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Journey Inside the Atom, Class 9 Science Chapter 8: NCERT question answers for every Think It Over, Pause & Ponder, Think as a Scientist, What If, Inline Question, Revise Reflect Refine and Quest Continues question (CBSE 2026-27), with full working for every numerical. Use it with your Journey Inside the Atom Class 9 notes to check each answer.
This chapter follows the atom from Dalton's solid sphere to Thomson's plum pudding model, Rutherford's gold foil experiment and Bohr's shells, and these Class 9 Science Chapter 8 Journey Inside the Atom NCERT solutions explain why each model was replaced. They also work through atomic number and mass number, isotopes and isobars, and electron distribution in K, L, M, N shells for the first 20 elements, with every numerical shown step by step.
Last updated 20 October 2026 · Reviewed by Suchita Arora, 13+ years teaching CBSE Maths & Science
Journey Inside the Atom takes the atom apart piece by piece — from Acharya Kanada's and Dalton's idea of an indivisible particle, through Thomson's plum pudding model, Rutherford's gold foil experiment and nuclear model, to Bohr's fixed energy shells. Along the way it introduces the electron, proton and neutron, and builds up to atomic number, mass number, isotopes and isobars. Every question is solved here, section by section, with full working for every numerical, as complete Class 9 Science Chapter 8 NCERT solutions with answers.
How each atomic model was built on — and eventually broke down against — new experimental evidence.
Why most alpha particles passed straight through gold foil, while a few bounced sharply back.
Working out protons, neutrons and electrons, and telling isotopes and isobars apart with numericals.
Dalton, Acharya Kanada and the Greek philosophers believed that atoms cannot be divided.
But in the late 19th century, scientists found that certain elements emit radiation (radioactivity). This showed that atoms are made of smaller particles.
In 1897, J. J. Thomson discovered the electron, a negatively charged particle that comes out of atoms.
Rutherford later discovered the nucleus and the proton, and Chadwick discovered the neutron.
In Rutherford's model, an electron moving in a circle keeps changing its direction, so it is accelerating and should lose energy and spiral into the nucleus.
Bohr solved this problem by proposing that electrons move only in fixed circular paths called stationary states (shells or energy levels).
While an electron stays in a fixed shell, its energy remains constant, so it does not lose energy.
An electron changes its shell only by absorbing or releasing a fixed amount of energy. It cannot exist in between two shells.
Each model could explain the experiments known at that time, but new experiments gave results that it could not explain.
Dalton's model (atom is indivisible) could not explain radioactivity and the discovery of the electron, so Thomson proposed his model.
Thomson's model could not explain the gold foil experiment, so Rutherford proposed the nuclear model.
Rutherford's model could not explain the stability of the atom, so Bohr proposed the model with fixed shells.
Even Bohr's model was later found to have limitations, and the quantum mechanical model followed.
The charge on the model is the sum of the positive charge of the clay and the negative charge of the beads.
Here the positive charge is less than the total negative charge.
So the net charge is negative.
A bit of negative charge on the clay reduces its positive charge.
Then the total positive charge becomes less than the total negative charge of the beads.
So the net charge is negative again.
How it matches:
The soft pulp is spread throughout the fruit, like the positive charge spread throughout Thomson's atom.
The seeds are embedded in the pulp, like the electrons embedded in the positive sphere.
Where it falls short:
The pulp is not electrically charged, and the seeds do not carry any negative charge, so the fruit does not show how the charges balance.
The seeds are big and few, and they are placed in separate segments, whereas the electrons are tiny and spread through the whole atom.
The fruit has a peel and segments, which an atom does not have.
Thomson passed an electric current through gases at very low pressure in a cathode ray tube and observed cathode rays.
He found that the cathode rays are streams of negatively charged particles (electrons).
He also found that the nature of the cathode rays was independent of the material of the cathode and of the gas in the tube.
So the same particles came out from every kind of material, which means that electrons are present in the atoms of all elements.
The nucleus of a gold atom is positively charged.
A negatively charged particle is attracted by the positive nucleus, whereas an α-particle is repelled by it.
Most of the atom is empty space, so most of the negatively charged particles would still pass straight through the foil.
The few particles passing close to a nucleus would bend towards the nucleus, instead of being pushed away from it.
In Thomson's model, the positive charge is spread evenly over the whole atom.
