NCERT Solutions Class 9 Science Chapter 8 — Journey Inside the Atom
NCERT Class 9 Science — Journey Inside the Atom, Chapter 8 Revise, Reflect, Refine. All 15 questions solved.
Also here. Solutions to the in-text questions for this chapter are on their own page.
The exercise divides cleanly in two. Questions 1 to 7 and 10 are about the history: what each experiment actually showed, what each model could and could not explain, and in what order the models came. Questions 8, 9 and 11 to 15 are arithmetic on a single relation and its consequences. Four things carry all fifteen:
- An atom is neutral, so the number of electrons equals the number of protons.
- $A = Z + n$ — mass number = atomic number + number of neutrons — read in whichever direction the question needs.
- Shells fill by Bohr and Bury’s rules: at most $2n^2$ in the $n$th shell, at most 8 in the outermost, filled K, L, M, N in order.
- Isotopes have the same $Z$ and different $A$; isobars have the same $A$ and different $Z$.
Key insight. For the history questions, keep separate what an experiment observed from what a scientist proposed. The gold foil experiment observed scattering, and from it Rutherford proposed a nucleus; it observed nothing at all about neutrons or about how electrons move. Questions 1, 4, 7 and 10 all turn on that distinction, and it is what makes two of the four options in Question 1 wrong.
Question 1
Choose the correct options and explain the reason for the correct and incorrect options in the context of Ernest Rutherford’s gold foil experiment:
Solution. Take each in turn, and ask what the experiment could actually have detected.
(i) Incorrect. Alpha particles are deflected by charge, and the neutron has none. The experiment could say nothing about it. The neutron was discovered much later, by Chadwick, and the chapter introduces it separately for exactly that reason.
(ii) Correct. Thomson’s plum pudding model has its positive charge spread thinly through the whole atom, which could never turn a fast alpha particle right around. The observed large deflections are impossible on that model, so it was ruled out, and a small central concentration of positive charge — the nucleus — put in its place.
(iii) Correct. To reverse a heavy, fast alpha particle, the thing it meets must be both highly charged and massive enough not to be brushed aside. Since only a very few particles were deflected, that object must also be very small. Together: most of the mass and all the positive charge in a tiny centre.
(iv) Incorrect. The scattering pattern is produced by the nucleus. It gives no information about where the electrons are or what they do. That electrons revolve around the nucleus was Rutherford’s proposal to complete the picture, not something the deflections showed.
(ii) and (iii) are correct.
(i) is wrong because alpha particles respond to charge and the neutron is uncharged — it was discovered much later. (iv) is wrong because the deflections were caused by the nucleus and revealed nothing about the electrons; their motion was proposed, not observed.
Question 2
Which of the following statements are correct or incorrect according to the Bohr’s atomic model? Give a reason for each statement.
Solution. Bohr’s model was built to fix one specific failure of Rutherford’s, and each statement here is either that fix or the failure it replaced.
(i) Incorrect. This is precisely the problem Bohr’s model was invented to solve, not a feature of it. On Rutherford’s model an orbiting electron would radiate energy and spiral inwards; Bohr’s postulate is that an electron in a permitted orbit does not radiate.
(ii) Incorrect. Bohr allows only certain discrete orbits, each with a definite energy. “Anywhere, with no fixed energy” is the opposite of what the model says.
(iii) Correct. This is Bohr’s central postulate, and it is what makes the atom stable: electrons revolve in special orbits of fixed energy without losing energy.
(iv) Incorrect. The permitted energy levels are discrete, and an electron is never found between them. It can jump from one level to another by absorbing or emitting a fixed amount of energy, but it does not occupy the space in between.
Only (iii) is correct — it is Bohr’s postulate that electrons revolve in fixed orbits of definite energy without radiating, which is what makes the atom stable.
(i) is the failure of Rutherford’s model that Bohr’s was built to remove. (ii) contradicts the discrete orbits Bohr postulated. (iv) is wrong because the energy levels are discrete: an electron jumps between them and is never found in between.
Question 3
The composition of the nuclei of three atomic species X, Y, and Z are given as follows.
| X | Y | Z | |
|---|---|---|---|
| Number of protons | 18 | 17 | 17 |
| Number of neutrons | 19 | 18 | 20 |
Explain the relation between the following:
(i) Y and Z
(ii) Z and X
Solution. Work out the atomic number and mass number of each first; everything follows from comparing those two numbers.
| Protons ($Z$) | Neutrons | Mass number $A = Z + n$ | |
|---|---|---|---|
| X | 18 | 19 | 37 |
| Y | 17 | 18 | 35 |
| Z | 17 | 20 | 37 |
(i) Y and Z. Both have 17 protons, so both have atomic number 17 — they are the same element, chlorine. Their mass numbers differ, 35 against 37. Atoms of the same element with different mass numbers are isotopes.
