong force acts on them.
• Speed: about $$10^{7}\,\mathrm{m\,s^{-1}}$$, therefore they can cross a thin metal foil without being absorbed.

Because of these three properties the only sizeable electric force they feel inside an atom is the repulsive force of the tiny positive nucleus, which makes the pattern of deflections easy to interpret.

Step 2 · Force on a negatively charged particle inside the atom

The electrostatic force between two charges is

For an $$\alpha$$-particle, $$q_1 = +2e$$ and for the gold nucleus, $$q_2 = +79e$$ so $$q_1q_2 > 0$$ and the force is repulsive.

Let the incident particle be an electron. Now $$q_1 = -e$$ while $$q_2 = +79e$$, so $$q_1q_2 < 0$$ and the force given by (1) is attractive. Thus electrons would be pulled towards the nucleus rather than pushed away from it.

Step 3 · Consequences of attraction combined with very small mass

1. Curved paths instead of slight deflections
   Because the electron is 1836 times lighter than a proton, its momentum $$p = mv$$ is tiny. Even a moderate electric force bends it strongly, so its trajectory inside the atom would be a tight curve or even a spiral toward the nucleus rather than a gentle bend.
2. Very short range in matter
   Before the electron ever feels the nuclear attraction it keeps colliding with the atomic electrons of the gold foil. In every collision it loses energy because its mass is comparable with that of the target electrons. After a few such collisions it would be stopped inside the foil and never reach the fluorescent screen.
3. No back-scattering
   The spectacular $$1\text{ in }20\,000$$ “bounce-back” events seen with $$\alpha$$-particles were due to head-on repulsion from the massive positive nucleus. A negative particle would be pulled toward the nucleus, so there could be no rebound through angles close to $$180^{\circ}$$.

Step 4 · What the experimentalist would actually observe

• The zinc-sulphide screen would show very few scintillations because most electrons never emerge from the foil.
• The few that do emerge would be scattered through many small angles, producing a washed-out halo rather than the clear pattern Rutherford saw.
• Without any large-angle or back-scattered events the conclusion that ‘almost the whole mass and positive charge of the atom is concentrated in a very small nucleus’ could not be drawn.

Step 5 · Final inference

Replacing positively charged, heavy $$\alpha$$-particles with light, negatively charged particles would therefore destroy the very features that made Rutherford’s experiment decisive. Almost all of the incoming particles would be absorbed or only weakly scattered, and the presence of a small, massive, positively charged nucleus would remain hidden.

Step 1 ‒ Recall the two rival pictures of the atom

• Thomson’s “plum-pudding” model (1904) – the whole atom (radius $$R\approx10^{-10}\,\mathrm m$$) is a soft, positively charged sphere; the much lighter electrons are embedded in it like raisins in a pudding. The positive charge as well as almost the whole mass are uniformly spread out.
• Rutherford’s nuclear model (1911) – the positive charge $$+Ze$$ and practically the entire mass are packed into a tiny central nucleus of radius $$r_n\approx10^{-14}\,\mathrm m$$. Electrons move around this nucleus; most of the atom is empty space.

Step 2 ‒ What the gold-foil experiment actually saw

• Millions of fast $$\alpha$$-particles (charge $$q_\alpha=+2e$$) were aimed at a very thin gold foil.
• More than $$99\%$$ went straight through with no deflection.
• About $$1$$ in $$2\times10^{4}$$ bounced back through an angle greater than $$90^{\circ}$$ – some even retraced their path (≈ $$180^{\circ}$$).

Step 3 ‒ Could Thomson’s model deflect an $$\alpha$$ so strongly?

The only force that can turn an $$\alpha$$-particle is the electrostatic (Coulomb) force:

$$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_\alpha\,q_\text{gold}}{r^2}$$

In Thomson’s picture the gold atom is a uniform positive sphere. Inside such a sphere the electric field is (result from class IX physics):

$$E(r)=\dfrac{1}{4\pi\varepsilon_0}\,\dfrac{Z e\,r}{R^3}\quad(0\le r\le R).$$

The field – and hence the force $$F=q_\alpha E$$ – rises gradually and is zero at the centre. The largest force it can ever exert is at the surface:

$$F_{\max}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2e\,Z e}{R^2}.$$ Because $$R\approx10^{-10}\,\mathrm m$$, this force acts only for about that same distance. A short calculation (left to students) shows that such a gentle push can change the path of a multi-MeV $$\alpha$$ by at most a few degrees – never by $$90^{\circ}$$ or more. Therefore Thomson’s atom predicts no large-angle scattering.