Such a spread-out positive charge can exert only a weak force on a fast α-particle, so the α-particle should pass straight through or bend only slightly.
To bounce an α-particle straight back, a very strong repulsive force is needed. This is possible only if the positive charge and the mass are concentrated in a very small region.
Thomson's model has no such region.
Sample question: How did you find that the nucleus is so small compared with the atom?
Answer (in Rutherford's words):
Most of the α-particles passed straight through the gold foil, so most of the atom must be empty space.
Only a very few α-particles were deflected through large angles, so the nucleus that deflects them must occupy only a tiny part of the atom.
From these observations, I calculated the diameter of the nucleus as about 10⁻¹⁵ m and that of the atom as about 10⁻¹⁰ m.
So the nucleus is about 10⁵ times smaller than the atom.
Assertion (A): It is true. Rutherford concluded that the nucleus is dense and contains most of the mass of the atom.
Reason (R): It is also a true statement. Thomson's model had electrons embedded in a uniformly distributed positive sphere.
But Thomson's model had no nucleus. So R does not explain how Rutherford reached the conclusion in A.
Rutherford's conclusion came from the gold foil experiment, in which the results did not match Thomson's model.
Example: Suppose the scientist is a student named Aarav. The new element is named Aaravium and its symbol is Aa.
Checking the IUPAC rules:
1. The symbol is formed from the first letter or the first two letters of the name of the element. Here, it is the first two letters, A and a.
2. The first letter is written as a capital letter and the second letter as a small letter. So the symbol is Aa, not AA or aa.
3. The symbol has only two letters, like Al, Co and Ne.
Scientists from different countries would not be able to understand each other's work.
The same element could be mistaken for different elements, and different elements could be mistaken for the same one.
Chemical formulae and equations would be confusing, and mistakes could be made in experiments and in industry.
Language barriers would increase, and the study of chemistry would become difficult.
Given: atomic number \(Z = 26\), number of nucleons = mass number \(A = 56\)
To find: number of electrons, protons and neutrons
Nucleons are the protons and neutrons in the nucleus, so the mass number \(A = 56\).
Number of protons = atomic number = 26
The atom is neutral, so the number of electrons = number of protons = 26
Mass number = number of protons + number of neutrons
Given: number of protons = 20, mass number \(A = 41\)
To find: number of neutrons
Mass number = number of protons + number of neutrons
Given: number of neutrons = 18, atomic number \(Z = 17\)
To find: mass number
Atomic number = number of protons = 17
Mass number = number of protons + number of neutrons
Given: mass number \(A = 23\), number of electrons = 11
To find: number of neutrons
The atom is neutral, so the number of protons = number of electrons = 11.
So the atomic number \(Z = 11\).
Atomic number = 6, so the atom has 6 electrons.
The electrons fill the shells in the order K, L, M: K = 2, L = 4.
The electronic configuration is 2, 4.
Atomic number = 9, so the atom has 9 electrons.
K = 2, L = 7.
The electronic configuration is 2, 7.
Atomic number = 14, so the atom has 14 electrons.
K = 2, L = 8, M = 4.
The electronic configuration is 2, 8, 4.
The maximum number of electrons in a shell is \(2n^2\): K = 2, L = 8, M = 18. The outermost shell can hold at most 8 electrons.
The atom is neutral, so the number of electrons = atomic number. The electrons fill K first, then L, then M.
12 electrons: K = 2, L = 8, remaining = 12 − 2 − 8 = 2, so M = 2.
16 electrons: K = 2, L = 8, remaining = 16 − 2 − 8 = 6, so M = 6.
18 electrons: K = 2, L = 8, remaining = 18 − 2 − 8 = 8, so M = 8.
Given: mass number \(A = 23\), number of protons = 11
To find: the element and its number of neutrons
Atomic number = number of protons = 11.
The element with atomic number 11 is sodium (Na). Sodium is a soft metal and it reacts vigorously with water.
A riddle of your own: I am an atom with a mass number of 35 and 17 protons. I am a yellow-green gas. Who am I and how many neutrons do I have?
Atomic number = 17, so I am chlorine (Cl). Number of neutrons = 35 − 17 = 18.
Atomic number = number of protons = 11 for both atoms, so the atomic numbers are the same.
Mass number of the first atom = 11 + 12 = 23
Mass number of the second atom = 11 + 13 = 24
So the mass numbers are different.
