Skip to content 
NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom
NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom are created here for helping the class 11 students in clearing their doubts on chapter 2 of the chemistry NCERT textbook. All NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom are very important for class 11 students and for the candidates who are pursuing the preparation of competitive entrance exams like NEET, AIEEE, JEE, NDA, CDS, and other entrance exams. All questions are explained here scientifically in a step-by-step way so every student can understand the solutions.
Class 11 Physics -Chapter 3 NCERT Solution
NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom
Click for online shopping
Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc
Q1.(i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Ans. Mass of an electron is = 9.1 × 10-28 g
The number of electrons in 1 g = 1/(9.1 × 10-28 g) = 1.098 × 1027electrons
(ii)1 mole of electrons = 6.022 × 1023 electrons
The mass of 1 electron =9.1 × 10-31 kg
The mass of 6.022 × 1023 electrons = 6.022 × 1023 ×9.1 × 10-31 =5.48× 10-7kg
The charge of 1 electron = 1.602 × 10-19 coulomb
Charge on 1 mole electrons = 1.602 × 10-19 ×6.022 × 1023 =9.65× 104 coloumbs
Q2.(i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14 C(Assume that mass of a neutron = 1.675 × 10-27kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3at STP.
Will the answer change if the temperature and pressure are changed ?
Ans. (i) 1 mole of the methane(CH4) = 6.022 × 1023 molecules
The number of electrons present in 1 molecule of methane = number of electrons in C + number of e’s in H =6+4 =10 e’s
Therefore number of electrons in 6.022 × 1023 molecules =10× 6.022 × 1023 = 6.022 × 1024electrons
(ii) Molar mass of 14 C = 14 g
14 g =14000 mg(i.e mass of 1 mole 14 C)
Number of neutrons in 1 atom of 14 C = 8
1 mole = 6.022 × 1023
Number of neutrons in 14 000 mg =8 × 6.022 × 1023=48.176 × 1023
Therefore number of electrons in 7 mg of 14 C = (48.176 × 1023×7)/(14000)= 2.40× 1021
Since mass of neutron is given 1.675 × 10-27kg
The mass of 2.40× 1021 neutrons = 1.675 × 10-27kg × 2.40× 1021 =4.02 ×10-6
(iii) Let’s find out the number of protons and mass of total protons in 34 mg of NH3
No. of protons present in one molecule of NH3
Molar mass of NH3 is = 14 + 3 = 17 g
17 g = 17 × 103 mg
No. of protons present in one molecule of NH3 = 7 + 3 = 10
No. of molecules of NH3 present in 17 × 103 mg of ammonia = 6.022×1023
No. of molecules present in 34 mg of ammonia = (6.022×1023×34)/ (17 × 103) = 12.044× 1020
No. of protons present in 12.044× 1020 molecules of ammonia = 10×12.044× 1020=120.44× 1020=1.2044× 1022
The mass of a proton is = 1.67 × 10-27Kg
Since the no. of protons in 34 mg are 1.2044× 1022
Mass of protons in 34 mg of ammonia =Mass of 1.2044× 1022 protons =1.2044× 1022×1.67 × 10-27=2.011× 10-5Kg
Q3.How many protons and neutrons are present in the following nuclei
Ans.
No. of protons in C = 6 and numer of neutrns in C= 13 -6 =7
No. of protons in O = 8 and numer of neutrns in O= 16 -8 =8
No. of protons in Mg = 12 and numer of neutrns in Mg= 24 -12 =12
No. of protons in Sr = 38 and numer of neutrns in Sr= 88 -38 =50
Q4.Write the complete symbol for the atom (X) with the given atomic number (Z) and atomic mass (A)
(i) Z = 17,A = 35
(ii) Z = 92, A = 233
(in) Z = 4, A = 9.
Ans.(i) The atomic number =17 and atomic mass = 35,therefore complete symbol for the atom is
(ii) The atomic number =92 and atomic mass = 233,therefore complete symbol for the atom is
(iii) The atomic number =4 and atomic mass = 9,therefore complete symbol for the atom is
Q5.Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (v) and wave number (v) of yellow light.
Ans. Frequency(v) of yellow light = C/λ
Where λ= is wavelength of yellow light=580 mm and C is the velocity of the light
C = 3× 108 m/s, λ = 580 nm = 580× 10-9 m
v = (3× 108 )/(580× 10-9) = 5.17× 1014 Hertz
The wave number , is given as
Hence the wave number of yellow light is
Q6.Calculate the energy of each of the photons which
(i) correspond to light of frequency 3 × 1015 Hz
(ii) have a wavelength of 0-50 Å.