Step 4 ‒ How a tiny nucleus makes bouncing back possible

Suppose instead all the charge $$+Ze$$ sits inside a nucleus of radius $$r_n\approx10^{-14}\,\mathrm m$$. Then, at a closest approach $$r\approx r_n$$, the Coulomb force is

$$F_{\text{nucleus}}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2e\,Ze}{r_n^2}.$$

Compare the two forces:

$$\dfrac{F_{\text{nucleus}}}{F_{\max}}=\left(\dfrac{R}{r_n}\right)^2\approx\left(\dfrac{10^{-10}}{10^{-14}}\right)^2=10^{8}.$$

The nuclear force is about $$10^8$$ times stronger – easily enough to reverse an $$\alpha$$-particle’s motion over the tiny distance $$r_n$$, giving the observed backward scattering.

Step 5 ‒ The single “bounce-back” observation kills the plum pudding

• If even a few $$\alpha$$-particles come straight back, there must be something very small, very massive, and very positively charged inside the atom.
• The plum-pudding model has none of these features; its gentle, diffuse positive charge can never supply the required impulsive force.

Hence the moment Rutherford saw large-angle (≈ $$180^{\circ}$$) scattering, Thomson’s model was decisively ruled out and the nuclear model became necessary.

Understanding what the question is really asking
The textbook invites you to imagine a personal conversation with Ernest Rutherford. Rather than merely writing any random query, you should formulate a thoughtful scientific question that reveals (i) your grasp of the alpha–particle–scattering experiment, and (ii) an awareness of the problems that still remained unsolved immediately after Rutherford announced his nuclear model (1911).

Step 1: Recap Rutherford’s main experimental facts

• He bombarded an ultra-thin gold foil (≈ $$10^{-7}\,\text{m}$$ thick) with $$\alpha$$-particles coming from $$\mathrm{^{214}Po}$$.
• Most $$\alpha$$-particles passed straight through  $$\Rightarrow$$ the atom is mainly empty space.
• About 1 in $$10^{20}$$ (roughly) were deflected by angles $$\ge 90^{\circ}$$  $$\Rightarrow$$ the entire positive charge and almost the whole mass are packed in a tiny central region he called the nucleus, whose radius he estimated to be $$\approx 10^{-14}\,\text{m}$$, while the atomic radius is $$\approx 10^{-10}\,\text{m}$$.

Step 2: Identify the puzzles that Rutherford’s model left open

1. Stability of the nucleus: Positive protons should repel one another by Coulomb’s law: $$F_{\mathrm{electrostatic}} = k\,\dfrac{q_1q_2}{r^2}$$. Why, then, does a nucleus not fly apart?
2. Stability of the atom itself: According to classical electrodynamics, an electron moving in a circular orbit should constantly radiate energy and spiral into the nucleus in about $$10^{-8}\,\text{s}$$.
3. Atomic spectra: Rutherford’s model could not explain the discrete line spectra of hydrogen and other elements.

Step 3: Choose the most penetrating question
Among the above, point 1 (intra-nuclear stability) addresses brand-new physics that would not be answered until the 1930s with the discovery of the neutron and the idea of the strong nuclear force. Hence asking Rutherford about this aspect would both acknowledge his achievement and probe the deepest mystery still open in 1911.

Step 4: Formulate the question clearly
Remember that Rutherford thought in terms of protons only (the neutron was unknown). The question should therefore focus on how the repulsive electrostatic force could possibly be overcome within $$\approx10^{-14}\,\text{m}$$.

Proposed question to Rutherford

“Sir Ernest, once your scattering data had convinced you that almost the whole mass and the entire positive charge of the atom are squeezed into a nucleus only about $$10^{-14}\,\mathrm{m}$$ wide, what mechanism do you envisage that keeps so many like-charged protons packed together without the nucleus flying apart?”

Why this is a good question

• It shows you understand the experimental basis: the astonishingly small nuclear radius derived from scattering angles.
• It pinpoints a genuine gap in 1911 physics.
• It paves the way to later discoveries (neutron, strong nuclear force).

Note for students: Any thoughtful question that singles out an unsolved issue—electron collapse, spectral lines, source of $ $$(+)$ $ charge—would also be acceptable. The key is to connect the question to Rutherford’s own findings, not to something unrelated.

Step 1 – Check the truth of the Assertion (A)
Rutherford performed the gold-foil α-particle scattering experiment and noted three observations: (a) about 99 % of the α-particles passed straight through the foil, (b) a small fraction were deflected through measurable angles, and (c) an extremely small number (roughly 1 in 20 000) were back-scattered, i.e. they rebounded almost straight back along the incoming direction. The only way to account for such sharp, large-angle deflections is to assume that almost all the positive charge and almost the entire mass of an atom are packed into an extremely small, dense central core. Rutherford named this core the nucleus. Hence the Assertion is true.

Step 2 – Check the truth of the Reason (R)
According to J. J. Thomson's 1904 plum-pudding (or water-melon) model, an atom is imagined as a positively charged, uniformly spread sphere in which the negatively charged electrons are embedded like raisins or seeds. This is exactly what the Reason states, so R is also true.