The atomic number decides the identity of an element, so both atoms belong to the same element (sodium). They are isotopes.
Given: isotope of mass 79 u: 49.7%; isotope of mass 81 u: 50.3%
To find: average atomic mass
The average atomic mass is the weighted average of the isotopic masses, taking their relative abundances into account.
A thicker foil has more layers of gold atoms, so there are more nuclei in the path of the α-particles.
So the chance that an α-particle passes close to a nucleus, or hits it, increases.
Expected observations with the thicker foil:
1. Fewer α-particles pass straight through without deflection.
2. More α-particles are deflected, and some may be deflected more than once.
3. More α-particles bounce back.
Rutherford found that the diameter of an atom is about 10⁻¹⁰ m and that of its nucleus is about 10⁻¹⁵ m.
Nearly all the mass of an atom is in the nucleus, and the rest of the atom is mostly empty space.
If there were no empty space, each atom would shrink to about the size of its nucleus, i.e., about 10⁵ times smaller in diameter.
So the length of every object would become about 10⁵ times smaller.
For example, a person of height 1.5 m would become about
The same mass would be packed in a far smaller volume, so objects would be extremely dense.
In Fig. 8.10, the nucleus of the lithium atom has 3 protons and 4 neutrons (Table 8.3), and 3 electrons move around it.
The atomic number is the number of protons in the nucleus.
Each isotope of hydrogen has 1 proton.
Every atom is electrically neutral, so the number of electrons = the number of protons = 1.
| Isotope | Protons | Neutrons | Electrons |
|---|---|---|---|
| Protium \({}^{1}_{1}\text{H}\) | 1 | 0 | 1 |
| Deuterium \({}^{2}_{1}\text{H}\) | 1 | 1 | 1 |
| Tritium \({}^{3}_{1}\text{H}\) | 1 | 2 | 1 |
The combining capacity of an element is the number of hydrogen or chlorine atoms with which one atom of the element combines.
One nitrogen atom combines with 3 hydrogen atoms.
One magnesium atom combines with 2 chlorine atoms.
Eight electrons in the outermost shell make a complete octet (two electrons in the case of helium).
An atom with a complete octet is stable and largely unreactive.
Such atoms have no need to lose, gain or share electrons.
For example, neon (2, 8) and argon (2, 8, 8) have complete octets. Their valency is zero.
Valency is the number of electrons gained, lost or shared to complete the octet.
If the valence electrons are fewer than 4, the atom loses them, so valency = number of valence electrons.
If the valence electrons are more than 4, the atom gains electrons, so valency = 8 − number of valence electrons.
If the shell is already complete (2 for helium or 8), the valency is 0. Carbon (4 valence electrons) shares 4 electrons, so its valency is 4.
| Element | Configuration | Valence e⁻ | Valency |
|---|---|---|---|
| Hydrogen (H) | 1 | 1 | 1 |
| Helium (He) | 2 | 2 | 0 |
| Lithium (Li) | 2, 1 | 1 | 1 |
| Beryllium (Be) | 2, 2 | 2 | 2 |
| Boron (B) | 2, 3 | 3 | 3 |
| Carbon (C) | 2, 4 | 4 | 4 |
| Nitrogen (N) | 2, 5 | 5 | 8 − 5 = 3 |
| Oxygen (O) | 2, 6 | 6 | 8 − 6 = 2 |
| Fluorine (F) | 2, 7 | 7 | 8 − 7 = 1 |
| Neon (Ne) | 2, 8 | 8 | 0 |
| Sodium (Na) | 2, 8, 1 | 1 | 1 |
| Magnesium (Mg) | 2, 8, 2 | 2 | 2 |
| Aluminium (Al) | 2, 8, 3 | 3 | 3 |
| Silicon (Si) | 2, 8, 4 | 4 | 4 |
| Phosphorus (P) | 2, 8, 5 | 5 | 8 − 5 = 3 |
| Sulfur (S) | 2, 8, 6 | 6 | 8 − 6 = 2 |
| Chlorine (Cl) | 2, 8, 7 | 7 | 8 − 7 = 1 |
| Argon (Ar) | 2, 8, 8 | 8 | 0 |
The K shell (n = 1) can hold a maximum of \(2 \times 1^2 = 2\) electrons.
Electrons fill the shell closest to the nucleus first.
So both electrons of helium go into the K shell.
