 # Chapter 2 Structure of Atom

In this chapter, you will know about the discovery of electrons, protons and neutrons alongwith their characteristics. This chapter also contains the description of Thomson,Rutherford and Bohr atomic model. Through this chapter you will also understand the important features of the quantum mechanical model of atoms. You will also understand the nature of electromagnetic radiation and Planck's quantum theory. It also explains the photoelectric effect and describes features of atomic spectra. It also states the de Broglie relation and Heisenberg uncertainty principle. It also defines an atomic orbital in terms of quantum numbers. The chapter also states aufbau principle, Pauli exclusion principle and Hund's rule of maximum multiplicity and writes the electronic configurations of atoms.

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### Exercise 1 ( Page No. : 73 )

•  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: (i) Mass of an electron = 9.1 x 10-28 g Or 9.1 x 10-28 g contains = 1 electron Therefore 1g contains = 1/9.1 x 10-28 *1 = 1.098 x 1027 electrons   (ii) We know, one mole of electron = 6.022 x 1023 electron Mass of one electron = 9.1 × 10–28 g Or mass of 6.022 x 1023 electron = 9.1 x 10-28 x 6.022 x 1023 = 5.48 x 10-4 g Charge on one electron = 1.6 × 10–19 coulomb Charge on one mole of electron = (1.6 × 10–19 C x 6.022 × 1023) = 9.63 × 104 C 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 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 answer change if the temperature and pressure are chang Ans: (i) Molecule of CH4 (methane) contains electron = 10 Therefore 1 mole (6.022 x 1023 atoms) contains electron = 6.022 x 1024 (ii)  a) 1g atom of 14C = 14g = 6.022 x 1023 atoms = 6.022 x 1024 x 8 neutrons Thus 14g or 14000 mg have 6.022 x 1024 x 8 neutrons Therefore 7 mg will have neutrons = 6.022 x 1024 x 8 / 14000 x 7 = 2.4088 x 1022 b) mass of 1 neutron = 1.675 x 10-27 kg Therefore mass of 2.4088 x 1021 neutrons = 2.4088 x 1021 x 1.67 x 10-27 = 4.0347 x 10-6 kg (iii)  a) 1 mol of NH3 = 17g NH3 = 6.022 x 1023 molecules of NH3 = (6.022x1023)(7 + 3) proton = 6.022 x 1024 protons Therefore 34 mg i.e 0.034 g NH3 = 6.022 x 1024 x 0.034/1 = 1.2044 x 1022 protons b) mass of 1 proton = 1.6726 x 10-27 kg Therefore mass of 1.2044 x 1022 protons = (1.6726 x 10-27)(1.2044 x 1022) kg = 2.0145 x 10-5 kg No, the answer will not change with change in temperature & pressure. Q3 How many neutrons and protons are there in the following nuclei? Ans: We know atomic number(z) = no of protons = no of electrons Also mass number(A) = no of protons + no of neutrons 136C: Here mass number = 13 Atomic number = Number of protons = 6 Number of neutrons = (mass number) – (Atomic number) = 13 – 6 = 7 : Here mass number = 16 Atomic number = 8 Number of protons = 8 Number of neutrons = (mass number) – (Atomic number) = 16 – 8 = 8 : Here Mass number = 24 Atomic number = Number of protons = 12 Number of neutrons = (mass number) – (Atomic number) = 24 – 12 = 12 : Here mass number = 56 Atomic number = Number of protons = 26 Number of neutrons = (mass number ) – (Atomic number) = 56 – 26 = 30 : Here mass number = 88 Atomic number = Number of protons = 38 Number of neutrons = (mass number) – (Atomic number) = 88 – 38 = 50 Q4 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 Ans: 1) the element with atomic number(Z) 17 & mass number (A) 35 is chlorine = 2) the element with atomic number(Z) 92 & mass number (A)233 is uranium = 3) the element with atomic number(Z) 4 & mass number (A) 9 is berellium = Q5 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. Ans: Here λ = 580 nm(given) We have to find out ν & Therefore from the equation we can write, Where, ν = frequency of yellow light c = velocity of light in vacuum = 3 × 108 m/s λ = wavelength of yellow light = 580 nm = 580 × 10–9 m Substituting the values in we get Thus, frequency of yellow light emitted from the sodium lamp = 5.17 × 1014 s–1 Wave number = it is defined as the number of wavelengths which can be accommodated