Step 3 – Does R explain A?
The idea of a nucleus came from Rutherford's experimental observations of large-angle α-particle scattering. It has no logical connection with Thomson's embedded-electron picture; in fact, Thomson's model would have predicted no large-angle scattering at all. Even though both sentences are factually correct, the statement in R is not the cause or logical explanation for Rutherford's conclusion in A.

Step 4 – Choose the correct option
Both A and R are true, but R is not the correct explanation of A. Hence option (ii) is correct.

Step 1 – Choosing an appropriate name

Suppose my name as the discoverer is Dr. Chat GPT. To honour this discovery I coin the element name

Chatgptium

The chosen ending “-ium” is the standard suffix recommended by IUPAC for newly discovered metallic elements (just as in curium, einsteinium, etc.).

Step 2 – Proposing its chemical symbol

I suggest the two-letter symbol

$$\mathrm{Ch}$$

Step 3 – Verifying the symbol against each IUPAC rule

Step 4 – Conclusion

Because the proposed symbol $$\mathrm{Ch}$$ observes every IUPAC convention enumerated above, it is a valid and acceptable chemical symbol for my newly discovered element Chatgptium.

Step 1 — Recall the purpose of a symbol
A chemical symbol is a universally accepted shorthand for the name of an element. For example, the symbol $$\mathrm{O}$$ represents the element oxygen everywhere in the world. By convention, the symbol contains either one capital letter (e.g. $$\mathrm{H}$$ for hydrogen) or a capital letter followed by a small letter (e.g. $$\mathrm{Na}$$ for sodium).

Step 2 — Imagine each scientist inventing a personal symbol
Suppose:

• Scientist A writes copper as $$\mathrm{C}$$,
• Scientist B writes the same element as $$\mathrm{Cp}$$,
• Scientist C stays with the IUPAC symbol $$\mathrm{Cu}$$.

Now look at the simple reaction in universal notation

$$\mathrm{Cu} + 2\,\mathrm{AgNO_3} \;\longrightarrow\; 2\,\mathrm{Ag} + \mathrm{Cu(NO_3)_2}$$

If the three scientists publish their own versions, we might see:

• Scientist A: $$\mathrm{C} + 2\,\mathrm{AgNO_3} \to \ldots$$
• Scientist B: $$\mathrm{Cp} + 2\,\mathrm{AgNO_3} \to \ldots$$

Readers could easily mistake $$\mathrm{C}$$ for carbon or wonder whether $$\mathrm{Cp}$$ is a new, undiscovered element. The same reaction would appear to involve different substances.

Step 3 — List the concrete problems that follow

1. Communication breakdown  Journal articles, textbooks and data tables would all use dissimilar symbols, forcing every reader to learn dozens of private codes before understanding any chemical statement.
2. Calculation errors  While balancing equations or computing molar masses, a student might treat $$\mathrm{C}$$ as carbon (molar mass $$12\,\mathrm{g\,mol^{-1}}$$) instead of copper ($$63.5\,\mathrm{g\,mol^{-1}}$$), giving completely wrong numerical results.
3. Safety hazards  Misidentifying chemicals in the laboratory can lead to dangerous reactions. For instance, mistaking $$\mathrm{Na}$$ (sodium) for $$\mathrm{N}$$ (nitrogen) when reading a procedure could cause an explosion on contact with water.
4. Trade and industry confusion  Chemical names and formulae appear on container labels, import–export documents and safety data sheets. Non-standard symbols would make international trade unreliable and legally complicated.
5. Loss of cumulative knowledge  Databases such as the periodic table, thermodynamic tables and spectral libraries rely on unique symbols. Multiplying symbols would fragment information and slow scientific progress.

Step 4 — State the conclusion
Therefore, using different symbols for the same element would introduce ambiguity, errors and safety risks, defeating the very purpose of chemical symbolism. A single, standard set of symbols—adopted worldwide—prevents these problems.

Given data

• Atomic (proton) number: $$Z = 26$$
• Total number of nucleons (mass number): $$A = 56$$

We must deduce three quantities for a neutral atom:

1. Number of protons
2. Number of electrons
3. Number of neutrons

1. Protons

The atomic number $$Z$$ is by definition the number of protons present in the nucleus.

Therefore

\[\text{Number of protons} = Z = 26\]

2. Electrons

For an electrically neutral atom, the positive charge of the protons must be exactly balanced by an equal number of negatively charged electrons. Hence

$$\text{Number of electrons} = \text{Number of protons} = 26$$

3. Neutrons

The mass number $$A$$ is the total count of nucleons (protons + neutrons). Thus

$$A = \text{protons} + \text{neutrons}$$

Substituting the known values:

$$56 = 26 + \text{neutrons}$$

Re-arranging gives

$$\text{neutrons} = 56 - 26 = 30$$

Final counts

\[\boxed{\text{Protons} = 26,\; \text{Electrons} = 26,\; \text{Neutrons} = 30}\]

Step 1 — Recall the definitions

• The atomic number is the number of protons in the nucleus. We denote it by $$Z$$.
• The mass number is the total number of nucleons (protons + neutrons). We denote it by $$A$$ and write $$A = Z + N$$, where $$N$$ is the number of neutrons.