(ii) Z and X. Their atomic numbers differ, 17 against 18, so they are different elements (chlorine and argon). But both have mass number 37. Atoms of different elements with the same mass number are isobars.
(i) Y and Z are isotopes — both have $Z = 17$ (chlorine) but mass numbers 35 and 37.
(ii) Z and X are isobars — both have mass number 37 but different atomic numbers, 17 and 18, so they are different elements.
Question 4
What conclusion did Rutherford draw about the position and characteristics of the atom’s positively charged part based on the few alpha particles that bounced back or were deflected at large angles in the gold foil experiment?
Solution. Three features of the observation each force a separate conclusion, and it is worth separating them.
That some particles were turned back at all means they met something positively charged — an alpha particle is positive, and only a repulsion could reverse it. So the atom’s positive charge is not spread thinly but gathered together.
That the reversal was sharp and complete means the object was not pushed aside by the impact. It must therefore carry most of the atom’s mass.
That only a very few particles out of many were so deflected means this object occupies a very small part of the atom’s cross-section — most of the atom is empty space. Rutherford put the size at about $10^{-15}$ m against the atom’s $10^{-10}$ m.
Putting the three together: the positive charge sits at the centre of the atom, in a tiny, dense, massive region he named the nucleus.
That the positive charge of an atom is not spread throughout it but concentrated at the centre, in an extremely small, dense region he called the nucleus, which also contains most of the mass of the atom. Since only a few α-particles were deflected, the nucleus must be very small compared with the atom — about $10^{-15}$ m against $10^{-10}$ m — so most of the atom is empty space.
Question 5
Explain and arrange the following statements in the correct chronological order to show how atomic models have evolved over time.
(i) Bohr’s model proposed that electrons move in fixed orbits around the nucleus, each with a definite energy.
(ii) Thomson’s model depicted the atom as a ‘plum pudding’ with electrons embedded in a sphere of positive charge.
(iii) Rutherford’s model proposed that atoms have a dense central nucleus.
(iv) Dalton’s model described atoms as indivisible particles.
Solution. Each model was built to explain something its predecessor could not, so the order is also the logic.
1. (iv) Dalton. The atom as a tiny indivisible particle — the starting point, before anything was known to be inside it.
2. (ii) Thomson. The discovery of the electron showed the atom is not indivisible, so a model was needed with electrons inside it: negatively charged electrons embedded in a sphere of positive charge, the whole thing neutral.
3. (iii) Rutherford. The gold foil experiment showed Thomson’s spread-out positive charge could not produce the large deflections observed, so the positive charge was concentrated into a tiny dense nucleus with the electrons outside it.
4. (i) Bohr. Rutherford’s revolving electron should radiate energy and spiral into the nucleus, so atoms should collapse. Bohr fixed this by allowing only certain fixed orbits, each of definite energy, in which the electron does not radiate.
The correct chronological order is (iv) → (ii) → (iii) → (i):
(iv) Dalton — the atom as an indivisible particle; (ii) Thomson — electrons embedded in a positive sphere, after the electron was discovered; (iii) Rutherford — a dense central nucleus, after the gold foil experiment disproved Thomson’s; (i) Bohr — fixed orbits of definite energy, to explain why the atom does not collapse.
Question 6
Electrons move around the nucleus in orbits. Why do they not fly away from the atom? Explain what keeps them attracted to the nucleus.
Solution. An electron moving in a circle is not moving in a straight line, so by Newton’s first law some force must be acting on it, directed towards the centre. Without such a force it would indeed fly off along a tangent.
That force is electrostatic attraction. The nucleus is positively charged because of its protons, and electrons are negatively charged; unlike charges attract. The attraction points from the electron towards the nucleus — exactly the direction required — and so it provides the centripetal force that keeps the electron in its orbit rather than escaping.
Bohr’s contribution was to add that this can happen without the electron losing energy, provided it stays in one of the permitted orbits. That is why the arrangement is stable and the electron neither flies away nor spirals in.
Because of the electrostatic force of attraction between the positively charged nucleus and the negatively charged electrons. This attraction is directed towards the nucleus and so supplies the centripetal force that keeps each electron in its orbit instead of flying off along a tangent. By Bohr’s postulate an electron in a permitted orbit does so without losing energy, which is why the arrangement is stable.
Question 7
Assertion (A): The discovery of subatomic particles helped in understanding the atomic structure.
