The rays in a gas discharge.
Goldstein saw the positively charged canal rays in 1886. The proton opened from them.
Class 9 · Science · Chapter 4 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Structure of the Atom
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आवेशित कण — क्रियाकलाप 4.1 · Bihar 4.1 · Activity 4.1
Dalton had called the atom indivisible. By the end of the century static electricity showed that the story was incomplete. On rubbing, an object becomes charged. The charge does not arrive as new matter from outside. It comes from particles inside the atom.
By 1900 one sub-atomic particle had been identified, the electron. The identification belongs to J. J. Thomson. The symbol is e−. The charge is taken as minus one and the mass as negligible.
Goldstein saw canal rays in 1886. They were positively charged radiations. The proton opened from them. The proton charge equals the electron charge with the opposite sign. The mass is about 2000 times the electron mass. The symbol is p+. The mass is taken as one unit and the charge as plus one. An atom with one proton and one electron is neutral.
This activity has two parts. Do not turn them into Activity 4.2.
A. Comb dry hair. Does the comb then attract small pieces of paper?
B. Rub a glass rod with a silk cloth and bring the rod near an inflated balloon. Watch what happens.
When two objects are rubbed they take up an electric charge. The question is where the charge came from. The answer is that the atom is divisible and contains charged particles. An electron comes away more easily. The proton stays inside.
Question: An atom has one electron and one proton. What charge does it carry?
Formula: total charge = charge of the proton + charge of the electron.
Substitution: (+1) + (−1) = 0.
Unit: this is the same charge unit in which the proton is +1 and the electron is −1. The atom is neutral.
Join canal rays and the proton in one sentence. Only “a positive particle” stays incomplete.
The factor 2000 belongs to the proton and the electron. The neutron mass is about equal to the proton mass.
The rays in a gas discharge.
Goldstein saw the positively charged canal rays in 1886. The proton opened from them.
False — the proton mass is about 2000 times the electron mass. The electron mass is taken as negligible.
The 1906 Nobel is linked with this work.
J. J. Thomson. The proton opened from canal rays and the neutron belongs to Chadwick.
Add (+1) and (−1).
(+1) + (−1) = 0. The atom is neutral.
A comb run through dry hair attracts paper. A glass rod rubbed with silk shows an effect near a balloon. The charge comes from particles inside the atom, because the atom is divisible.
थॉमसन का मॉडल · Bihar 4.2.1 · positive sphere
Thomson was the first to give a model of the structure of the atom. Electrons are embedded in a sphere of positive charge, like currants in a Christmas pudding. The watermelon picture says the same thing. The red part is the positive charge and the seeds are the electrons.
Two points sit in the model. The atom is a positive sphere and the electrons are sunk in it. The negative and positive charges are equal, so the whole atom is neutral.
There is no nucleus in this model. Neutrality is explained. The later scattering experiment is not. It is not a Thomson statement that the nucleus holds only nucleons.
The gold foil and the nucleus belong to Rutherford. If the question asks only for the neutral sphere of Thomson, writing the nucleus changes the answer.
Question: Why is the atom neutral in the Thomson model?
Formula: total charge = positive charge + negative charge.
Substitution: the two charges are equal in size, so the sum is 0.
Unit: the result is 0 in the charge unit. The sphere is positive, the electrons are negative, and the whole atom is neutral.
Write the two points in order: the sphere, then the equality of the charges.
Keep the limit in one line too. The model does not explain the scattering.
False — the positive charge is spread through the sphere. The nucleus came from the Rutherford experiment.
The picture of watermelon seeds.
The electrons are embedded in the positive sphere.
The atom is a positive sphere and the electrons are embedded in it. Equal positive and negative charges make the atom neutral. This model does not explain the scattering of alpha particles.
The red part of the watermelon.
In a positive sphere. The electrons are embedded in it.
