NCERT solutions of class 9 science chapter 4-Structure of Atom
NCERT solutions of class 9 science chapter 4-Structure of Atom
Here are integral NCERT solutions of Class 9 Science Chapter 4-Structure of Atom are created for helping the class 9 students. These solutions are explained here by using the proper scientific method throgh pictures ,tables and formula by a step by step clarifications of unsolved questions of Chapter 4 named Structure of the Atom of Science NCERT textbook class 9. On the off chance that you are an student of class 9 who is utilizing NCERT Textbook to contemplate Science coarse for the preparation of exam, at that point you should go over Chapter 4 Structure of the Atom. After you have examined exercise, you should be searching for answers of unsolved questions. Here you can get total NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom in one spot. For a superior comprehension of this part, you ought to likewise see Chapter 4 Structure of the Atom Class 9 science notes.
Click for online shopping
Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc
Download PDF-NCERT solutions of class 9 science chapter 4-Structure of Atom
PDF-NCERT solutions of class 9 science chapter 4-Structure of Atom
NCERT solutions of class 9 science chapter 4-Structure of Atom
Page no 47
Q1.What are canal rays ?
Ans. Canal rays is a beam of positive charged particles (positive ions),which originates from anode and directed towards cathode under the influence of high voltage.It is discovered by Goldstein, it lead him to discover protons after this experiment.
Q2.If an atom contains one electron and one proton. Will it carry any charge or not ?
Ans. charge in one electron = – 1.6 x 10-19C
Charge in one proton = +1.6 x 10-19C
The net charge of atom = 0, so an atom containing the same number of protons and electrons will always be neutral.
Page no 49
Q1.On the basis of Thomson’s model of an atom, explain how the atom is neutral as a whole.
Ans. According to J.J Thomson’s model, the atom is spherical in shape and negative charge, electrons are embedded on positive charge on the same way as seeds are on the edible part of the watermelon, In atom total positive charge is equal to total negative charge, so net charge on an atom is zero or atom is neutral.
Q2.On the basis of Rutherford’s model of an atom, which subatomic particle is present in the nuleus of an atom ?
Ans. On the basis of Rutherford’s model protons are present in the nucleus of an atom.
Q3.Draw a sketch of Bohr’s model of an atom with three shells.
Ans.
Q4.What do you think would be the observation of the α- particle scattering experiment is carried out using a foil of a metal other than gold.
Ans. If the observation of the α particle scattering experiment carried out using a foil of a metal other than gold , there wouldn’t be any change in the observation .As it is required thinest foil so that α particles could be penetrated.Gold has the greatest malleability so such a thin foil which is equivalent to the width of few layers of atoms can be made by the gold only.
Page no 49
Q1. Name the three sub-atomic particles of an atom.
Ans.Three sub-atomic particles of an atom are
(i)Protons (ii) Neutron (iii) Electron
Q2. The helium atom has an atomic mass of 4 u and two protons in its nucleus. How many neutrons does it have?
Ans. Atomic mass = number of protons + number of electrons
Atomic mass of He = 4 u
Number of protons in it =2
Therefore, number of neutrons = atomic mass – number of protons = 4 – 2 = 2
Page no 50
Q1.Write the distribution of electrons in carbon and sodium atoms.
Ans. No of the electrons in carbon and sodium atom has 6 and 11 respectively.
Therefore, electrons distribution of them will be as follows
C ⇒ 2, 4
Na ⇒2, 8, 1
Q2. If K and L shells of an atom are full , then what would be the total number of electrons in the atom ?
Ans. Maximum electrons K shell can accommodate = 2n2 = 2 x12 = 2
Maximum electrons L shell can hold = 2 x 22=8
Therefore, total number of electrons the atom will have =2 + 8 = 10
Q3.Have will you find the valency of chlorine, sulphur and magnesium?
Ans. Atomic numbers of chlorine, sulphur, and magnesium are 17, 16, and 12 respectively ,so their electronic structure will be as follows. All the elements have a tendency of making an octet in their outermost orbit and whatever exchange of electrons is taken place in making it is known as valency.
