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.
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NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom
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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 × 10^{27}electrons
(ii)1 mole of electrons = 6.022 × 10^{23 }electrons
The mass of 1 electron =9.1 × 10^{-31} kg
The mass of 6.022 × 10^{23 }electrons = 6.022 × 10^{23 }×9.1 × 10^{-31} =5.48× 10^{-7}kg
The charge of 1 electron = 1.602 × 10^{-19 }coulomb
Charge on 1 mole electrons = 1.602 × 10^{-19 }×6.022 × 10^{23 }=9.65× 10^{4 }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^{-27}kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH_{3}at STP.
Will the answer change if the temperature and pressure are changed ?
Ans. (i) 1 mole of the methane(CH_{4}) = 6.022 × 10^{23 }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 × 10^{23 }molecules =10× 6.022 × 10^{23 }= 6.022 × 10^{24}electrons
(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 × 10^{23}
Number of neutrons in 14 000 mg =8 × 6.022 × 10^{23}=48.176 × 10^{23}
Therefore number of electrons in 7 mg of ^{14} C = (48.176 × 10^{23}×7)/(14000)= 2.40× 10^{21}
Since mass of neutron is given 1.675 × 10^{-27}kg
The mass of 2.40× 10^{21 }neutrons = 1.675 × 10^{-27}kg × 2.40× 10^{21 }=4.02 ×10^{-6}
(iii) Let’s find out the number of protons and mass of total protons in 34 mg of NH_{3}
No. of protons present in one molecule of NH_{3}
Molar mass of NH_{3} is = 14 + 3 = 17 g
17 g = 17 × 10^{3} mg
No. of protons present in one molecule of NH_{3} = 7 + 3 = 10
No. of molecules of NH_{3} present in 17 × 10^{3} mg of ammonia = 6.022×10^{23}
No. of molecules present in 34 mg of ammonia = (6.022×10^{23}×34)/ (17 × 10^{3}) = 12.044× 10^{20}
No. of protons present in 12.044× 10^{20 }molecules of ammonia = 10×12.044× 10^{20}=120.44× 10^{20}=1.2044× 10^{22}
The mass of a proton is = 1.67 × 10^{-27}Kg
Since the no. of protons in 34 mg are 1.2044× 10^{22}
Mass of protons in 34 mg of ammonia =Mass of 1.2044× 10^{22 }protons =1.2044× 10^{22}×1.67 × 10^{-27}=2.011× 10^{-5}Kg
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× 10^{8 }m/s, λ = 580 nm = 580× 10^{-9 }m
v = (3× 10^{8 })/(580× 10^{-9}) = 5.17× 10^{14 }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 × 10^{15 }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 × 10^{15} Hz = 3 × 10^{15}s^{-1}
∴ E = (6.626 × 10^{-34} J s) × (3 × 10^{15} s^{-1}) = 1.9878 × 10^{18} J
The energy of each photon is 1.9878 × 10^{18} J
(ii) Wavelength, λ = 0.50 Å =0.50×10^{-10}m
The energy of a photon is E = hν
The velocity of the light, C = 3×10^{8}m/s
We have C = νλ ⇒ν=C/λ
E = hC/λ = (6.626 × 10^{-34} ×3×10^{8})/(0.50×10^{-10})=3.98×10^{-15}Joule
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× 10^{8})/(0.5 × 10^{-10}) = 6× 10^{-2}m
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^{-12}m=4× 10^{-9}m
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× 10^{8 }m/s
E = (6.626 × 10^{-34} J s×3× 10^{8 }m/s)/(4× 10^{-9}m) = 4.97× 10^{-17}J
The energy of a photon is 4.97× 10^{-17}J
Total energy = 1 Joule
The number of photons that provides 1 Joule of energy = 1/Total energy = 1/(4.97× 10^{-17}) =2.012× 10^{16 }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 × 10^{8}m/s
E = hC/ λ
E = (6.626 × 10^{-34} J s×3× 10^{8 }m/s)/(4× 10^{-7}m) = 4.97× 10^{-19}J
Since we are given , 1 eV = 1.6020 × 10^{-19} J
E=(1×4.97× 10^{-19}J)/(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^{-19}J
1.55× 10^{-19}J = (1/2)9.109 × 10^{-31} Kg ×v²
v² = (2×1.55× 10^{-19}J)/(9.109 × 10^{-31}) =.3406× 10^{12}
v = √(.3406× 10^{12}) = 5.83× 10^{5 }m/s
Hence the velocity of the photoelectron is 5.83× 10^{5 }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× 10^{8 }m/s)/(242 × 10^{-9}m) = 0.0821× 10^{-17}J
1KJ = 1000J
E =[(0.0821× 10^{-17})/1000]KJ = 0.0821× 10^{-20}KJ
1 mole of Na atom = 6.022× 10^{23}atoms
Ionization energy of Na in k-J mol^{-1 }is = 6.022× 10^{23}× 0.0821× 10^{-20 }=494 k-J mol^{-1}
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