The K shell is then complete (a duplet), so helium is stable and unreactive.
Given: thickness of the sheet = 0.1 mm \(= 10^{-4}\) m; diameter of one atom \(\approx 10^{-10}\) m
To find: the number of atoms stacked one over the other to make the thickness
0.1 mm = 0.1 × 10⁻³ m = 10⁻⁴ m
Number of atoms = thickness of the sheet ÷ diameter of one atom
The experiment was done in 1911, and neutrons were discovered by Chadwick in 1932.
The scattering of α-particles gave no information about neutrons.
In the plum pudding model the positive charge is spread evenly, so all α-particles should have passed with little deflection.
A few α-particles were sharply deflected, which this model could not explain.
This led Rutherford to the idea of a small, dense nucleus at the centre.
Large deflections and bounce-backs of the positive α-particles need a strong repulsion from a heavy, positively charged centre.
Most α-particles went straight, so this centre must be very small.
Electrons are very light and cannot deflect heavy α-particles noticeably.
The deflections came from the nucleus, so the experiment did not show how electrons move.
The planetary picture was Rutherford's proposal, not a result of the scattering.
According to Bohr, an electron in a fixed orbit (stationary state) has constant energy and does not lose energy.
So it does not fall into the nucleus.
Electrons revolve only in certain allowed shells.
Each shell has a definite energy.
This is Bohr's postulate of stationary states. The energy of an electron stays constant while it is in a shell.
Electrons can revolve only in the allowed shells and not in between them.
An electron moves from one shell to another by absorbing or releasing a fixed amount of energy.
| X | Y | Z | |
|---|---|---|---|
| Number of protons | 18 | 17 | 17 |
| Number of neutrons | 19 | 18 | 20 |
Atomic number = number of protons, and mass number = number of protons + number of neutrons.
| X | Y | Z | |
|---|---|---|---|
| Atomic number (Z) | 18 | 17 | 17 |
| Mass number (A) | 18 + 19 = 37 | 17 + 18 = 35 | 17 + 20 = 37 |
Y and Z have the same atomic number (17) but different mass numbers (35 and 37).
They are atoms of the same element (chlorine) with different numbers of neutrons.
Z and X have different atomic numbers (17 and 18) but the same mass number (37).
They are atoms of different elements (chlorine and argon).
Only a very few α-particles were deflected through large angles or bounced back, while most passed straight through.
A strong repulsion on a positive α-particle can come only from a heavy, concentrated, positive charge.
Conclusions of Rutherford:
1. Position: the positive charge is not spread throughout the atom. It is concentrated at the centre of the atom, in a region called the nucleus.
2. Size: the nucleus is extremely small, about 10⁵ times smaller than the atom.
3. Mass and charge: the nucleus is dense and contains all the positive charge and most of the mass of the atom.
Atoms are indivisible particles, the building blocks of matter.
After the discovery of the electron in 1897, the atom was pictured as a ‘plum pudding’: electrons embedded in a sphere of positive charge.
The gold foil experiment led to the nuclear model, with a dense central nucleus.
Bohr explained the stability of the atom with electrons moving in fixed orbits of definite energy.
The nucleus contains protons and is positively charged. Electrons are negatively charged.
Unlike charges attract each other, so there is a force of electrical attraction between the nucleus and every electron.
This attractive force acts towards the nucleus. It keeps the electron moving in its circular path, in the same way as the Sun's attraction keeps the planets in their orbits.
So the electrons cannot fly away.
Assertion (A): It is true. The discovery of the electron, the proton and the neutron helped scientists to build models of the atom.
Reason (R): It is also true. A neutral atom has equal numbers of electrons and protons.
But this fact only tells us why the atom is neutral. It does not explain why the discovery of the subatomic particles helped in understanding the structure.
Given: mass number \(A = 24\), atomic number \(Z = 12\)
Number of protons = atomic number = 12
Number of neutrons = mass number − atomic number
The atom is neutral, so the number of electrons = number of protons = 12
Arrangement of electrons:
K shell: 2 electrons, L shell: 8 electrons, M shell: 12 − 2 − 8 = 2 electrons.
