NCERT Solutions Class 9 Science Chapter 8 In-text Questions — Journey Inside the Atom

NCERT Class 9 Science — Journey Inside the Atom, the seven Pause and Ponder boxes on pages 143 to 156. All 18 questions solved.

Also here. Solutions to the end-of-chapter questions for this chapter are on their own page.

The boxes in this chapter follow the history as the chapter tells it: Thomson’s model, the gold foil experiment that broke it, Rutherford’s nucleus, then the bookkeeping of protons, neutrons and electrons that the nucleus makes necessary. Four facts carry nearly every question:

• An atom is electrically neutral, so the number of protons equals the number of electrons.
• Atomic number $Z$ = number of protons; mass number $A$ = number of protons + neutrons (the nucleons). So the number of neutrons is $A – Z$.
• Electrons fill shells by Bohr and Bury’s rules: at most $2n^2$ in the $n$th shell, at most 8 in the outermost, and filled in order K, L, M, N.
• Average atomic mass is a weighted average over the isotopes, each mass multiplied by its own abundance.

Key insight. Almost every numerical question here is the one relation $A = Z + n$ read in a different direction — given two of the three, find the third. Write down which two you have been given before reaching for anything else, and half the chapter’s arithmetic disappears.

Pause and Ponder, page 143. The box that follows Section 8.2.1, on Thomson’s model of the atom.

Question 1

Suppose you made up your own ‘atom’, as Thomson described, using clay for the positive charge and small beads for the electrons spread through it. What will happen if:

(i) the positive charge on the clay is lesser than the total negative charge of the beads?

(ii) by mistake, the clay itself carries a bit of negative charge? Would your model still represent a neutral atom?

Solution. Thomson’s picture is a sphere of positive charge with the electrons stuck through it like seeds, and the whole thing is neutral because the two charges are equal and opposite. Both parts of this question break that balance and ask what is left.

(i) If the clay’s positive charge is less than the beads’ total negative charge, the two no longer cancel. There is negative charge left over, so the model as a whole carries a net negative charge — it is not a neutral atom but a negatively charged ion (an anion). A real atom becomes one exactly this way, by gaining electrons.

(ii) If the clay itself is negative, then there is no positive charge anywhere in the model: the clay and the beads are both negative, and they add instead of cancelling. The model is strongly negatively charged, and worse, it no longer represents Thomson’s idea at all — the whole point of his model was a positive sphere in which the electrons sit. So no, it would not represent a neutral atom.

Question 2

Could an orange or a lemon, which also contain seeds inside soft pulp, be a good comparison? In what ways does it match Thomson’s idea and where does it fall short?

Solution. The comparison is worth taking seriously in both directions, which is what the question asks.

Where it matches. The pulp is a single continuous body filling the whole fruit, as Thomson’s positive charge fills the whole atom, and the seeds are small objects embedded in it here and there, as his electrons are embedded in the positive sphere. The seeds are also much smaller than the pulp, just as electrons are far lighter than the atom they sit in.

Where it falls short. In a fruit the seeds are clustered near the centre rather than distributed throughout, whereas Thomson’s electrons are spread evenly. The pulp carries no charge, so nothing in the fruit corresponds to the attraction that holds the electrons in place — and it is that electrical balance, not the geometry, that made the model a model of an atom. A fruit also has a definite skin, while an atom has no such boundary; and the seeds cannot be removed and put back, whereas an atom can lose or gain electrons.

Question 3

Why did Thomson conclude that electrons are present in all atoms?

Solution. The argument rests on the electrons being the same whatever they came from. In his discharge-tube experiments the particles obtained had the same charge and the same very small mass no matter which gas filled the tube or which metal the electrodes were made of.

If the particle had depended on the material, it would have been a property of that material. Because it did not, the conclusion follows: this same particle must be a component of the atoms of every one of those materials — that is, electrons are present in all atoms.

Pause and Ponder, page 144. The box that follows Section 8.2.2, on the gold foil experiment and Rutherford’s model.