Step 2 — Write down the known data

• Protons in the nucleus: $$20$$. Hence $$Z = 20$$.
• Mass number: $$A = 41$$.

Step 3 — Use the relation between A, Z and N

From the definition, $$A = Z + N$$. Rearranging for neutrons $$N$$ gives

$$N = A - Z$$

Step 4 — Substitute the numerical values

$$N = 41 - 20$$

$$N = 21$$

Conclusion

The nucleus contains $$21$$ neutrons.

Given data

• Number of neutrons in the atom, n_n  =  18
• Atomic number (proton number) of the element, $$Z = 17$$

Step 1 : Recall the definition of mass number

The mass number $$A$$ of an atom is the total number of heavy nuclear particles, i.e. the sum of its protons and neutrons:

$$A = \text{number of protons} + \text{number of neutrons}$$

Step 2 : Identify the required quantities

The atomic number $$Z$$ gives the number of protons. Hence:

Number of protons = $$Z = 17$$

Number of neutrons = 18 (already given).

Step 3 : Substitute and calculate

Plug the values into the formula for $$A$$:

$$A = 17 + 18$$

$$A = 35$$

Result

The mass number of the atom is therefore

\[A = 35\]

The nuclear symbol $$^{23}\mathrm{A}$$ tells us two separate facts:

• The superscript 23 is the mass number: $$A = 23$$.
• The chemical symbol’s (missing) subscript is the atomic number $$Z$$, which equals the number of protons and, in a neutral atom, also equals the number of electrons.

Because the atom is neutral and has 11 electrons,

$$Z = 11$$

The number of neutrons $$N$$ in any nuclide is obtained from

$$N = A - Z$$

Substituting the known values,

$$N = 23 - 11$$

\[N = 12\]

Hence, the atom contains 12 neutrons.

Concept recalled – 2n² rule
For the n-th shell the maximum electrons it can hold is $$2n^{2}$$. Therefore

• K–shell (n=1): $$2\times 1^{2}=2$$ electrons
• L–shell (n=2): $$2\times 2^{2}=8$$ electrons
• M–shell (n=3): $$2\times 3^{2}=18$$ electrons

We successively fill the shells from the nucleus outward until the atomic number Z is reached.

(i) Z = 12

1. K–shell gets its maximum 2 electrons ⇒ still $$12-2 = 10$$ left.
2. L–shell now takes its maximum 8 electrons ⇒ still $$10-8 = 2$$ left.
3. M–shell takes the remaining 2.

Shell notation : 2, 8, 2
Sub-shell (orbital) notation : $$1\mathrm{s}^{2}\;2\mathrm{s}^{2}\;2\mathrm{p}^{6}\;3\mathrm{s}^{2}$$

(ii) Z = 16

1. K–shell → 2 electrons ⇒ $$16-2 = 14$$ left.
2. L–shell → 8 electrons ⇒ $$14-8 = 6$$ left.
3. M–shell → 6 electrons.

Shell notation : 2, 8, 6
Sub-shell notation : $$1\mathrm{s}^{2}\;2\mathrm{s}^{2}\;2\mathrm{p}^{6}\;3\mathrm{s}^{2}\;3\mathrm{p}^{4}$$

(iii) Z = 18

1. K–shell → 2 electrons ⇒ $$18-2 = 16$$ left.
2. L–shell → 8 electrons ⇒ $$16-8 = 8$$ left.
3. M–shell → 8 electrons.

Shell notation : 2, 8, 8
Sub-shell notation : $$1\mathrm{s}^{2}\;2\mathrm{s}^{2}\;2\mathrm{p}^{6}\;3\mathrm{s}^{2}\;3\mathrm{p}^{6}$$

Step 1 — Identify the given data

• Mass (nucleon) number, $$A = 23$$
• Number of protons (atomic number), $$Z = 11$$

Step 2 — Recall the key relations

• The atomic number $$Z$$ equals the number of protons.
• The mass number $$A$$ equals the total number of nucleons: $$A = \text{protons} + \text{neutrons}$$.
• Therefore, $$\text{neutrons} = A - Z$$.

Step 3 — Find the neutron number

Substitute the given values:

$$\text{neutrons} = A - Z = 23 - 11 = 12$$

Hence, the atom contains 12 neutrons.