Counting the electrons in each shell of Fig. 8.17:
The atom is neutral, so the number of protons = the number of electrons = the atomic number.
| Atom | K shell | L shell | M shell | Total electrons |
|---|---|---|---|---|
| (a) | 2 | 1 | – | 3 |
| (b) | 2 | 5 | – | 7 |
| (c) | 2 | 8 | 3 | 13 |
| (d) | 2 | 7 | – | 9 |
The atomic number gives the element (Table 8.4): 3 is lithium, 7 is nitrogen, 13 is aluminium and 9 is fluorine.
From Table 8.4:
Adding the electrons of all the shells:
(a) 2 + 1 = 3, (b) 2 + 5 = 7, (c) 2 + 8 + 3 = 13, (d) 2 + 7 = 9
The valence electrons are the electrons in the outermost shell:
(a) 1 valence electron: loses 1 electron, valency 1
(b) 5 valence electrons: gains 8 − 5 = 3 electrons, valency 3
(c) 3 valence electrons: loses 3 electrons, valency 3
(d) 7 valence electrons: gains 8 − 7 = 1 electron, valency 1
Number of protons = number of electrons
Atomic number = number of protons
Rutherford's model:
An electron moving in a circular path keeps changing its direction, so it is accelerating.
An accelerating charged particle should lose energy continuously, so the electron should spiral inwards and fall into the nucleus.
Then atoms would collapse, but in reality atoms are stable. So the model failed to explain the stability of the atom.
Bohr's model:
Bohr proposed that electrons revolve only in certain fixed circular paths called stationary states (shells), each with a definite energy.
While an electron stays in a shell, its energy remains constant, so it does not lose energy and does not fall into the nucleus.
Given: mass number \(A = 70\), number of electrons = 31
To find: number of neutrons
The atom is neutral, so the number of protons = number of electrons = 31.
So the atomic number \(Z = 31\).
Given: number of protons = 79, mass number \(A = 197\)
Number of neutrons = mass number − number of protons
The atom is neutral, so the number of electrons = number of protons.
| Atomic number | Mass number | Number of neutrons | Number of protons | Number of electrons | Name of the element |
|---|---|---|---|---|---|
| 5 | – | 6 | – | – | – |
| – | 14 | – | – | 7 | Nitrogen |
| – | 24 | – | 12 | – | – |
| 15 | – | 16 | – | – | – |
| – | 1 | 0 | – | – | – |
Use: number of protons = atomic number = number of electrons, and mass number = protons + neutrons.
Row 1: \(Z = 5\), so protons = electrons = 5. Mass number = 5 + 6 = 11. The element with \(Z = 5\) is boron.
Row 2: electrons = 7, so protons = \(Z\) = 7. Neutrons = 14 − 7 = 7. The element is nitrogen.
Row 3: protons = 12, so \(Z = 12\) and electrons = 12. Neutrons = 24 − 12 = 12. The element with \(Z = 12\) is magnesium.
Row 4: \(Z = 15\), so protons = electrons = 15. Mass number = 15 + 16 = 31. The element with \(Z = 15\) is phosphorus.
Row 5: mass number 1 with 0 neutrons means protons = 1 − 0 = 1. So \(Z = 1\) and electrons = 1. The element is hydrogen.
| Atomic number | Mass number | Number of neutrons | Number of protons | Number of electrons | Name of the element |
|---|---|---|---|---|---|
| 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 |
Given: mass number \(A = 35\), number of neutrons = 18
Number of protons = mass number − number of neutrons = 35 − 18 = 17
Number of protons = 35 − 18 = 17.
The atom is neutral, so the number of electrons = number of protons = 17.
Atomic number = number of protons
The element with atomic number 17 is chlorine.
17 electrons: K = 2, L = 8, M = 17 − 2 − 8 = 7.
The outermost (M) shell has 7 electrons.
New number of neutrons = 18 + 2 = 20.
New mass number = 17 + 20 = 37.
Both atoms have the same atomic number (17) but different mass numbers (35 and 37).
Given: 12 protons, 12 neutrons; each electron is replaced by a particle of the same charge but 500 times heavier
The atomic number is the number of protons in the nucleus.
The protons are not changed.
The mass of an electron is about \(\dfrac{1}{1840}\) u, which is negligible.
Each new particle has a mass of about \(500 \times \dfrac{1}{1840} \approx 0.27\) u.
Atomic mass before ≈ 12 + 12 = 24 u, and after ≈ 24 + 3.3 ≈ 27.3 u.
The mass number = number of protons + number of neutrons = 12 + 12 = 24.
Electrons are not counted in the mass number.