in 1cm length along the direction of propagation. Wave number of yellow light is given by the equation We have λ = 580 nm(given),putting the value in the equation we get Q6 Find energy of each of the photons which (i) correspond to light of frequency 3× 1015 Hz. (ii) have wavelength of 0.50 Å. Ans: Planck’s quantum theory = Based on the assumption that all atoms on the surface of the heated solid vibrate at the frequency, Planck developed a model that came to be known as Planck’s equation. Through experiments of frequencies and temperature, Planck was able to generate a constant, Planck’s constant                                                            h = 6.62607 x 10-34 J s  Using this constant he was able to restate his theory: energy was directly proportional to frequency. He wrote his equation as                                                            E=hν where E is energy, h is Planck’s constant, and v is frequency. (i) Energy (E) of a photon is given by the expression, E=hν Where, h = Planck’s constant = 6.626 × 10–34 Js ν = frequency of light = 3 × 1015 Hz Substituting the values in the given expression E = hv we get 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, where , h = Planck’s constant = 6.626 × 10–34 Js c = velocity of light in vacuum = 3 × 108 m/s Substituting the values in the given expression of E: Q7 Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10–10 s. Ans: Given period = 2.0 x 10-10s We know, Frequency (ν) of light ,therefore putting the values .we get We also know Where,c = velocity of light in vacuum = 3×108 m/s Substituting the value in the given expression of λ, we get Wave number of light  Q8 What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy? Ans: Energy (E) of a photon = hc/λ Where,λ = wavelength of light = 4000 pm = 4000 ×10–12 m = 4x 10-9 m c = velocity of light in vacuum = 3 × 108 m/s h = Planck’s constant = 6.626 × 10–34 Js thefore the energy of photon (E) = 6.626 × 10–34 Js X 3 × 108 m/s / 4x 10-9 m = 4.965 x 10-16 J Now 4.965 x 10-16 J is the energy of = 1 photon Therefore 1 J will be the energy of = 1/4.965 x 10-16 J = 2.014 x 1015 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 (eV),   (ii) the kinetic energy of the emission, and   (iii) the velocity of the photoelectron (1 eV= 1.6020 × 10–19 J). Ans: (1)We know λ = 4 × 10–7 m(given) C = 3 x 108 From the equation E= hv or hc/ λ Where, h = Planck’s constant = 6.626 × 10–34 Js c = velocity of light in vacuum = 3 × 108 m/s λ = wavelength of photon = 4 × 10–7 m Substituting the values in the given expression of E: Hence, the energy of the photon is 4.97 × 10–19 J.   (ii) The kinetic energy of emission Ek is given by = (3.1020 – 2.13) eV = 0.9720 eV Hence, the kinetic energy of emission is 0.97 eV.   (iii) The velocity of a photoelectron (ν) can be calculated by the expression, Where, (hv-hv0) 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: v = 5.84 × 105 ms–1 Hence, the velocity of the photoelectron is 5.84 × 105 ms–1. Q10 Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol–1. Ans: From E= hv or hc/ λ, we have λ = 242nm or 242 x 10-9m C = 3 x 108 m/s h = 6.62 x 10-34 Js Now putting these values in the equation we get E= (6.62 x 10-34 Js x 3 x 108 m/s) / (242 x 10-9m) = 0.0821 x 10-17 J/atom Or E = (0.0821 x 10-17 ) / (1000 x 6.02 x 1023) = 494 KJ/mol Q11 A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second. Ans: (Given)Power of bulb, P = 25 Watt = 25 Js–1 We know, Energy of one photon, E = hν λ Substituting the values in the given expression of E: E = 34.87 × 10–20 J Rate of emission of quanta per second is given by R = P/ E, Where R is the rate of emission, P is the power & E is the energy Substituting the values in the equation we get Q12 Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (v0) and work function (W0) of the metal. Ans: Given λ = 6800 amgstrom or 6800x 10-10 m = 6.8 x 10-7m C = 3 x 108 m/s Also λ = c/v or v = c/ λ Therefore putting the values in the equation, we get =3 x 108 / 6.8 x 10-7m = 4.41 x 1014 s-1 Hence, work function (W0) of