Cl ⇒ 2,8,7
The valency of chlorine = – 1
S ⇒2, 8, 6
The valency of sulphur = – 2
Mg⇒2, 8, 2
The valency of magnesium = +2
Q4. If the number of electrons in an atom is 8 and the number of protons is also 8, then (i) what is the atomic number of the atom? and (ii)What is the charge on the atom?
Ans.(i)Atomic number of the atom = number of protons= number of electrons in it
Therefore, atomic number of the given atom =8
(ii) Charges on one electron and on one proton are equal and opposite to each other,there numbers in atom are equal so net charge on atom will be zero.
Q5.With the help of Table 4.1. Find out the mass number of oxygen and sulphur atom.
Ans. In the table the number of protons in oxygen=8 and number of neutrons = 8, so mass number of oxygen = 8 + 8 = 16 u. Number of protons and neutrons in sulphur are each of 16, so mass number of sulphur = 16 + 16 = 32 u
Page no-53
Q1. For the symbol H,D and T tabulate three sub – atomic particles found in each of them.
Ans. H, D and T are the symbols used for three isotopes of the hydrogen of the name protium, duetirium and tritium respectively, The atomic number of each one is 1 and their atomic masses are 1,2 and 3 respectively . So number of subatomic particles in them will be as following.
H→ neutrons =Atomic mass -Atomic number= 1 – 1 = 0, protons =1, electrons=1
D→ neutrons = Atomic mass -Atomic number=2- 1 = 1, protons = 1, electrons =1
T→nuetrons =Atomic mass -Atomic number= 3 – 1 = 2, protons = 1, electrons =1
Q2.Write the electronic configuration of any one pair of isotopes and isobars.
Ans- Electronic configuration of two isotopes of chlorine are
35Cl17 →2, 8, 7 and 37Cl17 →2,8,7
Electronic configuration of a pair of isobars (Ca and Ar) are
40Ca20 →2,8,8,2 and 40Ar18→ 2, 8, 8
EXERCISES
Q1.Compare the properties of electrons, protons and neutrons.
Ans.
| Properties | Proton | Nuetron | Electron |
| Charge | +1.6× 10-19 | Nuetral | -1.6× 10-19 |
| Mass | 200 times of the mass of electron | Almost equal to proton | Negligible |
| Location | Located at nucleus | Located at nucleus | Revolves around the nucleus |
Q2.What are the limitations of J.J.Thompson’s model of the atom ?
Ans. According to J.J Thomson’s atomic model electrons are embedded on the positive charge but later it was found that electrons revolves around the nucleus and positive charge protons resides in the nucleus of the atom.
Q3.What are the limitations of Rutherford’s model of the atom ?
Ans.According to Rutherford,s model of the atom, the electrons revolves around the nucleus in fixed and circular orbital where it will be accelerated continuously ,it will loss energy in the form of radiations as electron is a charged particle as a result ultimately will collapse at the nucleus and atom will be finished,while atom is a stabilized particle.
Q4.Describe Bohr’s model of the atom.
Ans. According to Bohr’s model, all the electrons revolve in discrete orbitals, the electrons filled up in these orbits as per rule 2n2, where n is the number of orbitals. He named these orbitals, first orbital, K, second orbital, L, third orbital, M, and so on.., the number of electrons in these orbitals are fixed. These are discrete orbitals where electrons revolve but do not radiate their energy.
Q5.Compare all the proposed models of an atom given in this chapter.
| J.J.Thompson’s Atomic Model | Ruthurford’s Atomic Model | Neils Bohr’s Atomic Model |
| J.J.Thomson compared the atom with watermelon,according to him atom is spherical in shape and electrons are embedded on the surface of atom just like seeds on the watermelon,these electrons constitute negative charge and edible part as the positive charge,the whole of the atom is neutral. | According to Rutherford’s atomic model, the negatively charged electrons revolves around the nucleus in fixed orbits and positve charge is concentrated at the centre of atom.The number of positive charge is equal to the number of negative charge,so atom is neutral. | Neils Bohr’s followed Ruderford’s atomic model and fulfills the deficiency in Rutherford’ atomic model .According to him the negatively charged electrons revolves around the nucleus in descreet orbitals. In these orbitals electrons do not radiate energy and didtributed in the form of 2n.He named these orbitals K,L,M…etc. |
Q6.Summarise the rules for writing of distribution of electrons in various shells for the first eighteen elements.
