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# 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

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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 × 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× 10coloumbs

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

$\fn_cm \left ( i \right )_{6}^{13}\textrm{C}\; \; \left ( ii \right )_{8}^{16}\textrm{O}\; \; \left ( iii \right )\; \; _{12}^{24}\textrm{Mg}\; \; \left ( iv \right )_{38}^{88}\textrm{Sr}$

Ans.

$\fn_cm \left ( i \right )_{6}^{13}\textrm{C}$

No. of protons in C = 6 and numer of neutrns in C= 13 -6 =7

$\fn_cm \; \; \left ( ii \right )_{8}^{16}\textrm{O}$

No. of protons in O = 8 and numer of neutrns in O= 16 -8 =8

$\fn_cm \; \; \left ( iii \right )\; \; _{12}^{24}\textrm{Mg}$

No. of protons in Mg = 12 and numer of neutrns in Mg= 24 -12 =12

$\fn_cm \left ( iv \right )_{38}^{88}\textrm{Sr}$

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  $\fn_cm _{17}^{35}\textrm{X}$

(ii) The atomic number =92 and atomic mass = 233,therefore complete symbol for the atom is   $\fn_cm _{92}^{233}\textrm{X}$

(iii) The atomic number =4 and atomic mass = 9,therefore complete symbol for the atom is$\fn_cm _{4}^{9}\textrm{X}$

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 ,$\fn_cm \bar{\nu }$ is given as

$\fn_cm \bar{\nu }=\frac{1}{\lambda }=\frac{1}{580\times 10^{-9}}=1.724\times 10^{6}$

Hence the wave number of yellow light is $\fn_cm 1.724\times 10^{6}m^{-1}$

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 $\fn_cm \bar{\nu }$ is given by

$\fn_cm \bar{\nu }=\frac{1}{\lambda }=\frac{1}{6\times 10^{-2}}=16.66m^{-1}$

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 = (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

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