the metal = hν0 = (6.626 × 10–34 Js) (4.41 × 1014 s–1) = 2.922 × 10–19 J Q13 What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2? Ans: According to Balmer formula Wave no = RH [1/n12 – 1/n22] Here n1 = 2 ,       n2 = 4,     RH = 109678 Putting these values in the equation we get Wave number =109678(1/22 - ¼2) = 109678 x 3/16 Also λ = 1/ wave number Therefore λ = 16 / 109678 x 3 = 486 nm Q14 How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit). Ans: The expression of energy is given by, Where, Z = atomic number of the atom n = principal quantum number For ionization from n1 = 5 to , Therefore ΔE = E2- E1 = - 21.8 X10-19 (1/n22-1/n12) = 21.8 X10-19(1/n22-1/n12) = 21.8 X10-19 (1/52-1/∞) = 8.72 x 10-20 J For ionization from 1st orbit, n1= 1, Therefore ΔE’ = 21.8x10-19(1/12-1/∞) = 21.8x10-19 J Now ΔE’/ ΔE = 21.8x10-19 / 8.72x10-20 = 25 Thus the energy required to remove electron from 1st orbit is 25 times than the required to electron from 5th orbit. Q15 What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state? Ans: Number of lines produced when electron from nth shell drops to ground state = n(n-1)/2 Now according to question lines produced when electron drops from 6th shell to ground state = 6(6-1) / 2 = 15 These are produced due to following transition 6 to 5      5 to 4    4 to 3     3to 2     2 to 1 (5 lines) 6 to 4     5 to 3    4 to 2     3 to 1             (4 lines)   6 to 3     5 to 2    4 to 1                          (3 lines) 6 to 2     5 to 1                                      (2 lines) 6 to 1                                                    (1 lines) Q16 (i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom. Ans: Energy of an electrons = -(2.18x 10-18)/n2 Where n= principal quantum number Now the Energy associated with the fifth orbit of hydrogen atom is \begin{align}E_{5}=\frac{-(2.18\times10^{-18})}{(5)^{2}}=\frac{-2.18\times10^{-18}}{25}\end{align}                                 E5 = –8.72 × 10–20 J   (ii) Radius of Bohr’s nth orbit for hydrogen atom is given by, rn = (0.0529 nm) n2 For, n = 5 r5 = (0.0529 nm) (5)2 r5 = 1.3225 nm Q17 Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen. Ans: According to Balmer formula ṽ=1/λ = RH[1/n12-1/n22] For the Balmer series, ni = 2. Thus, the expression of wavenumber(ṽ) is given by, Wave number (ṽ) is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, ṽ has to be the smallest. For ṽ to be minimum, nf should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking nf = 3,we get: ṽ= 1.5236 × 106 m–1 Q18 What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10–11 ergs. Ans: Ground state energy (E1) = – 2.18 × 10–11 ergs (given)  = –2.18 × 10–11 × 10–7 J  = – 2.18 × 10–18 J Also the energy required to shift the electron from n = 1 to n = 5 is given as ΔE = E5 – E1 The expression for the energy of hydrogen of electron is En = -2π2me4Z2/n2h2 Where m= mass of electrons Z=atomic mass of atom  e = charge of electron h = planck’s constant Now putting the values in the equation ΔE = E5 – E1 we get  Q19 The electron energy in hydrogen atom is given by En = (–2.18 × 10–18)/n2 J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition? Ans: The expression for the energy of hydrogen of electron is En = -2π2me4Z2/n2h2 Where m= mass of electrons Z=atomic mass of atom e = charge of electron h = planck’s constant When n = 1 then En = - (2.18X10-18 ) (GIVEN) When n = 2 then En= - (2.18X10-18 )/4 = 0.5465X10-18 J   Therefore the energy required for ionization from n = 2 is 5.45 x 10-19 J Now wavelength of light needed = E=hv = hc/λ Or λ = [{(6.62x10-24)(3x108)} / 5.45] x 10-19 = 3647 Q20 Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1. Ans: V= 2.05 x107 (given) According to de Broglie’s equation, Where, λ = wavelength of moving particle m = mass of particle (9.10939 x 10-31) v = velocity of particle h = Planck’s constant (6.62 x 10-34) Substituting the values in the expression of λ: Hence, the wavelength of the electron moving with a velocity of 2.05 × 107 ms–1 is 3.548 × 10–11 m. Q21 The mass of an electron is 9.1 × 10–31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength. Ans: We know , KE = ½ mv2 Or v = (2KE/M)1/2 Or v = 811.579 m/s Now putting the values in the de Broglie’s equation We get Hence, the wavelength of the electron is 8.9625 × 10–7 m. Q22 Which of the following are isoelectronic species i.e., those having the same number of electrons? Na+, K+, Mg2+, Ca2+, S2–, Ar Ans: Isoelectronic species = they are the species belonging to different atoms or ions which have same number of electrons but different magnitude of nuclear charge. Now, A positive charge denotes the loss of an electron & A negative charge denotes the gain of an electron by a species.   1) Number of electrons in sodium (Na) = 11 Therefore, Number of electrons in (Na+) = 10 2) Number of electrons in K+ = 18 3) Number of electrons in Mg2+ = 10 4) Number of electrons in Ca2+ = 18 5) Number of electrons in sulphur (S) = 16 ∴ Number of electrons in S2- = 18 6) Number of electrons in argon (Ar) = 18   Hence, the following are isoelectronic species: 1) Na+ and Mg2+ (10 electrons each) 2) K+, Ca2+, S2– and Ar (18 electrons each) Q23 :(i) Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2–(d) F– (ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5? (iii) Which atoms are indicated by the following configurations? (a) [He] 2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1. Ans: Electronic configuration of an atom is defined as the representation of the position of electrons in the various energy shell & subshells. Now A negative charge on the species indicates the gain of an electron by it & A positive charge denotes the loss of an electron   (i) (a) H– ion The electronic configuration of H atom is 1s1.(atomic number = 1) ∴ Electronic configuration of H– = 1s2   (b) Na+ ion The electronic configuration of Na atom is 1s2 2s2 2p6 3s1.(atomic number = 11) ∴ Electronic configuration of Na+ = 1s2 2s2 2p6 3s0  Or  1s2 2s2 2p6   (c) O2– ion The electronic configuration of 0 atom is 1s2 2s2 2p4.(atomic number = 8) ∴ Electronic configuration of O2– ion = 1s2 2s2 p6   (d) F– ion The electronic configuration of F atom is 1s2 2s2 2p5.(atomic number = 9) ∴ Electron configuration of F– ion = 1s2 2s2 2p6   (ii) (a) 3s1 Completing the electron configuration of the element as 1s2 2s2 2p6 3s1. ∴ Number of electrons present in the atom of the element = 2 + 2 + 6 + 1 = 11 ∴ Atomic number of the element = 11(sodium)   (b) 2p3 Completing the electron configuration of the element as 1s2 2s2 2p3. ∴ Number of electrons present in the atom of the element = 2 + 2 + 3 = 7 ∴ Atomic number of the element = 7(nitrogen)   (c) 3p5 Completing the electron configuration of the element as 1s2 2s2 2p5. ∴ Number of electrons present in the atom of the element = 2 + 2 + 5 = 9 ∴ Atomic number of the element = 9(fluorine)   (iii) (a) [He] 2s1 The electronic configuration of the element is [He] 2s1 = 1s2 2s1. ∴ Atomic number of the element = 3 (lithium , a p-block element)   (b) [Ne] 3s2 3p3 The electronic configuration of the element is [Ne] 3s2 3p3= 1s2 2s2 2p6 3s2 3p3. ∴ Atomic number of the element = 15(phosphorous, a p block element)   (c) [Ar] 4s2 3d1 The electronic configuration of the element is [Ar] 4s2 3d1= 1s2 2s2 2p6 3s2 3p6 4s2 3d1. ∴ Atomic number of the element = 21(scandium , a d block element) Q24 What is the lowest value of n that allows g orbitals to exist? Ans: Quantum numbers = the set of four number which gives a complete information about the electrons in an atom. Here n = principal quantum number As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1). For n = 1 (K shell) has l = 0 (one subshell) For n = 2 (L shell) has l= 0,1 (2 subshell) For n = 3 (M shell) has l = 0,1,2 (3 subshell) For n = 4 (N shell) has l = 0,1,2,3 (4 subshell) For n = 5 (O shell) has l = 0,1,2,3,4 (5subshell) ∴ For l = 4, minimum value of n = 5 Q25 An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron. Ans: When l = 2 the value of m = -2,-1, 0 ,+1,+2 Now, for the 3d orbital: Principal quantum number (n) = 3 Azimuthal quantum number (l) = 2 Magnetic quantum