रदरफोर्ड और सोने की पत्ती · Bihar 4.2.2 · scattering · unnumbered picture
Rutherford wanted to know how the electrons sit in the atom. Fast alpha particles were sent at a thin gold foil. The foil was about 1000 atoms thick, and still as thin as possible. Alpha particles are doubly charged helium ions. The mass is 4 u, so a fast particle carries a good deal of energy. Because they are heavier than protons, large deflections were not expected.
The results went the other way. Most particles went straight through. Some bent by small angles. About one in every 12000 bounced back. Rutherford found that rebound astonishing.
There are three conclusions. Most of the space is empty, because most particles did not bend. The positive charge occupies a very small space, because few particles bent. All the positive charge and the mass sit in a very small volume, because very few particles turned by about 180° and came back. The radius of the nucleus came out about 105 times smaller than the radius of the atom.
The nuclear model: there is a positive nucleus and nearly all the mass is there. Electrons revolve in circular paths. The nucleus is very small compared with the atom. A revolving electron can lose energy and fall into the nucleus, so the model does not explain the stability of the atom.
This picture has no activity number. Do not write 4.3.
A child with closed eyes who throws stones at a wall hears a sound every time. Thrown at a barbed-wire fence, most stones pass through the gaps and no sound comes. The gold foil is like that fence. Most alpha particles pass through empty space. A particle that comes back has hit the small dense place called the nucleus.
Question: About how many of 12000 alpha particles rebound, and how much smaller is the nuclear radius than the atom?
Formula: rebounding particles = total particles / 12000. The radius ratio is about 1 : 105.
Substitution: 12000 / 12000 = 1 particle. The radius is about 105 times smaller.
Unit: the count is one particle. The radius comparison is a factor, not grams.
Write the three observations and the three conclusions apart. Mixing them in one sentence costs marks.
The drawback is a separate question. The discovery of the nucleus is right. The explanation of stability is not.
Thomson is the electron, Chadwick the neutron.
The atomic nucleus. Most particles went straight and very few rebounded.
False — they are doubly charged helium ions and the mass is 4 u.
The foil was about 1000 atoms thick. This number is the other one.
12000.
The gaps in the fence.
Most of the space is empty.
A charged electron revolving in a circular path would lose energy and should fall into the nucleus. The atom would not stay stable, while atoms are stable.
True — gold was chosen to get a thin foil. The nuclear structure is not special to gold.
बोहर की कक्षा और न्यूट्रॉन · Bihar 4.2.3 and 4.2.4
To remove the drawback of Rutherford, Niels Bohr stated two points. Only certain special orbits are allowed inside the atom. These are called discrete orbits. While revolving in these orbits the electron does not lose energy.
These orbits are also called energy levels. They are named K, L, M, N or n = 1, 2, 3, 4. When the shells are complete the atom is more stable and less reactive. That point opens later in valency.
In 1932 J. Chadwick found another particle. It has no charge and its mass is nearly equal to the proton mass. The name is neutron. The symbol is n. Every atom except hydrogen has neutrons in the nucleus. The mass of an atom is the sum of the masses of the protons and the neutrons. Both sit in the nucleus, so they are also called nucleons.
There are three particles: the electron negative, the proton positive, the neutron uncharged. A neutron is not made by sticking an electron and a proton together.
Question: A helium atom has atomic mass 4 u and two protons in its nucleus. How many neutrons does it have?
Formula: neutrons = mass number − protons. The mass number here is 4.
Substitution: 4 − 2 = 2.
Unit: 2 neutrons. About 2 u of the mass belongs to them, because each neutron is about 1 u.
Write both Bohr points in full. Only “an orbit” does not separate him from Rutherford.
In the helium question show 4 − 2 = 2. Only “two”, with no subtraction, stays incomplete.
This is the answer to the Rutherford drawback.
In a discrete orbit the electron does not lose energy.
False — the neutron is a separate particle. The charge is zero, but it is not a stuck-together pair.
The particle with zero charge.
J. Chadwick.
Subtract 2 from 4.