The new particles have the same charge as electrons, and their number is the same (12).
So the negative charge still balances the positive charge of 12 protons.
Choose three elements: carbon (Z = 6), sodium (Z = 11) and chlorine (Z = 17).
Write the electronic configuration of each element:
| Element | K shell | L shell | M shell |
|---|---|---|---|
| Carbon (C) | 2 | 4 | 0 |
| Sodium (Na) | 2 | 8 | 1 |
| Chlorine (Cl) | 2 | 8 | 7 |
Take the energy levels (K, L, M) on the x-axis and the number of electrons on the y-axis.
Scale: 1 division = 1 electron on the y-axis.
For each shell, draw one bar for each element, with the height equal to the number of electrons.
Our understanding of the atom has changed again and again: from Dalton's indivisible atom to the models of Thomson, Rutherford and Bohr.
Even Bohr's model was later found to be not entirely correct. Electrons do not follow well-defined paths like fixed Bohr orbits.
Today, we understand that electrons exist as ‘electron clouds’ around the nucleus. We can predict the regions where they are most likely to be, but not exactly where they are.
This quantum mechanical model will be learnt in higher grades, and the exploration of the atom is still going on.
Each atomic model — Dalton's indivisible sphere, Thomson's plum pudding, Rutherford's tiny nucleus, Bohr's fixed energy shells — was built to explain what the previous model could not, and the atom's true structure of protons, neutrons and electrons, arranged as atomic number, mass number, isotopes and isobars, is the product of that century-long chain of revision.
Seven extra questions in the style of the textbook's own activities and examples, for independent practice once you've gone through everything above. Attempt each one on paper first, then tap to check.
Given: atomic number \(Z = 13\), mass number \(A = 27\)
Number of protons = atomic number = 13.
The atom is neutral, so the number of electrons = 13.
Number of neutrons = mass number − atomic number = 27 − 13 = 14.
K = 2, L = 8, M = 13 − 2 − 8 = 3.
The outermost shell has 3 electrons, which is fewer than 4, so the atom loses 3 electrons to get an octet.
They have the same atomic number (17) and different mass numbers (35 and 37).
Neutrons in X = 35 − 17 = 18.
Neutrons in Y = 37 − 17 = 20.
Chemical properties depend mainly on the number of valence electrons.
Both have 17 electrons with the same configuration 2, 8, 7.
They have different atomic numbers (20 and 18) and the same mass number (40).
Neutrons in P = 40 − 20 = 20.
Neutrons in Q = 40 − 18 = 22.
Given: isotope of mass 28 u: 92%; isotope of mass 30 u: 8%
To find: average atomic mass of silicon
Electronic configuration of magnesium = 2, 8, 2. It has 2 valence electrons, which is fewer than 4, so it loses 2 electrons to get an octet. Its valency is 2.
Electronic configuration of chlorine = 2, 8, 7. It has 7 valence electrons, so it needs 1 more electron to complete its octet. Its valency is 1.
One magnesium atom has 2 electrons to give, and one chlorine atom can take only 1 electron.
So 2 chlorine atoms are needed to take the 2 electrons given by one magnesium atom.
Given: isotopes of mass 63 u and 65 u; average atomic mass = 63.5 u
To find: percentage abundance of each isotope
Let the abundance of the isotope of mass 63 be \(x\)%. Then the abundance of the isotope of mass 65 is \((100 - x)\)%.
Average atomic mass = weighted average of the isotopic masses:
So the abundance of the isotope of mass 65 = 100 − 75 = 25%.
Check: 63 × 0.75 + 65 × 0.25 = 47.25 + 16.25 = 63.5 u
Given: mass number \(A = 31\), number of neutrons = 16; K shell = 2, L shell = 8
Number of protons = mass number − number of neutrons = 31 − 16 = 15.
Atomic number = number of protons = 15.
The atom is neutral, so the number of electrons = 15.
2 + 8 + \(x\) = 15, so \(x = 5\).
The outermost shell has 5 electrons, so the atom gains 8 − 5 = 3 electrons to complete its octet.
The element with atomic number 15 is phosphorus (P).
Next up is Chapter 9 — Atomic Foundations of Matter. Or revisit Chapter 7, browse the full chapter list, or revise with the Class 9 Science Notes, or test yourself with the free Class 9 Science Sample Paper and Marking Scheme.
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