number (ml) = – 2, – 1, 0, 1, 2 Q26 An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element. Ans: An atom consists of protons, electrons & neutrons.A positive atom will contain higher number of proton(H+), a negative atom will have higher number of electrons(e-) & a neutral atom will contain same number of protons & electrons.Therefore, (i) For an atom to be neutral, the number of protons is equal to the number of electrons. ∴ Number of protons in the atom of the given element = 29 (ii) The electronic configuration of the atom is 1s2 2s2 2p6 3s2 3p6 4s1 3d10. Which is the electronic configuration of copper Q27 Give the number of electrons in the species , H2 and Ans: Now A negative charge on the species indicates the gain of an electron by it & A positive charge denotes the loss of an electron : Number of electrons present in hydrogen molecule (H2) = 1 + 1 = 2 ∴ Number of electrons in = 2 – 1 = 1 H2: Number of electrons in H2 = 1 + 1 = 2 : Number of electrons present in oxygen molecule (O2) = 8 + 8 = 16 ∴ Number of electrons in = 16 – 1 = 15 Q29 Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n = 1, l = 0; (b) n = 3; l =1 (c) n = 4; l = 2; (d) n = 4; l =3. Ans: Here n= principal quantum number, l= azimuthal quantum number   (a) when n = 1, l = 0 (Given) The orbital is 1s. (can have maximum of 2 electron)   (b) For n = 3 and l = 1 The orbital is 3p. (can have maximum of 6 electrons).   (c) For n = 4 and l = 2 The orbital is 4d. (can have maximum of 10 electrons)   (d) For n = 4 and l = 3 The orbital is 4f. (can have maximum of 14 electrons) Q30 Explain, giving reasons, which of the following sets of quantum numbers are not possible. a n = 0 l = 0 ml = 0     b n = 1 l = 0 ml = 0     c n = 1 l = 1 ml = 0      d n = 2 l = 1 ml = 0     e n = 3 l = 3 ml = – 3   f n = 3 l = 1 ml = 0 Ans: (a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.   (b) The given set of quantum numbers is possible.   (c) The given set of quantum numbers is not possible when n = 1, l is not equal to 1   (d) The given set of quantum numbers is possible.   (e) The given set of quantum numbers is not possible when n = 3 ,l = is not equal to 3   (f) The given set of quantum numbers is possible. Q31 How many electrons in an atom may have the following quantum numbers? (a) n = 4,   (b) n = 3, l = 0 Ans: Total number of electrons in an atom for a value of n = 2n2   a) Total no of electrons when n=4 ,then n= 2x42 = 32 & half of them have ms = -1/2   b) When n= 3, l=0 means 3s orbital which can have 2 electrons . Q32 Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit. Ans: The angular momentum of an electron = mvr = nh/2π --------------1 Also according to de Broglie equation λ = h/mv Or mv = h/v -----------2 Putting 2 in 1 h/v =nh/2π or 2λr= nλ Since ‘2πr’ represents the circumference of the Bohr orbit (r), it is proved by equation (3) that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit. Q33 What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum? Ans: For an atom = 1/λ = RHZ2 (1/n12-1/n22) For He+ spectrum Z = 2,  n2=4,  n1= 2 Therefore = 1/λ = RH x 4(1/22-1/42) = 3RH/4 For hydrogen spectrum = 3RH/4,  Z = 1 Therefore = 1/λ = RH X 1(1/n12-1/n22) Or RH (1/n12-1/n22) = 3RH/4 Or 1/n12-1/n22  = 3/4 Which can be so for n1=1 & n2 = 2, i.e the transition is from n = 2 to n=1 Q34 Calculate the energy required for the process The ionization energy for the H atom in the ground state is 2.18 ×10–18 J atom–1 Ans: For H like particles, En = - 2π2mZ2e4 / n2h2 For H atom  I.E = E – E1 = 0 – (- 2π2me4 / 12h2) = 2π2me4 / h2 = 2.18x 10-18J /atom (given) For the given process, energy required = En – E1 = 0 -  ( - 2π2me4 / 12h2) = 4 x 2π2me4 / h2 = 4 x 2.18 x 10-18J = 8.72 x 10-18 J Q35 If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long. Ans: Diameter of carbon atom = 0.15nm = 0.15 x 10-9m = 1.5 x 10-10m Length along which atoms are to be placed = 20cm = 20x10-2m = 2 x 10-1m Therefore no of carbon atoms which can be placed along the length = (2 x 10-1m) / (1.5x10-10 ) =1.33x109 Q36 2 × 108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm. Ans: Total Length = 2.4 cm Total number of atoms along the length = 2 × 108 Therefore diameter of each atom = 2.4 / (2 × 108) = 1.2 x 10-8 cm And radius of the atom = (1.2 x 10-8) / 2 = 0.60 x 10-8cm = 0.060 x 10-9m = 0.060nm Q37 The diameter of zinc atom is 2.6 Å . Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise. Ans: (a) radius = 2.6/2 = 1.3Å = 1.3 x 10-10m = 130pm (b) given Length of the arrangement = 1.6 cm = 1.6 × 10–2 m Diameter of zinc atom = 2.6 × 10–10 m ∴ Number of zinc atoms present in the arrangement Q38 A certain particle carries 2.5 × 10–16C of static electric charge. Calculate the number of electrons present in it. Ans: Charge carried by one electron = 1.6022 × 10–19 C Therefore electrons present in particle carrying 2.5x10-16C charge = Q39 In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10–18C, calculate the number of electrons present on it. Ans: Charge carried by the oil drop = 1.282 ×10–18C Charge carried by one electron = 1.6022 × 10–19C ∴electrons present on the oil drop carrying 1.282x10-18C charge Q40 In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the α-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results? Ans: In 1911, Rutherford performed alpha rays scattering experiment to demonstrate the structure of atom. Heavy atoms have a heavy nucleus carrying a large amount of positive charge.Hence,some alpha particles are easily deflected back on hitting the nucleus.Also a number of alpha particles are deflected through small angles because of large positive charge on the nucleus.If light atoms are use,their nuclei will be light & moreover,they will have small positive charge on the nucleus.Hence, the number of particles deflected back & those deflecte through some angle will be negligible. Q41 Symbols can be written, whereas symbols  are not acceptable. Answer briefly. Ans: The general convention of representing an element along with its atomic mass (A) and atomic number (Z) is Atomic number of an element is fixed.However,mass number is not fixed as it depends upon the isotope taken.Hence it is essential to indicate mass number Q42 An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol. Ans: Mass number = 81 i.e p + n = 81 Let us suppose number of protons = x and neutrons = x+31.7Xx /100 = 1.317x Therefore x + 1.317x = 81 Or 2.317x = 81 Or x = 81/2.317 = 35 T hus number of protons = 35 ,i.e atomic number = 35 Hence the symbol is 8135 Br Q43 An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion. Ans: Let the number of electrons in the ion carrying a negative charge be x. Then, Number of neutrons present = x + 11.1% of x = x + 0.111 x = 1.111 x Number of electrons in the neutral atom = (x – 1) (When an ion carries a negative charge, it carries an extra electron) ∴ Number of protons in the neutral atom = x – 1 Therefore 37 = 1.111x + x -1 Or 2.111 x = 38 Or x = 18 Therefore no of protons = atomic no = x -1 = 18 – 1 = 17 ∴The symbol of the ion is Q44 An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion. Ans: Let the number of electrons present in ion ,S3+ = x ∴ Number of neutrons in it = x + 30.4% of x = 1.304 x Since the ion is tripositive, ⇒ Number of electrons in neutral atom = x + 3 ∴ Number of protons in neutral atom = x + 3 Now mass number = no of protons + no of neutrons 56 = x + 3 + 1.304x Or 2.304x = 53 Or x = 23 Therefore no of protons = atomic no = x + 3 = 23 + 3 = 26 ∴ The symbol of the ion Q45 Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays. Ans: Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio Q46 Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser. Ans: E = Nhv = Nhc/λ Where N = number of photons emitted h = Planck’s constant c = velocity of radiation λ = wavelength of radiation Substituting the values in the given expression of Energy (E): = 3.33 × 106 J Hence, the power of the laser is 3.33 × 106 J. Q47 Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy. Ans: λ = 616nm  or = 616x10-9m a) Frequency , v=c/ λ = 3.8x108 / 616x10-9 = 4.87x1014 s-1 b) Velocity of the