Ans. The energy of a photon is E = hν
Where h is Planck constant, h = 6.626 × 10-34 J s
Frequency of light,v = 3 × 1015 Hz = 3 × 1015s-1
∴ E = (6.626 × 10-34 J s) × (3 × 1015 s-1) = 1.9878 × 1018 J
The energy of each photon is 1.9878 × 1018 J
(ii) Wavelength, λ = 0.50 Å =0.50×10-10m
The energy of a photon is E = hν
The velocity of the light, C = 3×108m/s
We have C = νλ ⇒ν=C/λ
E = hC/λ = (6.626 × 10-34 ×3×108)/(0.50×10-10)=3.98×10-15Joule
Q7.Calculate the wavelength, frequency, and wavenumber of lightwave whose period is 2.0 × 10-10 s.
Ans.The frequency,ν = 1/T
Where T is the time period, T= 2.0 × 10-10 s
ν = 1/(2.0 × 10-10 ) = 0.5 × 10-10
The frequency of given light wave is 0.5 × 10-10 Hertz
The wavelength,λ = C/ν = (3× 108)/(0.5 × 10-10) = 6× 10-2m
The wave number is given by
Q8.What is the number of photons of light with a wavelength of 4000 pm which provides 1 Joule of energy?
Ans. The wavelength of the light given is, λ=4000 pm = 4000× 10-12m=4× 10-9m
The energy of the photon, E = hν = hC/λ
Where h is Planck constant,h=6.626 × 10-34 J s, C is the velocity of the light, C =3× 108 m/s
E = (6.626 × 10-34 J s×3× 108 m/s)/(4× 10-9m) = 4.97× 10-17J
The energy of a photon is 4.97× 10-17J
Total energy = 1 Joule
The number of photons that provides 1 Joule of energy = 1/Total energy = 1/(4.97× 10-17) =2.012× 1016 Photons
Q9.A photon of wavelength 4 × 10-7 m strikes on metal surface ; the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon,
(ii) the kinetic energy of the emission
(iii) the velocity of the photoelectron. (Given 1 eV = 1.6020 × 10-19 J).
(i) The energy of the photon,E = hν
The wavelength of photon,λ = 4 × 10-7 m
C = νλ ⇒ν= C/ λ ,Where C is the velocity of the light = 3 × 108m/s
E = hC/ λ
E = (6.626 × 10-34 J s×3× 108 m/s)/(4× 10-7m) = 4.97× 10-19J
Since we are given , 1 eV = 1.6020 × 10-19 J
E=(1×4.97× 10-19J)/(1.6020 × 10-19 J) =3.1 eV
(ii) Kinatic Energy of the emission = E – work function(K.E of emitted electron after interaction with photon)=3.1 – 2.13 =0.97 eV
(iii) Velocity of the photoelectron = Velocity of the emission
K.E of the emission = (1/2)mv²
Mass ,m of the electron = 9.109 × 10-31 Kg
E=3.1 eV =(0.97× 1.6020 × 10-19 )J =1.55× 10-19J
1.55× 10-19J = (1/2)9.109 × 10-31 Kg ×v²
v² = (2×1.55× 10-19J)/(9.109 × 10-31) =.3406× 1012
v = √(.3406× 1012) = 5.83× 105 m/s
Hence the velocity of the photoelectron is 5.83× 105 m/s
Q10.Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom. Calculate the ionization energy of sodium in k-J mol-1
Ans. The wavelength,λ of electromagnetic radiation is = 242 nm = 242 × 10-9 m
Energy,E = hC/λ
E = (6.626 × 10-34 J s×3× 108 m/s)/(242 × 10-9m) = 0.0821× 10-17J
1KJ = 1000J
E =[(0.0821× 10-17)/1000]KJ = 0.0821× 10-20KJ
1 mole of Na atom = 6.022× 1023atoms
Ionization energy of Na in k-J mol-1 is = 6.022× 1023× 0.0821× 10-20 =494 k-J mol-1
You can compensate us for our survival in the web
Paytm number 9891436286
The money collected by us will be used for the education of poor students who leaves their study because of a lack of money.
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
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