Step 4 — Identify the element

An atom with $$Z = 11$$ is sodium, symbol $$\mathrm{Na}$$. Sodium is well known to be

• a soft, silvery metal, and
• reacts vigorously with water to give $$\mathrm{NaOH}$$ and $$\mathrm{H_2}$$ gas.

Therefore, the riddle describes the sodium atom $$\prescript{23}{11}\mathrm{Na}$$, which has 11 protons and 12 neutrons.

Step 5 — A new riddle for practice

"I am an atom whose mass number is 19 and I have 9 protons. I am a pale-yellow gas that is added to toothpaste to prevent cavities. Who am I and how many neutrons do I have?"

Answer to the new riddle (for the teacher): The element is fluorine $$\prescript{19}{9}\mathrm{F}$$, and it contains $$19 - 9 = 10$$ neutrons.

Step 1 : Recall the definitions

• Atomic number, written $$Z$$, equals the number of protons present in the nucleus: \[Z = \text{number of protons}\]
• Mass number, written $$A$$, equals the sum of protons and neutrons: \[A = \text{protons} + \text{neutrons}\]

Step 2 : Write the given nuclear information

Step 3 : Calculate the atomic number for each atom

Because both atoms have 11 protons, for each one

$$Z = 11$$

Hence the atomic number is 11 for both atoms.

Step 4 : Calculate the mass number for each atom

• Atom I: $$A_1 = 11 + 12 = 23$$
• Atom II: $$A_2 = 11 + 13 = 24$$

Thus the mass numbers are different: one is 23 and the other is 24.

Step 5 : Decide whether they are the same element

Elements are identified by their atomic number $$Z$$. Since both atoms have $$Z = 11$$, they belong to the same element, namely sodium ($$\mathrm{Na}$$).

Atoms of the same element that differ only in their number of neutrons (and therefore in mass number) are called isotopes.

Conclusion

Both atoms are isotopes of sodium: one is $$\mathrm{^{23}_{11}Na}$$ and the other is $$\mathrm{^{24}_{11}Na}$$. Their atomic numbers are identical (11), but their mass numbers are 23 and 24, respectively.

Step 1 – Recall the formula for average (relative) atomic mass

If an element exists as two or more isotopes, the average atomic mass (in atomic mass units, u) is the weighted mean of the isotopic masses, each multiplied by its percentage abundance and then divided by 100 %. Symbolically, for two isotopes 1 and 2:

$$\text{Average atomic mass}=\dfrac{\left(m_1\times\%_1\right)+\left(m_2\times\%_2\right)}{100}$$

where

• $$m_1,m_2$$ are the mass numbers of the two isotopes, and
• $$\%_1,\%_2$$ are their percentage abundances.

Step 2 – Insert the data for bromine

For bromine we have:

• Isotope 1: $$^{79}_{35}\mathrm{Br}$$ – mass number $$m_1=79$$, abundance $$\%_1=49.7\%$$
• Isotope 2: $$^{81}_{35}\mathrm{Br}$$ – mass number $$m_2=81$$, abundance $$\%_2=50.3\%$$

Substituting these values into the formula gives

$$\text{Average atomic mass}=\dfrac{\left(79\times49.7\right)+\left(81\times50.3\right)}{100}$$

Step 3 – Perform the multiplications

Compute each product separately:

$$79\times49.7 = 3926.3$$

$$81\times50.3 = 4074.3$$

Step 4 – Add the two products

$$3926.3 + 4074.3 = 8000.6$$

Step 5 – Divide by 100 % to get the mean

$$\text{Average atomic mass}=\dfrac{8000.6}{100}=80.006$$

Step 6 – Round off to an appropriate number of significant figures

To match the precision of the data (three significant figures), we round to $$80.0$$.

\[\boxed{\text{Average atomic mass of bromine}\;\approx\;80.0\,\mathrm{u}}\]

Step 1 — Rutherford’s key observation

• When a narrow beam of $$\alpha$$-particles (helium nuclei with charge $$+2e$$) was directed at an extremely thin gold foil, almost all particles passed straight through without any noticeable change in their paths.
• However, a very tiny fraction (about 1 in $$10\,000$$) either:
    • bounced straight back (deflection $$\approx 180^{\circ}$$), or
    • were scattered through very large angles (close to $$90^{\circ}$$).

Step 2 — What this tells us about the atom’s interior

• If the positive charge were spread uniformly throughout the atom (as the earlier “plum-pudding” model suggested), the repulsive electric field that an $$\alpha$$-particle felt while crossing the atom would be very weak and diffuse. Large-angle deflections would then be practically impossible.
• The fact that a few $$\alpha$$-particles were turned back implies they experienced an intense repulsive force in a very tiny region. From Coulomb’s law the electric force rises sharply when the distance $$r$$ from a positive charge becomes very small $$\bigl(F \propto 1/r^{2}\bigr)$$. Therefore the entire positive charge of the atom must be concentrated in a minuscule core rather than smeared out.
• Because only about $$0.01\%$$ of the incoming particles met this strong field, the high-charge region must occupy only about $$10^{-4}$$ (or even less) of the atom’s volume — i.e. its linear size is roughly $$10^{-2}$$ of the atomic radius.