radiation = 3x108 m/s Therefore distance travelled in 30s = 30x3x108 = 9.0x109m c) E = hv = hc/ λ = (6.626x10-34)(3.0x108) / (616x10-9 ) = 32.27x10-20J d) No of quanta in 2J of energy = 2/32.27 x 10-28 = 6.2x1018 Q48 In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10–18 J from the radiations of 600 nm, calculate the number of photons received by the detector. Ans: Energy of 1 photon = hv= hc/ λ = (6.626x10-34)(3x108) / 600x10-9 = 3.313 x 10-19J Total energy received = 3.15 x10-8J Therefore no of photons received = 3.15x10-18 / 3.313x10-19 = 9.51 Q49 Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source. Ans: Frequency = 1/2.x10-9 = 0.5x109 / sec Energy = Nhv = (2.5x105)(6.626x10-34)(0.5x109) = 8.28x10-10J Q50 The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states. Ans: λ1 = 589 nm = 5.89X10-9m therefore v1=c/ λ1 = 3x108/589x10-9 = 5.093x1014/sec λ2 = 589.6nm = 589.6x10-9m therefore v2 = c/ λ2 = 3x108/589.6x10-9 = 5.088x1014/sec ΔE=E2-E1 = h(v2-v1) = (6.626x10-34)(5.093x-5.088)x1014 = 3.31x10-22J Q51 The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron. Ans: A) Work function of caesium (WO) = hvo Therefore vo = WO/h = 1.9 x1.602 x10-19 / 6.626x10-34 = 4.59x1014/sec   B) λo =c/vo = 3x108 / 4.59x1014 = 6.54x10-7m   C) K.E of ejected electron = h(v-vo) = hc(1/λ – 1/ λo) =(6.626x3x10-26) (1/500x10-9 – 1/654x10-9) =(6.626x3x10-26) / 10-9(154/500x654) = 9.36x10-20J k.E = 1mv2/2 = 9.36x10-20J =9.1x10-31/2 = 9.36x10-20J Or v2 = 20.55x1010m2s-2 Or v = 4.53x105ms-1 Q52 Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.                λ (nm) 500 450 400        v × 10–5 (cm s–1) 2.55 4.35 5.35 Ans: (a) Let us suppose threshold wavelength to be λD nm (= λ0 x 10-9 m), the kinetic energy of the radiation is given as: h(v-v0)=1/2 mv2 Or hc(1/λ - 1/λD) = 1/2 mv2 Threshold wavelength (λD)= 540 nm  Substituting this value in equation iii,we get = h  x (3x108)/10-9 [1/400-1/540] = [(9.11x10-31)(5.20x106)2] / 2 =6.66x10-34Js Q53 The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal. Ans: From the principle of conservation of energy, the energy of an incident photon (E) is equal to the sum of the work function (W0) of radiation and its kinetic energy (K.E) i.e.,   Energy of incident radiation(E)= hc/λ = (6.626x10-34)(3x108)/(256.7x 10-9) = 7.74x10-19 J   Since the potential applied gives the kinetic energy to the radiation ,therefore K.E of the electron = 0.35Ev   Therefore work function = 4.83 – 0.35 = 4.48 eV Q54 If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 ms–1, calculate the energy with which it is bound to the nucleus. Ans: Energy of incident photon (E) is given by, = 10.2480 × 10–17 J = 1.025 × 10–16 J Energy with which the electron was bound to the nucleus = 13.25 × 10–16 J – 1.025 × 10–16 J = 12.225 × 10–16 J  = 12.225 × 10–16 J/1.602 x10-19 eV = 7.63x103eV Q55 Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) [1/32 – 1/n2] Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum. Ans: V = c/λ = 3x108/1285x10-9 = 3.29 x 1015(1/32 – 1/n2) = 1/n2 = 1/9 - 3x108/1285x10-9 x (1 / 3.29 x 1015) = 0.111- 0.071 = 0.04 or 1/25 Or n2 = 25, n= 5 The spectrum lies in the infra-red region. Q56 Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum. Ans: The radius of the nth orbit of hydrogen-like particles = 0.529n2/Z Å Now r1 = 1.3225 nm or 1322.5 pm = 52.9n12 And r2 = 211.6pm = 52.9n22/Z Therefore r1/r2 = 1322.5 / 211.6 = n12/n22 or n12/n22 = 6.25 or n1/n2 = 2.5 therefore n2 = 2 , n1 = 5. Thus the transition is from 5th orbit to 2nd orbit. It belongs Balmer series Wave number for the transition is given by, 1.097 × 107 m–1 (1/22-1/52) =1.097 x 107m-1 (21/100) = 2.303 × 106 m–1 Wavelength (λ) associated with the emission transition is given by, = 0.434 ×10–6 m λ = 434 nm the region is visible region Q57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 106 ms–1, calculate de Broglie wavelength associated with this electron. Ans: From de Broglie’s equation, λ = h/mv Substituting the values in this equation,we get 6.626x 10-34 / (9.11x10-31)(1.6x106) = 4.55 x 10-10m = 