Reason (R): The number of electrons is equal to the number of protons in an atom.
Choose the correct option:
Solution. A is true. The whole chapter is the demonstration: each subatomic particle discovered — electron, proton, neutron — forced a better model of the atom.
R is true. An atom is electrically neutral, so its electrons and protons must be equal in number.
But R does not explain A. R is one particular fact about atoms, itself a consequence of discovering electrons and protons. It does not say why discovering subatomic particles advanced our understanding of structure — that came from experiments like the discharge tube and the gold foil, which revealed where those particles sit. A true statement that happens to be about the same subject is not automatically the explanation.
(ii) Both A and R are true, but R is not the correct explanation of A. The equality of electrons and protons is a consequence of discovering those particles, not the reason the discoveries advanced our understanding of atomic structure — that came from what experiments revealed about where the particles sit.
Question 8
Magnesium is essential for many biological processes, including muscle contraction. For an atom of magnesium with a mass number of 24 and atomic number 12, determine the number of (i) protons, (ii) neutrons, (iii) electrons, and also illustrate the arrangement of electrons in a magnesium atom.
Solution. The atomic number is the number of protons, so $Z = 12$ gives 12 protons, and since the atom is neutral, 12 electrons. The neutrons follow from the mass number:
$$\text{neutrons} = A – Z = 24 – 12 = 12$$
Now arrange the 12 electrons by Bohr and Bury’s rules. The K shell takes at most $2 \times 1^2 = 2$; the L shell at most $2 \times 2^2 = 8$; that accounts for 10, leaving 2 for the M shell.
$$\text{K} = 2, \quad \text{L} = 8, \quad \text{M} = 2 \qquad\Longrightarrow\qquad 2,\ 8,\ 2$$
(i) Protons: 12 (ii) Neutrons: $24 – 12 = 12$ (iii) Electrons: 12
The electrons are arranged 2, 8, 2 — two in the K shell, eight in the L shell and two in the M shell, as drawn above.
Question 9
Find the following information for the elements shown in Fig. 8.17:
(i) Name of the element (ii) Symbol (iii) Total number of electrons (iv) Number of valence electrons (v) Valency of the element (vi) Number of protons (vii) Atomic number
Solution. Count the electrons shell by shell in each diagram; the total is the number of electrons, which for a neutral atom is also the number of protons and the atomic number. The valence electrons are those in the outermost shell, and the valency is the number of electrons the atom gains, loses or shares to complete its octet — so if the outermost shell has fewer than four, the valency is that number, and if more than four, it is $8$ minus that number.
(a) K = 2, L = 1 → 3 electrons, configuration 2, 1. Atomic number 3 — lithium. One valence electron, which it loses, so the valency is 1.
(b) K = 2, L = 5 → 7 electrons, configuration 2, 5. Atomic number 7 — nitrogen. Five valence electrons, so it gains $8 – 5 = 3$, and the valency is 3.
(c) K = 2, L = 8, M = 3 → 13 electrons, configuration 2, 8, 3. Atomic number 13 — aluminium. Three valence electrons, which it loses, so the valency is 3.
(d) K = 2, L = 7 → 9 electrons, configuration 2, 7. Atomic number 9 — fluorine. Seven valence electrons, so it gains $8 – 7 = 1$, and the valency is 1.
| (a) | (b) | (c) | (d) | |
|---|---|---|---|---|
| (i) Name | Lithium | Nitrogen | Aluminium | Fluorine |
| (ii) Symbol | Li | N | Al | F |
| (iii) Total electrons | 3 | 7 | 13 | 9 |
| (iv) Valence electrons | 1 | 5 | 3 | 7 |
| (v) Valency | 1 | 3 | 3 | 1 |
| (vi) Protons | 3 | 7 | 13 | 9 |
| (vii) Atomic number | 3 | 7 | 13 | 9 |
(a) Lithium, Li — 3 electrons (2, 1), 1 valence electron, valency 1, 3 protons, $Z = 3$.
(b) Nitrogen, N — 7 electrons (2, 5), 5 valence electrons, valency 3, 7 protons, $Z = 7$.
(c) Aluminium, Al — 13 electrons (2, 8, 3), 3 valence electrons, valency 3, 13 protons, $Z = 13$.
(d) Fluorine, F — 9 electrons (2, 7), 7 valence electrons, valency 1, 9 protons, $Z = 9$.
Question 10
Both Rutherford’s and Bohr’s models have electrons orbiting the nucleus. Why did Rutherford’s model fail to explain atomic stability, while Bohr’s model succeeded?