Step 3 — Rutherford’s conclusion

• The atom contains a small, dense, positively charged centre.
• This centre — later named the nucleus — holds almost the entire mass of the atom.
• The nucleus is located at the geometric centre of the atom, while the electrons move in the relatively vast empty space surrounding it.

Thus, the rare but dramatic large-angle or back-scattering events in the gold-foil experiment led Rutherford to propose the nuclear model of the atom: a tiny central nucleus carrying all the positive charge and nearly all the mass, with the rest of the atom being mostly empty space filled by electrons.

Step 1 — Understand each statement

• (iv) Dalton’s model: In his Atomic Theory (around 1808) John Dalton said that matter is made of tiny, hard, indivisible particles called atoms. At this stage there was no idea of sub-atomic particles, charge, or internal structure.
• (ii) Thomson’s model: After the discovery of the electron (1897), J. J. Thomson pictured the atom as a positively charged sphere with negatively charged electrons scattered inside it—known as the “plum-pudding” or “water-melon seed” model (≈ 1898–1904).
• (iii) Rutherford’s model: In 1911 Ernest Rutherford’s α-particle scattering experiment showed that almost all the mass and positive charge of an atom is concentrated in a very small, dense nucleus; electrons move in the empty space around it.
• (i) Bohr’s model: Niels Bohr (1913) kept Rutherford’s nuclear atom but added the idea that electrons revolve only in certain allowed circular orbits (called shells) without radiating energy. Each permitted orbit has a fixed energy, explaining the line spectra of elements.

Step 2 — Place them in historical order

The time-line of development is therefore:

1. (iv) Dalton’s indivisible solid-sphere model (1808)
2. (ii) Thomson’s plum-pudding model (≈ 1898–1904)
3. (iii) Rutherford’s nuclear model (1911)
4. (i) Bohr’s planetary, fixed-orbit model (1913)

Step 3 — Explain how each new model improved the previous one

• Dalton’s theory introduced atoms but could not account for sub-atomic particles or electrical properties.
• Thomson incorporated the newly discovered electron and explained overall electrical neutrality, but could not explain α-particle scattering.
• Rutherford’s model correctly localised positive charge and mass in a nucleus, explaining the scattering experiment, yet could not explain atomic stability or spectral lines (because moving electrons should radiate energy and spiral in).
• Bohr solved the stability and spectrum problem by quantising electron orbits, laying the foundation for modern quantum mechanics.

Hence, arranging the given statements chronologically gives the logical evolution of the atomic model from a solid, indivisible sphere to the quantised nuclear atom.

Step 1 – Understand the basic charges inside an atom

• The nucleus is made up of protons (+ve charge) and neutrons (neutral).
• Each proton carries a positive charge $$+e = +1.6 \times 10^{-19}\,\text{C}$$.
• Each electron in the surrounding space carries an equal but opposite charge $$-e$$.

Step 2 – Recall the law of electrostatic attraction

According to Coulomb’s law, two point charges $$q_1$$ and $$q_2$$ separated by a distance $$r$$ exert equal and opposite forces on each other of magnitude

\[F = k\frac{|q_1 q_2|}{r^2}\]

where $$k = 9 \times 10^9\,\text{N m}^2\,\text{C}^{-2}$$ is the electrostatic constant.

Because the signs of $$q_1$$ and $$q_2$$ are opposite (+e on the proton and –e on the electron), the force is attractive.

Step 3 – Link the attraction to circular motion of an electron

• In the Bohr model an electron moves around the nucleus in a nearly circular orbit of radius $$r$$.
• For any body executing uniform circular motion, a centripetal force $$F_c$$ is required and is given by $$F_c = m_e\,\dfrac{v^2}{r}$$, where $$m_e$$ is the mass of the electron and $$v$$ its speed.
• Bohr postulated that the needed centripetal force is provided exactly by the electrostatic force:

\[k\frac{e^2}{r^2} = m_e\,\frac{v^2}{r}\;\;\Rightarrow\;\;F_{\text{electrostatic}} = F_{\text{centripetal}}\]

Thus the attractive electrostatic force keeps pulling the electron toward the nucleus and supplies the inward force required to bend its path into an orbit instead of letting it fly off in a straight line.

Step 4 – Energy viewpoint (why it remains bound)

• The electron–nucleus system possesses potential energy $$U = -k\dfrac{e^2}{r}$$ (negative sign shows attraction).
• The electron’s kinetic energy is $$K = \dfrac{1}{2}m_e v^2$$.
• For the allowed Bohr orbits, the total energy $$E = K + U$$ is negative, meaning the electron is in a bound state. An external supply of energy equal to $$|E|$$ would be needed to remove it from the atom.