4 − 2 = 2 neutrons.
The electron is negative, Thomson. The proton is positive, from canal rays, Goldstein. The neutron has no charge, Chadwick, 1932.
इलेक्ट्रॉन बाँटना — क्रियाकलाप 4.2 · Bihar 4.3 · Activity 4.2 · 2n^2
Bohr and Bury showed how to share electrons into orbits.
First rule: the maximum number of electrons in a shell is 2n2. n is the orbit number.
| Shell | n | 2n^2 |
|---|---|---|
| K | 1 | 2 |
| L | 2 | 8 |
| M | 3 | 18 |
| N | 4 | 32 |
Second rule: the outermost shell does not hold more than 8 electrons. Third rule: electrons are not put in an outer shell until the inner shells are filled. The filling is stepwise.
The table of the first eighteen elements stops here. Argon is 2, 8, 8. The 2n2 maximum of M is 18, but in argon M is the outer shell, so it holds 8. The counts 2, 8, 8 in the picture are the outer filling. The full capacity of M is 18. Helium has 2 electrons in its outer shell and valency 0. The octet of 8 belongs to the later inert elements.
Make a static atomic model of the first eighteen elements that shows the electronic configuration. From hydrogen to argon, fill K, L and M. Fill the inner shell first. Carbon is 2, 4. Sodium is 2, 8, 1. Argon is 2, 8, 8. Do not write charges on the model. Write only the shell and the number of electrons.
Question: What is the 2n2 maximum for the M shell? How many does argon have in M?
Formula: maximum = 2n2. For M, n = 3. The outer-shell rule is separate.
Substitution: 2 × 32 = 2 × 9 = 18. In argon M is outermost, so it holds 8. The configuration is 2, 8, 8.
Unit: the capacity is 18 electrons. That shell in argon has 8 electrons.
Write both 2n^2 and the outer limit. Do not turn the capacity of M into 8 by writing only 2, 8, 8.
Sodium is 2, 8, 1. Both 2, 8 and 8, 2, 1 are the wrong order.
10-second revision
n = 2.
2 × 2^2 = 8. The K maximum is 2 and the M maximum is 18.
False — the outermost shell holds at most 8. 18 is the 2n^2 maximum of M when it is not the outer limit in use.
Fill K and L, and put the last one in M.
2, 8, 1. The atomic number is 11.
2 + 8.
2 + 8 = 10.
Carbon is 2, 4. Sodium is 2, 8, 1. The third rule says electrons do not go to an outer shell until the inner shell is filled.
True — the comb part is Activity 4.1.
संयोजकता और बाहरी कोश · Bihar 4.4 · octet
The electrons of the outer shell are the valence electrons. If the outer shell is completely full, the valency is zero. Helium has 2 outer electrons. The later inert elements have 8. An outer shell of 8 electrons is an octet.
An atom loses, gains or shares electrons to complete the octet. The number of electrons used in that work is the valency. Hydrogen, lithium and sodium have one electron in the outer shell, so the valency is 1. Magnesium is 2, aluminium is 3.
If the outer electrons are close to a full shell, the count turns around. Fluorine has 7 in the outer shell. Losing seven is hard and gaining one is easy. The valency is 8 − 7 = 1, not 7. Oxygen has 6, so the valency is 2. Chlorine is 1, sulphur is 2. Silicon has 4 in the outer shell, so the valency is 4. The table gives phosphorus both 3 and 5.
Question: Find the valency of oxygen, chlorine and magnesium.
Formula: near a full shell, valency = 8 − outer electrons. Far from it, valency = outer electrons.
Substitution: oxygen 8 − 6 = 2. Chlorine 8 − 7 = 1. Magnesium loses 2 outer electrons, valency 2.
Unit: the three answers are counts: 2, 1 and 2.
Do not write 7 for fluorine. The paper needs 8 − 7 on the page.
Do not call helium an octet. Its full shell is 2 and the valency is 0.
