Chapter 2
Structure of Atom
1. (i) Calculate the number of electrons which will together weigh one gram.(ii) Calculate the mass and charge of one mole of electrons.
Solution :
(i) Mass of one electron
= 9.10939 × 10–31 kg Number of electrons that weigh 9.10939 × 10–31 kg = 1Number of electrons that will weigh 1 g = (1 × 10–3 )
= 0.1098 × 10–3 + 31
= 0.1098 × 1028
= 1.098 × 1027
(ii) Mass of one electron
= 9.10939 × 10–31 kg Mass of one mole of electron
= (6.022 × 1023) × (9.10939 ×10–31 kg)= 5.48 × 10–7 kg Charge on one electron = 1.6022 × 10–19 coulomb Charge on one mole of electron = (1.6022 × 10–19 C) (6.022 × 1023)= 9.65 × 104 C
2. (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 14C. (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 NH3 at STP. Will the Sol. change if the temperature and pressure are changed?
Solution :
(i) Number of electrons present in 1 molecule of methane (CH4){1(6) + 4(1)} = 10Number of electrons present in 1 mole i.e., 6.023 × 1023 molecules of methane= 6.022 × 1023 × 10 = 6.022 × 1024
(ii) (a) Number of atoms of 14C in 1 mole= 6.023 × 1023Since 1 atom of 14C contains (14 – 6) i.e., 8 neutrons, the number of neutrons in 14 g of 14C is (6.023 × 1023) ×8. Or, 14 g of 14C contains (6.022 × 1023 × 8) neutrons.Number of neutrons in 7 mg = 2.4092 × 1021(b) Mass of one neutron = 1.67493 × 10–27 kgMass of total neutrons in 7 g of 14C= (2.4092 × 1021) (1.67493 × 10–27 kg)= 4.0352 × 10–6 kg
(iii) (a) 1 mole of NH3 = {1(14) + 3(1)} g of NH3= 17 g of NH3= 6.022× 1023 molecules of NH3 Total number of protons present in 1 molecule of NH3= {1(7) + 3(1)}= 10Number of protons in 6.023 × 1023 molecules of NH3= (6.023 × 1023) (10)= 6.023 × 1024⇒ 17 g of NH3 contains (6.023 × 1024) protons.Number of protons in 34 mg of NH3 = 1.2046 × 1022
(b) Mass of one proton = 1.67493 × 10–27 kgTotal mass of protons in 34 mg of NH3= (1.67493 × 10–27 kg) (1.2046 × 1022)= 2.0176 × 10–5 kgThe number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.
12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (NCERT Solution of Class-11 chemistry) and work function (W0) of the metal.
Solution :
Threshold wavelength of radian = 6800 × 10–10 m Threshold frequency of = 4.41 × 1014 s–1 Thus, the threshold frequency of the metal is 4.41 × 1014 s1.Hence, work function (W 0) of the metal = h ν 0= (6.626 × 10–34 Js) (4.41 × 1014 s–1)= 2.922 × 10–19 J
3. How many neutrons and protons are there in the following nuclei?136C, 168O , 2412Mg ,5626Fe, 8838Sr
Solution :
136 C:Atomic mass = 13Atomic number = Number of protons = 6Number of neutrons = (Atomic mass) – (Atomic number)= 13 – 6 = 7Chapter 2- Structure of atom :Atomic mass = 16Atomic number = 8Number of protons = 8Number of neutrons = (Atomic mass) – (Atomic number)= 16 – 8 = 8
:Atomic mass = 24 Atomic number = Number of protons = 12 Number of neutrons = (Atomic mass) – (Atomic number)= 24 – 12 = 12 :Atomic mass = 56 Atomic number = Number of protons = 26 Number of neutrons = (Atomic mass) – (Atomic number)= 56 – 26 = 30 :Atomic mass = 88 Atomic number = Number of protons = 38 Number of neutrons = (Atomic mass) – (Atomic number)= 88 – 38 = 504. Write the complete symbol for the atom with the given atomic number (Z) and Atomic mass
(A)(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9
Solution :
5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency
(ν) and wave number of the yellow light.Solution :From the expression,We get
(i)Where,ν = frequency of yellow ligh tc = velocity of light in vacuum = 3 × 108 m/sλ = wavelength of yellow light = 580 nm = 580 × 10–9 m Substituting the values in expression (i)Thus, frequency of yellow light emitted from the sodium lamp= 5.17 × 1014 s–1 Wave number of yellow light,
6. Find energy of each of the photons which(i) correspond to light of frequency 3× 1015 Hz.(ii) have wavelength of 0.50 Å.
Solution :
(i) Energy (E) of a photon is given by the expression,E Where,h = Planck’s constant = 6.626 × 10–34 Jsν = frequency of light = 3 × 1015 Hz Substituting the values in the given expression of E:E = (6.626 × 10–34) (3 × 1015)E = 1.988 × 10–18 J
(ii) Energy (E) of a photon having wavelength (λ) is given by the expression = Planck’s constant = 6.626 × 10–34 Jsc = velocity of light in vacuum = 3 × 108 m/sSubstituting the values in the given expression of E:
7. Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10–10 s.
Solution :
Frequency (ν) of light Wavelength (λ) of light Where,c = velocity of light in vacuum = 3×108 m/sSub stituting the value in the given expression of λ
8. What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Solution :
Energy (E) of a photon = hνEnergy (En) of ‘n’ photons = nhν Where,λ = wavelength of light = 4000 pm = 4000 ×10–12 mc = velocity of light in vacuum = 3 × 108 m/sh = Planck’s constant = 6.626 × 10–34 JsSubstituting the values in the given expression of n Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012 × 1016.
9. 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 (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J).Solution :(i) Energy (E) of a photon = hν /Where,h = Planck’s constant = 6.626 × 10–34 Jsc = velocity of light in vacuum = 3 × 108 m/sλ = wavelength of photon = 4 × 10–7 m Substituting the values in the given expression of E: Structure of atom Hence, the energy of the photon is 4.97 × 10–19 J.(ii) The kinetic energy of emission Ek is given = (3.1020 – 2.13) eV= 0.9720 e VHence, the kinetic energy of emission is 0.97 eV.(iii) The velocity of a photoelectron (ν) can be calculated by the expression,of is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:NCERT Solution of v = 5.84 × 105 ms–1Hence, the velocity of the photoelectron is 5.84 × 105 ms–1.
10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol–1.
Solution :
Energy of sodium (E)Nc= 4.947 × 105 J mol–1= 494.7 × 103 J mol–1= 494 kJ mol–1
11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second.
Solution :
Power of bulb, P = 25 Watt = 25 Js–1 Energy of one photon, E = h ν Substituting the values in the given expression of E:NCERT Solution of E = 34.87 × 10–20 J Rate of emission of quanta per second