Ans-The electrons accommodate in the shells of atom as per the rule 2n2(n=1,2,3,4,5….),where n is the number of orbitals or energy levels ,electrons are filled up in the orbitals in stepwise ways. Outer shell is not occupied by the electrons unless inner orbital is filled. The electrons are filled in first, second, third orbital…..and so on as per following ways.
First orbital→n=1, 2n2 = 2 ×12 = 2
Second orbital→n=2, 2n2 =2 × 22= 8
Third orbital→n=3, 2n2 = 2 ×32= 18
On the same way other orbitals also filled.
Q7.Define valency by taking examples of silicon and oxygen.
Ans.Valency- The tendency of atoms of an element to make an octet in its outermost orbit and transaction of electrons in making this octet is known as valency of the element. The valency of an element is decided by the electrons available in the outermost orbit known as valence electrons,if the valence electrons are more than 4 then valency is calculated by subtracting these from 8 and if valence electrons are less than 4 or equal to 4 then the valency will be the same as the valence electrons.
As an example, silicon(Si) has the atomic number of 14, so in its outermost orbit, there will be 4 electrons, so its valency will be 4.
Oxygen has the atomic number of 8, so its valence electrons are 6, then its valency = 8 -6 =2.
8-Explain with examples (i) Atomic number (ii) Mass number (iii) Isotopes and (iv)Isobars. Give any two uses of isotopes.
Ans.(i)Atomic number-The total number of positive charges at the nucleus is known as the atomic number which is equivalent to the number of protons, for example, the atomic number of hydrogen is 1 and the atomic number of oxygen is.
(ii)Mass number- The sum of the number of protons and neutrons at the nucleus is known as the mass number, as an example oxygen number of protons =8 and number of neutrons =8, so the mass number of oxygen atom = 8 +8 =16 u.
(iii)Isotopes- The pair of elements that has the same atomic numbers but have different atomic masses are known as isotopes, as for example, hydrogen has three isotopes, protium(1H1), deuterium (2H1), and tritium (3H1) which have the same atomic number,1 and different atomic masses 1,2 and 3 respectively.
(iv)Isobars-The pair of elements which have the same atomic masses but the different atomic number, as for example argon,40Ar18 and calcium, 40Ca20.
Uses of Isotopes- One of the isotopes of iodine is used for the treatment of goiter. One of the isotopes of uranium i.e U(235) is used for the treatment of cancer.
Q9.Na+ has completely filled K and L shells. Explain.
Ans. The electronic distribution of the Na atom as per Neils Bohr’s atomic model must be in the form of 2n², where n is the number of orbital
Electrons in first orbit,K =2×1² = 2
Electrons in second orbit,K =2×2² = 8
The remaining 1 electron exist in third orbit
K= 2, L= 8, M= 1
Na+ ion is formed after freeing one electron which exists in the outermost orbital(M)
Therefore electronic configuration of Na+
K = 2,L = 8
Therefore,Na+ has K and L shells completely filled
Q10.If bromine atom is available in the form of, say, two isotopes 79Br35 (49.7%) and 81Br35(50.3%), calculate the average atomic mass of bromine atom.
Ans.The average atomic mass of bromine
=49.7 % of 79 + 50.3% of 81
=(49.7 ×79)/100 + (50.3× 81)/100
=(3926.3 +4074.3)/100
= 80.006 u
Q11.The average atomic mass of a sample of an element X is 16.2 u. What are the percentages of isotopes 16X8 and18X8 in the sample?
Ans-Let the percentage of 16X8 = a %
Pecentage of another isotope 18×8 = (100 – a)%
Average atomic mass of X
= a % of 16X8 + (100 – a)% of 18X8
a % of 16X8 + (100 – a)% of 18X8 = 16.2
a % of 16 + (100 – a)% of 18 = 16.2
16a/100 – 18(100 – a)/100 = 16.2
a = 90
Therefore, percentage of 16X8 in the sample = 90 %
And percentage of 18X8 = 100 – 90 = 10 %
Q12.If Z = 3,what would be the valency of the element ? Also , name the element.