10-second revision
Subtract 7 from 8.
1. Gaining one electron is easier than losing seven.
The configuration is 2, 8, 4.
4. There are 4 electrons in the outer shell.
False — the outer shell is full, so the valency is 0. Two electrons are a duplet, not an octet.
Chlorine has 7 in the outer shell, valency 8 − 7 = 1. Sulphur has 6, valency 2. Magnesium has 2 and loses both, so the valency is 2.
परमाणु क्रमांक और द्रव्यमान संख्या · Bihar 4.5 · Z and A
The atomic number Z is the number of protons in the nucleus. Every atom of an element has the same Z. An element is identified by its protons. For hydrogen Z = 1. For carbon Z = 6. In a neutral atom the electrons also equal Z.
Almost the whole mass belongs to protons and neutrons. Both are in the nucleus and are called nucleons. Mass number = number of protons + number of neutrons. Carbon has 6 + 6 = 12, so the mass is 12 u. Aluminium has 13 protons and 14 neutrons, mass number 27.
The way to write it: mass number above, atomic number below, symbol in the middle. Nitrogen is 147N. This is not a chemical equation. The mass number of oxygen is 8 + 8 = 16. Sulphur is 16 + 16 = 32.
Question: An atom has 8 electrons and 8 protons. Find the atomic number and the charge.
Formula: Z = number of protons. Charge = protons − electrons, in the neutral unit.
Substitution: Z = 8. Charge = 8 − 8 = 0.
Unit: the atomic number is 8. The charge is 0. This is a neutral oxygen atom.
Keep the mass number and the unit of atomic mass apart. The number is 32, the mass is about 32 u.
Write the mass number above the symbol and the atomic number below. Reversing them changes the meaning of an isotope.
10-second revision
This is the identity of the element.
The number of protons. A neutral atom has the same number of electrons.
False — the mass number is the sum of protons and neutrons.
Protons 16 and neutrons 16.
32. 16 + 16 = 32.
Z is also 8.
0. The atom is neutral and the atomic number is 8.
The atomic number is the number of protons in the nucleus. For carbon Z = 6. The mass number is the sum of protons and neutrons. In carbon 6 + 6 = 12.
समस्थानिक और समभारिक · Bihar 4.6 · average mass
Some atoms of an element have the same atomic number and different mass numbers. These are isotopes. Hydrogen has three forms. Protium has 1 proton and 0 neutrons. Deuterium has 1 proton and 1 neutron. Tritium has 1 proton and 2 neutrons. 126C and 146C of carbon are isotopes. So are 35 and 37 of chlorine.
Each isotope is a pure substance. Chemical properties stay alike and physical properties change. The average mass comes from the percentages. In nature chlorine is 35 u and 37 u in the ratio 3:1. No single atom has a mass of 35.5 u.
The uses differ. One isotope of uranium is a fuel in nuclear reactors. One isotope of cobalt is used in the treatment of cancer. One isotope of iodine is used in the treatment of goitre. Tincture of iodine is not that isotope.
Isobars are atoms of different elements that have the same mass number and different atomic numbers. Calcium has atomic number 20 and argon has 18. Both have mass number 40. Isotopes share Z. Isobars share A.
The average belongs to the sample. Show 3:1 or 75% and 25% in the answer. Do not write tincture as the use of the isotope.
Question: The chlorine isotopes 35 u and 37 u are in the ratio 3:1. Find the average atomic mass.
Formula: average = (3 × 35 + 1 × 37) / 4.
Substitution: (105 + 37) / 4 = 142 / 4 = 35.5.
Unit: 35.5 u. This is an average. Do not write 35.5 u on one single atom.
In an average question show the sum with the percent or with 3:1. Writing 35.5 from memory alone is incomplete.
Write one pair of each: Cl-35 and Cl-37, and Ar-40 and Ca-40.
10-second revision
Z stays the same.
The number of neutrons differs, so the mass number changes.