Question 4

What do you think would happen if α-particles were replaced with negatively charged particles in Rutherford’s gold foil experiment?

Solution. The alpha particle is positive and the nucleus is positive, so the sharp deflections in the real experiment are repulsions — a positive particle turned away by a positive nucleus it comes close to.

Replace the alpha particles with negative ones and the sign of that force flips. A negative particle passing near a nucleus would be attracted towards it rather than pushed away, so it would be pulled inwards and deflected towards the nucleus instead of bounced back from it. None would come back towards the source in the way a few alpha particles did.

Most would still pass through undeflected, because the atom is still mostly empty space — so the conclusion that the atom is mostly empty would survive, but the striking backward scattering that made the experiment famous would not appear.

Question 5

Rutherford found that a few α-particles bounced back sharply. How does this single surprising result completely rule out Thomson’s ‘plum pudding model’ of the atom?

Solution. Ask what it takes to turn a fast, heavy, positively charged particle right around. It needs a very strong repulsive force, and that means meeting a large positive charge concentrated into a very small volume — something massive enough not to be pushed aside itself.

Thomson’s model has no such thing anywhere in it. Its positive charge is spread thinly through the whole atom, so wherever the alpha particle passes, it feels only a small push from the diffuse charge around it. That can nudge a particle slightly; it cannot reverse it. And the electrons are far too light to turn one back.

So Thomson’s model predicts that no alpha particle should bounce back. Even a few doing so is enough to rule it out — one observation the model cannot produce at all is fatal, however well it explains everything else. The result forces the positive charge to be concentrated in a tiny, massive centre: the nucleus.

Question 6

If you could ask Rutherford one question about his work, what would it be?

Solution. This one is open — any thoughtful question counts, and the best ones come from the gap the chapter itself leaves. Rutherford’s model explained the scattering beautifully but could not explain why atoms are stable, and that is the obvious thing to put to him.

Some questions worth asking:

• If an electron revolving around the nucleus is accelerating, it should radiate energy and spiral into the nucleus. Why, then, does the atom not collapse?
• What holds several positively charged protons together in so small a nucleus, when like charges repel?
• What occupies the enormous empty space between the nucleus and the electrons?
• What made you ask Geiger and Marsden to look for particles scattered backwards, when the model you were testing predicted none?

Pause and Ponder, page 145. A single assertion-and-reason question, after the limitations of Rutherford’s model.

Question 7

Assertion (A): Rutherford concluded that most of the mass of an atom is concentrated in a small region at the centre called the nucleus.

Reason (R): According to Thomson’s model, electrons are embedded in a uniformly distributed positive charge sphere.

Choose the correct option:

Solution. Check each statement, then check the link between them — which is where this question is decided.

A is true. It is exactly what Rutherford concluded from the gold foil experiment: the few alpha particles turned sharply back showed that the positive charge and most of the mass sit in a tiny central nucleus.

R is true. That is a correct statement of Thomson’s model.

But R does not explain A. R describes a different model — the one Rutherford’s result overturned. What actually explains A is the experimental observation that a few alpha particles were deflected through large angles. Thomson’s model is the thing being ruled out, not the reason for the conclusion.

So both statements are true, and the reason is not the explanation: option (ii).

Pause and Ponder, page 149. The box that follows Section 8.4, on the symbols of elements.

Question 8

Imagine you are a scientist who has discovered a new element. Name this element after yourself and justify that the symbol you have chosen follows the IUPAC rules.

Solution. The answer will differ for every reader, but the justification must follow the rules the chapter has just set out, and those are what is being marked.

The rules are: a symbol is one or two letters; the first letter is a capital; and if there is a second letter it is small. It is usually taken from the name of the element — and, as Table 8.2 shows, sometimes from a Latin, Greek or German name rather than the English one (Fe for ferrum, Na for natrium, W for wolfram).