455pm Q58 Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron. Ans: Mass of neutron = 1.675 x10-27 kg We know λ = h/mv or v = h/mλ Substituting the value ,we get 6.626x10-34 / (1.675x10-27)(800x10-12) = 4.94x104 m/sec Q59 If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1, calculate the de Broglie wavelength associated with it. Ans: According to de Broglie’s equation, λ = h/mv Where, λ = wavelength associated with the electron h = Planck’s constant m = mass of electron v = velocity of electron Substituting the values in the expression of λ: λ = 6.626x10-34 / (9.11x10-31)(2.19x106) =3.32x10-10 = 332 pm Q60 The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity. Ans: λ = h/mv = 6.626x10-34 / (0.1)(4.37x105) = 1.515x10-28m Q61 If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is h/4πm × 0.05 nm, is there any problem in defining this value. Ans: From Heisenberg’s uncertainty principle, Where, Δx = uncertainty in position of the electron Δp = uncertainty in momentum of the electron Δx = 0.002nm = 2x10-12m(given)  Therefore, Substituting the values in the expression of Δp: Δp = h/4π Δx or = 2.637 × 10–23 Jsm–1 Δp = 2.637 × 10–23 kgms–1 (1 J = 1 kgms2s–1) Actual momentum = h/4πx 0.05nm = 6.626x10-34/4 x3.14 x 5 x10-11 = 1.055 x 10-24 kg m/sec Q62 The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists: 1. n = 4, l = 2, ml = –2 , ms = –1/2 2. n = 3, l = 2, ml= 1 , ms = +1/2 3. n = 4, l = 1, ml = 0 , ms = +1/2 4. n = 3, l = 2, ml = –2 , ms = –1/2 5. n = 3, l = 1, ml = –1 , ms= +1/2 6. n = 4, l = 1, ml = 0 , ms = +1/2 Ans: Quantum number – they are the index numbers which gives the complete address of an electron Here n =principal quantum number l = azimuthal quantum number ms = Spin quantum number For n = 4 and l = 2, the orbital occupied is 4d. For n = 3 and l = 2, the orbital occupied is 3d. For n = 4 and l = 1, the orbital occupied is 4p. Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively. Therefore, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d). Q63 The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge? Ans: 4p electrons, being farthest from the nucleus,experience the lowest effective nuclear charge. Q64 Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p Ans: Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it. (i) 2s is closer to the nucleus than 3s.Hence 2s will experience larger effective nuclear charge. (ii) 4d will experience greater nuclear charge than 4f since 4d is closer to the nucleus than 4f. (iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3f because 3p is closer to nucleus than 3f. Q65 The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus? Ans: Nuclear charge is defined as the net positive charge experienced by an electron in a multi-electron atom. Silicon has greater nuclear charge(+14) than aluminium (+13). Hence the unpaired 3p electron in case of silicon will experience more effective nuclear charge. Q66 Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr. Ans: (a) Phosphorus (P): 1s2 2s2 2p6 3s2 3p3 No of unpaired electron = 3   (b) Silicon (Si): 1s2 2s2 2p6 3s2 3p2 No of unpaired electron = 2 (since p orbital can have maximum 6 electron )   (c) Chromium (Cr): 1s2 2s2 2p6 3s2 3p6 4s1 3d5 No of unpaired electrons = 6 (since 1 electron is to be added to 4s & 5 electron to be added to 3d orbital.   (d) Iron (Fe): 1s2 2s2 2p6 3s2 3p6 4s2 3d6 No of unpaired electrons = 4   (e) Krypton (Kr): 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 Since all orbitals are fully occupied, there are no unpaired electrons in krypton. Q67 (a) How many sub-shells are associated with n = 4? (b) How many electrons will be present in the sub-shells having ms value of –1/2 for n = 4? Ans: (a) n = 4 (Given) For a given value of ‘n’, ‘l’ can have values from zero to (n – 1). ∴ l = 0, 1, 2, 3 Thus, four sub-shells are associated with n = 4, which are s, p, d and f.   (b) Number of orbitals in the nth shell = n2 For n = 4 Number of orbitals = 16 Each orbital has one electron with ms = -1/2 Hence there will be 16 electrons with ms = -1/2