Solution. The two models place the electrons in the same way, so the difference lies entirely in what each says about the electron’s energy while it orbits.
Why Rutherford’s failed. An electron moving in a circle is constantly changing direction, and changing direction is acceleration. By the physics of the time, an accelerating charged particle continuously radiates energy. An electron that keeps losing energy cannot keep its orbit: it would spiral inwards and fall into the nucleus in a tiny fraction of a second, and every atom would collapse. Matter plainly does not behave that way, so the model could not be right as it stood.
Why Bohr’s succeeded. Bohr made the permitted orbits special. He postulated that certain discrete orbits are allowed, each with a definite energy, and that while an electron stays in one of them it does not radiate energy at all — regardless of its acceleration. With no energy lost, there is no spiralling inwards, and the atom is stable. An electron changes energy only by jumping from one permitted orbit to another, absorbing or emitting a fixed amount as it does so.
Because Rutherford’s model had no answer to radiation. A revolving electron is accelerating, and an accelerating charge should continuously radiate energy, so it would spiral into the nucleus and the atom would collapse. Bohr succeeded by postulating discrete orbits of fixed energy in which the electron does not radiate at all; with no energy lost the electron stays in its orbit, and it changes energy only by jumping between permitted levels.
Question 11
An atom ⁷⁰X has 31 electrons. How many neutrons are there in its nucleus?
Solution. The superscript is the mass number, so $A = 70$.
The atom is neutral, so it has as many protons as electrons: 31 electrons means 31 protons, and $Z = 31$.
$$\text{neutrons} = A – Z = 70 – 31 = 39$$
$$Z = 31 \text{ (from the 31 electrons)}, \qquad \text{neutrons} = 70 – 31 = \mathbf{39}$$
Question 12
An atom has 79 protons and a mass number of 197. Calculate (i) the number of neutrons, and (ii) the number of electrons.
Solution. The protons give the atomic number, $Z = 79$, and the mass number is $A = 197$.
(i) $\text{neutrons} = A – Z = 197 – 79 = 118$
(ii) The atom is neutral, so the number of electrons equals the number of protons: 79.
(The element is gold, the very metal used in Rutherford’s foil.)
(i) $\text{neutrons} = 197 – 79 = \mathbf{118}$
(ii) $\text{electrons} = \text{protons} = \mathbf{79}$
Question 13
Complete the Table 8.5.
Solution. Every blank comes from one of two facts: $A = Z + n$, and in a neutral atom the electrons equal the protons, which equal $Z$. Work each row from whichever two entries are given.
Row 1. $Z = 5$ and $n = 6$, so $A = 5 + 6 = 11$; protons 5, electrons 5. $Z = 5$ is boron.
Row 2. 7 electrons, so $Z = 7$ and 7 protons; $A = 14$, so $n = 14 – 7 = 7$. Named as nitrogen, which confirms $Z = 7$.
Row 3. 12 protons, so $Z = 12$ and 12 electrons; $A = 24$, so $n = 24 – 12 = 12$. $Z = 12$ is magnesium.
Row 4. $Z = 15$ and $n = 16$, so $A = 31$; protons 15, electrons 15. $Z = 15$ is phosphorus.
Row 5. $A = 1$ and $n = 0$, so $Z = 1 – 0 = 1$; protons 1, electrons 1. $Z = 1$ is hydrogen — the one atom with no neutron at all.
| Atomic number | Mass number | Neutrons | Protons | Electrons | Name |
|---|---|---|---|---|---|
| 5 | 11 | 6 | 5 | 5 | Boron |
| 7 | 14 | 7 | 7 | 7 | Nitrogen |
| 12 | 24 | 12 | 12 | 12 | Magnesium |
| 15 | 31 | 16 | 15 | 15 | Phosphorus |
| 1 | 1 | 0 | 1 | 1 | Hydrogen |
The completed table is above: boron (5, 11), nitrogen (7, 14), magnesium (12, 24), phosphorus (15, 31) and hydrogen (1, 1). Every blank follows from $A = Z + n$ and from the electrons equalling the protons in a neutral atom.
Question 14
Aman was discussing the structure of atom with his classmates. During the discussion, he learnt that an element X has a mass number of 35 and contains 18 neutrons. Based on this information, answer the following questions:
(i) How many electrons and protons does element X have?
(ii) What is its atomic number?
(iii) Identify the element X.
(iv) Write its electronic configuration.
(v) How many valence electrons does it have?
(vi) What will be the mass number if two neutrons are added to its nucleus?
(vii) What will be the relation of X with the new atom?