Take a name like Meeranium, after a scientist called Meera. A symbol of Me obeys every rule: two letters, capital M, small e. It should then be checked against the symbols already in use, since two elements cannot share one — and this is precisely why the rules exist.

Question 9

What problems could arise if every scientist used different symbols for the same element?

Solution. Symbols are shorthand for communication, so the cost of disagreement falls entirely on communication.

A chemical equation written by one chemist could not be read by another, or worse, could be read wrongly — the same two letters might mean different elements to different people, so a formula would be ambiguous rather than merely unfamiliar. Research papers, textbooks and databases from different countries could not be compared. Chemical names on medicines, fertilisers and industrial containers would not be reliable, which is a safety problem and not only an academic one. And every student would have to learn several systems instead of one.

This is why symbols are agreed internationally by IUPAC: they let scientists worldwide read one another’s work regardless of language.

Pause and Ponder, page 150. The box on atomic number and mass number — printed as a narrow column beside the body text.

Question 10

An atom with an atomic number of 26 has 56 nucleons. Find out its number of electrons, protons and neutrons.

Solution. Take the two definitions in turn.

The atomic number is the number of protons, so $Z = 26$ gives 26 protons. An atom is electrically neutral, so it has as many electrons as protons: 26 electrons.

“Nucleons” means protons and neutrons together, which is the mass number: $A = 56$. So

$$\text{neutrons} = A – Z = 56 – 26 = 30$$

(The element, incidentally, is iron.)

Question 11

The nucleus of an atom contains 20 protons. If its mass number is 41, find the number of neutrons in it.

Solution. The mass number counts protons and neutrons together, so the neutrons are what is left when the protons are taken away:

$$\text{neutrons} = A – Z = 41 – 20 = 21$$

Question 12

An atom has 18 neutrons and an atomic number of 17. What is its mass number?

Solution. Here the two parts of the nucleus are given and the total is wanted, so the same relation is used the other way round:

$$A = Z + \text{neutrons} = 17 + 18 = 35$$

(That is chlorine-35.)

Question 13

An atom ²³A has 11 electrons. Find the number of neutrons in it.

Solution. The superscript 23 in ²³A is the mass number, so $A = 23$.

The atom is neutral, so it has as many protons as electrons: 11 electrons means 11 protons, and therefore $Z = 11$.

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

Pause and Ponder, page 152. The box that follows Table 8.4, on electronic configuration.

Question 14

Identify the number of electrons in the outermost shell of the following elements:

(i) $^{12}_{\ 6}\text{C}$ (ii) $^{19}_{\ 9}\text{F}$ (iii) $^{28}_{14}\text{Si}$

Solution. Only the lower number, the atomic number, matters here: it gives the number of electrons, and the mass number plays no part in how they are arranged. Fill the shells by Bohr and Bury’s rules — at most 2 in K, then 8 in L, then M — and read off the last shell used.

(i) $^{12}_{\ 6}\text{C}$, $Z = 6$, so 6 electrons: K = 2, L = 4. Configuration 2, 4, outermost shell 4.

(ii) $^{19}_{\ 9}\text{F}$, $Z = 9$, so 9 electrons: K = 2, L = 7. Configuration 2, 7, outermost shell 7.

(iii) $^{28}_{14}\text{Si}$, $Z = 14$, so 14 electrons: K = 2, L = 8, M = 4. Configuration 2, 8, 4, outermost shell 4.

Question 15

Write the electronic configuration of the elements having atomic numbers 12, 16 and 18.

Solution. Fill the shells in order — K, then L, then M — with at most $2n^2$ electrons in the $n$th shell and never more than 8 in the outermost.

$Z = 12$ (magnesium): K = 2, L = 8, leaving 2 for M → 2, 8, 2

$Z = 16$ (sulfur): K = 2, L = 8, leaving 6 for M → 2, 8, 6

$Z = 18$ (argon): K = 2, L = 8, leaving 8 for M → 2, 8, 8

Argon’s outermost shell holds the full 8, which is why it is unreactive — a complete octet.