Because the electron does not have that much spare energy by itself, and because the electrostatic attraction is continuously acting, it stays tied to the nucleus instead of escaping.

Conclusion

The negatively charged electrons remain in orbit because the attractive electrostatic (Coulomb) force between them and the positively charged nucleus provides the necessary centripetal force and keeps their total energy negative; hence they do not fly away from the atom.

Step 1 – Check the truth of the Assertion (A)
The discovery of sub-atomic particles—electrons by J. J. Thomson, protons by E. Goldstein/Rutherford, and neutrons by Chadwick—made it possible to build realistic atomic models (Thomson’s ‘plum-pudding’, Rutherford’s nuclear model, Bohr’s model, etc.). Hence the discovery did help scientists to understand how an atom is arranged.
Therefore, Assertion (A) is true.

Step 2 – Check the truth of the Reason (R)
Inside a neutral atom the negative charge carried by $$e^-$$ must cancel the positive charge carried by $$p^+$$, so
\[\text{Number of electrons} = \text{Number of protons}\]
Thus Reason (R) is also true.

Step 3 – Does R explain A?
The fact that the two numbers are equal only states the condition for electrical neutrality; it does not explain how the discovery of sub-atomic particles helped us understand atomic structure (which involves locating these particles and proposing models).
Hence R is not the correct explanatory reason for A.

Conclusion
Both A and R are true, but R is not the correct explanation of A. So option (ii) is correct.

Given data

• Mass number (also called nucleon number) of magnesium: $$A = 24$$
• Atomic number (also called proton number) of magnesium: $$Z = 12$$

(i) Number of protons

The atomic number equals the number of protons present in the nucleus:

$$\text{No. of protons} = Z = 12$$

(ii) Number of neutrons

The nucleus contains protons and neutrons. Their total is the mass number:

$$A = Z + N$$ where $$N$$ is the number of neutrons.

Re-arrange for $$N$$:

$$N = A - Z$$

Substitute the given values:

$$N = 24 - 12 = 12$$

Hence a magnesium-24 nucleus has $$12$$ neutrons.

(iii) Number of electrons

A neutral atom has the same number of electrons as protons:

$$\text{No. of electrons} = 12$$

(iv) Arrangement of electrons (electronic configuration)

Electrons occupy successive energy shells (K, L, M, …). Each shell has a maximum capacity:

• K-shell: $$2$$ electrons
• L-shell: $$8$$ electrons
• M-shell: $$18$$ electrons (but we will need only $$2$$ here)

Total electrons to place: $$12$$.

1. First fill the K-shell: $$2$$ electrons remaining $$12-2 = 10$$
2. Next fill the L-shell up to its capacity $$8$$: remaining $$10-8 = 2$$
3. Place the last $$2$$ electrons in the M-shell.

The electronic configuration is therefore written as

$$2,\;8,\;2$$

Illustration to draw: Show a central nucleus labelled “12 p, 12 n”. Around it draw three concentric circles. On the innermost circle mark 2 small dots (K-shell), on the second circle mark 8 dots (L-shell), and on the outermost circle mark 2 dots (M-shell).

Given from Fig. 8.17 – Four schematic Bohr models are shown. Counting the dots (electrons) present in each shell we get:

For a neutral atom

$$\text{Atomic number }(Z) = \text{Number of protons} = \text{Number of electrons}$$

The valence electrons are the electrons present in the outer-most shell. The valency is obtained from the octet rule:

• If valence electrons ≤ 4,  valency = number of valence electrons.
• If valence electrons > 4,  valency = 8 − (valence electrons).

Using the Periodic Table to match each atomic number with its element, we obtain:

Thus all the required quantities for the four elements have been determined directly from their Bohr diagrams.

Background ideas

• From Coulomb’s law the electrostatic force between a nucleus of charge $$+Ze$$ and an electron of charge $$-e$$ at distance $$r$$ is $$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r^2}$$.
• For an electron moving in a circle this force supplies the required centripetal force, so classically we write $$\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r^2}=m\omega^2r$$ where $$m$$ is the electron mass and $$\omega$$ its angular speed.
• Physics you study in Class 9 also says: a charge that is being accelerated (its direction is changing even if the speed is constant) continuously emits electromagnetic radiation and therefore continuously loses energy.

Step 1 – What Rutherford’s model predicts

According to Rutherford, the electron is always in such a circular orbit. Because the electron is accelerating toward the centre, it must radiate energy.

Its total mechanical energy at distance $$r$$ is

\[E = K + U = \tfrac12 m\omega^2 r^2 - \dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r}\]

When radiation carries this energy away, $$E$$ decreases, so $$r$$ must also decrease (the electron spirals inward). Calculations using the Larmor radiation formula show that the time required to hit the nucleus is about $$10^{-8}\,\text{s}$$ – a hundred–millionth of a second!

