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 × 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

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