Ans- Atomic number,Z =3 So, number of electrons =3
Its electronic structure⇒2,1
So, its valency = 1
The element is lithium(Li)
Q13.Composition of the nuclei of two atomic species X and Y are given as under.
Give the mass numbers of X and Y .What is the relation between the two species ?
Ans- Mass number of X = Protons + Nuetrons
= 6 + 6 = 12 u
Mass number of Y = 6 + 8 = 14 u
Q14.For the following stataments,write T for True and F for false .
(a)J.J Thomson proposed that the nucleus of an atom contains only nucleons.
(b)A neutrons is formed by an electron and a proton combining together. Therefore ,it is neutral.
(c)The mass of an electron is about 1/2000 times that of proton.
(d)An isotope of iodine is for making tincture iodine. Which is used as a medicine.
Ans.(a)False (b)False (c) True (d) True
Put tick(√) against correct choice and cross (x) against wrong choice in questions 15, 16 and 17.
Q15.Rutherford’s alpha-particle scattering experiment was responsible for the discovery of
(a)Atomic nucleus (b)Electron
(c)Proton (d)Neutron
Ans.(a) Atomic nucleus
16-Isotopes of an element have
(a)the same physical properties
(b)different chemical properties
(c)different number of neutrons
(d)different atomic numbers
Ans. (c)different number of neutrons
Q17.Number of valence electrons in Cl– ion are:
(a)16 (b)8 (c)17 (d)18
Ans.(b) 8
Q18.Which one of the following is a correct electronic configuration of sodium ?
(a)2,8 (b)8,2,1 (c)2,1,8 (d)2,8,1
Ans.(d) 8,2,1
Q19.Complete the following table.
| Atomic number | Mass number | Number of nuetrons | Number of protons | Number of electrons | Name of atomic species |
| 9 | 10 | ||||
| 16 | 32 | sulphur | |||
| 24 | 12 | ||||
| 2 | 1 | ||||
| 1 | 0 | 1 | 0 |
Ans-
| Atomic number | Mass number | Number of nuetrons | Number of protons | Number of electrons | Name of atomic species |
| 9 | 19 | 10 | 9 | 9 | Potassium |
| 16 | 32 | 16 | 16 | 16 | sulphur |
| 12 | 24 | 12 | 12 | 12 | Magnesium |
| 1 | 2 | 1 | 1 | 1 | Helium |
| 1 | 1 | 0 | 1 | 0 | Hydrogen |
NCERT Solutions of Science and Maths for Class 9,10,11 and 12
NCERT Solutions for class 9 maths
NCERT Solutions for class 9 science
CBSE Class 9-Question paper of science 2020 with solutions
CBSE Class 9-Sample paper of science
CBSE Class 9-Unsolved question paper of science 2019
Study notes of Maths and Science NCERT and CBSE from class 9 to 12
NCERT Solutions for class 10 maths
CBSE Class 10-Question paper of maths 2021 with solutions
CBSE Class 10-Half yearly question paper of maths 2020 with solutions
CBSE Class 10 -Question paper of maths 2020 with solutions
CBSE Class 10-Question paper of maths 2019 with solutions
NCERT Solutions for Class 10 Science
NCERT Solutions for class 11 maths
| Chapter 1-Sets | Chapter 9-Sequences and Series |
| Chapter 2- Relations and functions | Chapter 10- Straight Lines |
| Chapter 3- Trigonometry | Chapter 11-Conic Sections |
| Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |
| Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |
| Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |
| Chapter 7- Permutations and Combinations | Chapter 15- Statistics |
| Chapter 8- Binomial Theorem | Chapter 16- Probability |
CBSE Class 11-Question paper of maths 2015
CBSE Class 11 – Second unit test of maths 2021 with solutions
NCERT solutions for class 12 maths
| Chapter 1-Relations and Functions | Chapter 9-Differential Equations |
| Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |
| Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |
| Chapter 4-Determinants | Chapter 12-Linear Programming |
| Chapter 5- Continuity and Differentiability | Chapter 13-Probability |
| Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |
| Chapter 7- Integrals | |
| Chapter 8-Application of Integrals |
Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2
Class 12 Maths Important Questions-Application of Integrals
Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22
Solutions of class 12 maths question paper 2021 preboard exam CBSE Solution