False — the isotope is used in the treatment of goitre. Tincture is a solution of ordinary iodine.
The atomic numbers 18 and 20 differ.
Isobars. Isotopes have the same atomic number.
(3×35 + 37)/4.
The 3:1 average is 35.5 u. A single atom is not 35.5 u.
All three have 1 proton and 1 electron. Neutrons are 0 in protium, 1 in deuterium and 2 in tritium.
True — Z is 6 for both. The mass numbers are 12 and 14. These are isotopes of carbon.
Pick a type. The 41 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
1886.
Goldstein. The rays were positively charged.
The electron is negligible.
About 2000 times.
The sphere model.
The positive and negative charges are equal.
Empty space.
They go straight through.
A charge that loses energy.
A revolving electron can lose energy and fall into the nucleus.
Zero charge.
Chadwick in 1932.
n = 3.
2 × 9 = 18. The outer rule says 8 separately.
K, L, then M.
2, 8, 1.
2, 8, 7 gains one and becomes 2, 8, 8.
8. The total electrons become 18, but the outer shell has 8.
Z.
The number of protons.
Same A, different Z.
Argon and calcium. The first two pairs are isotopes.
Configuration 2, 1.
Lithium, valency 1.
K = 2 and L = 8.
The atom was 2, 8, 1. After losing one electron, 2, 8 remains.
(49.7×79 + 50.3×81)/100 = 80.006.
80.006 u, that is about 80 u. 35.5 is the chlorine average.
False — the Thomson model has no nucleus at all.
True — the proton is about 2000 times heavier.
False — protium has 1 proton and 0 neutrons. The book sets hydrogen aside from the neutron statement.
True — the outer shell is full with 2 electrons.
False — 35.5 u is an average. The isotopes are 35 u and 37 u.
True — the physical properties differ.
True — it is a doubly charged helium ion.
False — the outermost shell holds at most 8. 18 is a capacity, not the outer limit.
1886.
Goldstein.
About 10^5 times.
The gold foil was about 1000 atoms thick. This number is the other one.
2.
n = 1, 2n^2 = 2.
6.
6 protons.
27.
13 protons + 14 neutrons.
2. 8 − 6 = 2.
There are 6 outer electrons.
35.5 u.
The 3:1 average.
Calcium.
Atomic number 20.
Thomson the electron, Goldstein canal rays, Rutherford the nucleus, Chadwick the neutron.
The hydrogen, chlorine and carbon pairs are isotopes. Argon and calcium are isobars of mass 40.
Assertion (A): Most of the space inside the atom is empty.
Reason (R): Most alpha particles went straight through the gold foil.
Both are true and R explains A.
Assertion (A): In a Bohr discrete orbit the electron does not lose energy.
Reason (R): This removes the falling drawback of the Rutherford model.
Both are true and R is the correct reason.
Assertion (A): Isotopes have the same chemical properties.
Reason (R): The physical properties of isotopes are always the same too.
A is true. R is false. The physical properties differ.
Assertion (A): The Thomson model has a positive nucleus.
Reason (R): The nucleus was found from alpha scattering.
A is false. R is true. The nucleus is the result of the Rutherford experiment.
Assertion (A): The average atomic mass of chlorine is 35.5 u.
Reason (R): Argon and calcium are isobars.
Both are true, but R does not explain the average 35.5. The average comes from 35 and 37 in the ratio 3:1.
The electron is negative. The proton and the alpha particle are positive. The neutron is zero.
Only argon and calcium are isobars. The others are isotopes.
The positively charged rays seen by Goldstein in 1886 are canal rays. The proton opened from them.
This model does not explain alpha-particle scattering, because it has no small positive nucleus.
The mass number is the total number of protons and neutrons in the nucleus.
Atoms with different atomic numbers and the same mass number are isobars. The pair is argon and calcium, both of mass number 40.
Valency is the combining capacity of an atom, the number of electrons lost, gained or shared to complete the octet.