Yet ordinary matter around us is perfectly stable for years. Hence Rutherford’s picture contradicts experience.

Step 2 – Bohr’s two postulates that remove the difficulty

1. Stationary orbits (non-radiating states): The electron is allowed to revolve only in certain special circular paths where, by postulate, it does not emit radiation in spite of being accelerated.
2. Quantisation of angular momentum: For those special paths, the electron’s angular momentum is restricted to the values\[mvr = n\hbar\quad (n = 1,2,3,\ldots)\]where $$\hbar = \dfrac{h}{2\pi}$$.

Because energy is lost or gained only when the electron jumps from one permitted orbit to another, it can stay forever in a single permitted orbit (especially the lowest one, $$n=1$$) without collapsing. Thus Bohr’s ideas give stable atoms and explain line spectra through the transition formula

\[\Delta E = E_{\text{high}} - E_{\text{low}} = h\nu\]

Conclusion

• Rutherford’s model is unstable because classical physics forces its orbiting electron to radiate continuously and collapse.
• Bohr assumed special “stationary” orbits and quantised angular momentum; inside those orbits the electron does not radiate, so the atom remains intact and its discrete energy differences match spectral lines.

For any electrically neutral atom:

• The number of electrons equals the number of protons (atomic number).
• The mass number $$A$$ equals the sum of protons and neutrons.

The symbol of the given isotope is $$^{70}\mathrm{X}$$. The superscript 70 is its mass number:

$$A = 70$$

The atom is neutral and has 31 electrons, therefore

$$Z = \text{number of protons} = \text{number of electrons} = 31$$

Let $$N$$ be the number of neutrons. By definition,

$$A = Z + N$$

Substitute $$A = 70$$ and $$Z = 31$$:

$$70 = 31 + N$$

Rearrange to find $$N$$:

$$N = 70 - 31$$

$$N = 39$$

Hence, the nucleus of $$^{70}\mathrm{X}$$ contains 39 neutrons.

Given data

• Number of protons (i.e. atomic number)  $$Z = 79$$
• Mass number  $$A = 197$$

Key ideas recalled

• The mass number $$A$$ of an atom is the total of its protons and neutrons:
  $$A = Z + N$$  where $$N$$ is the number of neutrons.
• An electrically neutral atom contains as many electrons as protons, so
  $$\text{number of electrons} = Z$$.

(i) Calculating the number of neutrons

Start from the definition of mass number:

$$A = Z + N$$

Re-arrange to make $$N$$ the subject:

$$N = A - Z$$

Substitute the given values $$A = 197$$ and $$Z = 79$$:

$$N = 197 - 79 = 118$$

Therefore the atom has \[N = 118\] neutrons.

(ii) Calculating the number of electrons

The atom is neutral, so

$$\text{Number of electrons} = Z = 79$$

Hence it contains \[79\] electrons.

Result

• Number of neutrons  = 118
• Number of electrons  = 79

To fill the missing entries we recall three basic facts for a neutral atom:

• Atomic number $$Z$$ = number of protons $$p$$.
• In a neutral atom, number of electrons $$e$$ equals number of protons, so $$e = Z$$.
• Mass number $$A$$ = protons + neutrons, so $$A = Z + N$$. Equivalently, $$N = A - Z\,.$$

We now treat each row.

Row 1

• Given: $$Z = 5,\; N = 6$$.
• Mass number: $$A = Z + N = 5 + 6 = 11$$.
• Protons: $$p = Z = 5$$.
• Electrons: $$e = Z = 5$$.
• Element with $$Z = 5$$ is Boron.

Row 2

• Given: $$A = 14,\; e = 7$$ and the element is Nitrogen.
• Nitrogen has $$Z = 7$$, therefore $$p = 7,\; e = 7$$ (matches the given).
• Neutrons: $$N = A - Z = 14 - 7 = 7$$.

Row 3

• Given: $$A = 24,\; p = 12$$.
• Atomic number: $$Z = p = 12$$.
• Electrons: $$e = Z = 12$$.
• Neutrons: $$N = A - Z = 24 - 12 = 12$$.
• Element with $$Z = 12$$ is Magnesium.

Row 4

• Given: $$Z = 15,\; N = 16$$.
• Mass number: $$A = Z + N = 15 + 16 = 31$$.
• Protons: $$p = Z = 15$$.
• Electrons: $$e = Z = 15$$.
• Element with $$Z = 15$$ is Phosphorus.

Row 5

• Given: $$A = 1,\; N = 0$$.
• Atomic number: $$Z = A - N = 1 - 0 = 1$$.
• Protons: $$p = Z = 1$$.
• Electrons: $$e = Z = 1$$.
• Element with $$Z = 1$$ is Hydrogen.

Completed Table 8.5