Solution. Start with the atomic number, since everything else follows from it.
(ii) $Z = A – n = 35 – 18 = 17$.
(i) $Z = 17$ means 17 protons, and the atom is neutral, so 17 electrons.
(iii) $Z = 17$ is chlorine, symbol Cl.
(iv) Fill the shells: K = 2, L = 8, leaving 7 for M → 2, 8, 7.
(v) The outermost shell holds 7 valence electrons — one short of a complete octet, which is why chlorine is so reactive and has valency 1.
(vi) Adding two neutrons changes only the neutron count, from 18 to 20. The atomic number stays 17, so
$$A_{\text{new}} = 17 + 20 = 37$$
(vii) The new atom has the same atomic number, 17, and a different mass number, 37 against 35. Atoms of the same element with different mass numbers are isotopes — and these two, chlorine-35 and chlorine-37, are exactly the pair the chapter uses to explain average atomic mass.
(i) 17 electrons and 17 protons (ii) $Z = 35 – 18 = 17$ (iii) Chlorine (Cl) (iv) 2, 8, 7 (v) 7 valence electrons (vi) $A = 17 + 20 = \mathbf{37}$ (vii) The two are isotopes — same atomic number, different mass numbers.
Question 15
In an atom, there are 12 protons and 12 neutrons in the nucleus. Now, imagine that all the electrons are replaced with some hypothetical particles that have the same charge as electrons but are 500 times heavier. What effect will this replacement have on the atom’s:
(i) Atomic number (ii) Atomic mass (iii) Mass number (iv) Overall charge
Solution. The atom is magnesium: $Z = 12$, $A = 24$, and 12 electrons. Only the electrons are replaced, so ask of each quantity whether it depends on the electrons at all.
(i) Atomic number — unchanged, 12. The atomic number counts protons, and not one proton has been touched.
(ii) Atomic mass — it increases. Atomic mass is the actual mass of the atom, and the electrons do contribute to it, even if only slightly: an electron is about $\tfrac{1}{1836}$ of a proton’s mass. Making each 500 times heavier adds
$$12 \times \frac{500}{1836} \approx 3.27\ \text{u}$$
which is no longer negligible — an increase of roughly 3 units on 24.
(iii) Mass number — unchanged, 24. The mass number counts nucleons only: protons plus neutrons, $12 + 12 = 24$. It takes no account of electrons however heavy they are. This is the pair worth noticing — (ii) changes and (iii) does not, because the two mean different things.
(iv) Overall charge — unchanged, still neutral. The replacements carry the same charge as electrons, and there are still 12 of them against 12 protons, so the positive and negative charges still cancel exactly. Mass has nothing to do with charge.
(i) Atomic number: unchanged at 12 — it counts protons only.
(ii) Atomic mass: increases, by about $12 \times 500/1836 \approx 3.3\ \text{u}$, since electrons do contribute to the real mass of an atom.
(iii) Mass number: unchanged at 24 — it counts nucleons (12 protons + 12 neutrons) and ignores electrons entirely.
(iv) Overall charge: unchanged — the atom is still neutral, because the particles carry the same charge as electrons and there are still 12 of them balancing the 12 protons.
Common mistakes
- Choosing (i) or (iv) in Question 1. The gold foil experiment detected charge, so it could say nothing about the uncharged neutron, and its deflections were caused by the nucleus, not the electrons. Both statements describe things the experiment did not show.
- Marking (i) correct in Question 2 because it sounds like something from the chapter. It is — but it is the flaw in Rutherford’s model, the very thing Bohr’s postulate was written to remove.
- Mixing up isotopes and isobars in Question 3. Isotopes share the atomic number (same element); isobars share the mass number (different elements). Comparing $Z$ first and $A$ second keeps them straight.
- Answering Question 15(iii) with a larger mass number. The mass number counts nucleons and nothing else, so heavier electrons cannot change it — while the atomic mass in (ii) does change. The question is testing exactly that distinction.
- Taking the superscript in ⁷⁰X as the atomic number in Question 11. It is the mass number; the atomic number has to come from the 31 electrons.
- Giving nitrogen a valency of 5 in Question 9. With five valence electrons it is easier to gain three than to lose five, so the valency is $8 – 5 = 3$. The same rule gives fluorine 1, not 7.
Practise next
- The in-text Pause and Ponder questions of this chapter, where the same models and the same $A = Z + n$ arithmetic are asked in shorter form, eighteen questions across seven boxes.
- Chapter 4, Describing Motion Around Us — its treatment of circular motion and acceleration is what Questions 6 and 10 rely on when they ask why a revolving electron should radiate.