Nearly all the mass is in the positive nucleus. Electrons revolve in circular paths. The drawback is that a revolving electron can lose energy and fall into the nucleus, while the atom is stable.
Carbon is 2, 4. Sodium is 2, 8, 1. Argon is 2, 8, 8. The outer shell does not hold more than 8 electrons. The 2n^2 maximum of M is still 18.
Silicon is 2, 8, 4. It shares 4 outer electrons, valency 4. Oxygen is 2, 6. It needs 2 electrons for the octet, valency 8 − 6 = 2.
Average = (49.7 × 79 + 50.3 × 81) / 100 = (3926.3 + 4074.3) / 100 = 8000.6 / 100 = 80.006 u.
Thomson: electrons are embedded in a positive sphere and the atom is neutral. Limit: it does not explain the scattering. Rutherford: a small positive nucleus and revolving electrons. Limit: the electron would lose energy and fall. Bohr: only discrete orbits, and they do not radiate. This repairs the Rutherford drawback.
The atomic number is the number of protons, 6 for carbon. The mass number is protons plus neutrons, 12 for carbon. Isotopes have the same atomic number and different masses, such as Cl-35 and Cl-37. Isobars have the same mass and different atomic numbers, such as Ar and Ca, both 40. Uses: an isotope of uranium is a nuclear fuel, and an isotope of cobalt is used in the treatment of cancer.
Let the percentage of 18X be x. (16(100 − x) + 18x) / 100 = 16.2. 1600 + 2x = 1620, so x = 10. 18X is 10% and 16X is 90%. Chlorine: (3×35 + 1×37)/4 = 142/4 = 35.5 u.
This model set is for practice. It is not a question from the annual examination of any year. Annual questions will be added only when a source page is available.
12000 is the rebound count.
About 1000 atoms.
Valency 0.
2, 8, 8.
2.
4 − 2.
On rubbing, the comb becomes charged and attracts paper. The charge comes from particles inside the atom, so the atom is not indivisible.
Sodium is 2, 8, 1. After losing one electron, 2, 8 remains. K has 2 and L has 8, so both are full. Cl− has the configuration 2, 8, 8, so the valence electrons are 8.
The maximum is 2n^2. The outer shell holds no more than 8. Inner shells fill first. Carbon is 2, 4. Sodium is 2, 8, 1. If K and L are full, the electrons are 2 + 8 = 10. The maximum of M is 2 × 3^2 = 18.
These are competency-based practice questions. They are not copies of a CBSE paper.
The outer rule and 2n^2 are separate.
The capacity is 18. In argon M is outermost, so there are 8 electrons.
1/12000.
About one will rebound. Most will go straight through the empty space.
Assertion (A): Cl− has 8 valence electrons.
Reason (R): Chlorine 2, 8, 7 gains one electron and becomes 2, 8, 8.
Both are true and R explains A. The total electrons are 18, but the valence electrons are 8.
Assertion (A): Tincture of iodine is made from an isotope of iodine.
Reason (R): One isotope of iodine is used in the treatment of goitre.
A is false. R is true. Tincture is an ordinary solution.
Zero charge does not prove that it is a pair. The neutron is a separate particle. Chadwick found it in 1932. The mass is about equal to the proton and it stays in the nucleus of every atom except hydrogen.
From (16(100 − x) + 18x)/100 = 16.2, x = 10. The heavier isotope is 10%. The 35.5 of chlorine is also not the mass of one atom. It is the average of two isotopes.
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What you learned
| What | Keep this |
|---|---|
| Proton | charge +1, mass about 1 u |
| Electron | charge −1, mass about 1/2000 u |
| Neutron | charge 0, mass about 1 u, none in protium |
| Shell rule | 2n^2, and at most 8 in the outer shell |
| Chlorine average | 35.5 u from 35 and 37 in the ratio 3:1 |
| Isobar pair | argon and calcium, both mass number 40 |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.