Question 16

Solve this riddle: I am an atom with a mass number of 23 and 11 protons. I am a soft metal and react vigorously with water. Who am I and how many neutrons do I have? You can also create one such riddle.

Solution. The number of protons identifies the element, because the atomic number is what fixes which element an atom is. With $Z = 11$, Table 8.4 gives sodium (Na) — and the description confirms it: sodium is a soft metal that reacts vigorously with water.

The neutrons then follow from the mass number:

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

To write a riddle of your own, give a mass number and an atomic number and one property, and let the atomic number do the identifying — for example: “I have 17 protons and a mass number of 35. I am a greenish-yellow gas used to disinfect water. Who am I?” (Chlorine, with 18 neutrons.)

Pause and Ponder, page 156. The last box, after Section 8.9.1 on isotopes and average atomic mass.

Question 17

Two different atoms have 11 protons each, but one has 12 neutrons, and the other has 13 neutrons. How do their atomic numbers and mass numbers compare? Are they the same element or different elements?

Solution. Take the two numbers separately.

Atomic numbers. The atomic number is the number of protons, and both atoms have 11. So $Z = 11$ for both — the atomic numbers are the same.

Mass numbers. $A = Z + n$, so

$$A_1 = 11 + 12 = 23, \qquad A_2 = 11 + 13 = 24$$

The mass numbers are different.

Same element or different? What decides the element is the atomic number, not the mass number — it fixes the number of electrons and so the chemical behaviour. Both have $Z = 11$, so both are sodium: the same element, in two forms differing only in neutron count. Such atoms are isotopes.

Question 18

If a bromine atom is available in the form of, say two isotopes, $^{79}_{35}\text{Br}$ (49.7%) and $^{81}_{35}\text{Br}$ (50.3%), calculate the average atomic mass of the bromine atom.

Solution. The average has to be weighted by how common each isotope is — a plain average of 79 and 81 would be right only if the two occurred equally, and they do not.

Multiply each mass by its own percentage and divide by 100:

$$\text{average atomic mass} = \frac{(79 \times 49.7) + (81 \times 50.3)}{100}$$

$$= \frac{3926.3 + 4074.3}{100} = \frac{8000.6}{100} = 80.006\ \text{u}$$

The answer sits just above 80 because the heavier isotope is slightly the more abundant of the two. And, as the chapter says of chlorine, no single bromine atom has this mass — it is the average over a large collection of them.

Common mistakes

• Averaging 79 and 81 to get 80 in Question 18. That ignores the abundances and is right only by luck here, because the two are nearly equal. The weighted average is 80.006 u, and with abundances like chlorine’s 75% and 25% the plain average would be badly wrong.
• Using the mass number to find the electronic configuration in Question 14. The top number plays no part; only the atomic number gives the number of electrons. Using 12 instead of 6 for carbon gives 2, 8, 2 — magnesium’s configuration.
• Choosing (i) in Question 7 because both statements are true. They are, but the reason describes a different model — the one the observation overturned — so it cannot be the explanation.
• Confusing nucleons with neutrons in Question 10. Nucleons are protons and neutrons together, so 56 nucleons is the mass number, and the neutrons are $56 – 26 = 30$, not 56.
• Saying the two atoms in Question 17 are different elements because their masses differ. The element is fixed by the atomic number alone. Same $Z$, different $A$ means isotopes of one element.
• Putting more than 8 electrons in the outermost shell in Question 15. The M shell can hold 18 in general, but never more than 8 while it is the outermost — which is why $Z = 18$ is 2, 8, 8 and not 2, 8, 8 with more to follow.

Practise next

• The end-of-chapter questions, Revise, Reflect, Refine, where the same relation $A = Z + n$ is applied to whole tables of atomic species and to isotopes and isobars.
• Chapter 4, Describing Motion Around Us — its in-text questions cover the accelerated circular motion that Section 8.3 uses to argue that Rutherford’s atom should collapse.