NCERT Solutions of Atomic Structure will provide the answers of all the questions being asked in NCERT chemistry book of class 11. This is the 2nd chapter of 11th chemistry. The answer of all the questions have been written in simple and comprehensive manner. The main focus in this solution notes has been given on the interest of the students which will help to have strong hold on the concepts of this chapter without any complications and doubts. Numericals have been solved in proper manner so that students can easily understand and learn to solve the questions in own way.
Students read throughout all the contents of any chapter but feels hesitation in solving the questions of any exercise. To breakdown like these problems, our team decided to upload the NCERT Solutions of Atomic Structure. We will have to read and apply all the important laws and formulas which have prescribed in structure of atoms. This is the basic chapter of chemistry therefore, every students must practice to solve all the questions of atomic structure. Let’s start to solve the questions of atomic structure.
NCERT Solutions of Atomic Structure
Atomic structure is the 2nd chapter of chemistry 11th. This chapter gives an exclusive idea about the structure of the atom of any element. NCERT Solutions of Atomic Structure include the following topics of this chapter. They are discovery of fundamental particles like electron, proton and neutron. Important terms as atomic number, mass number, isotopes, isobars, isoelectronic. Important models which are Thomson’s model, Rutherford model, Bohr’s model and their limitations. Some important principles such as electromagnetic waves theory, Planck’s quantum theory, de- Broglie theory, Aufbau principle, Heisenberg’s uncertainty principle, Pauli’s exclusion principle, Hunds rule. Definition of shell, subshell, electronic Configuration and its stability.
NCERT Solutions of Atomic Structure ⇒ PDF DOWNLOAD
NCERT Solutions of Atomic Structure are here:
1. (i) Calculate the number of electrons which will together weigh one gram.
Ans:- Mass of 1 electron = 9.11 × 10-31 kg or 9.11 × 10-31 kg = 1 electron
∴1 g or 10-3 kg = 1/9.11 × 10-31 kg × 10-3 electrons = 1.098 × 1027 electrons.
(ii) Calculate the mass and charge of one mole of electrons.
Ans:- Mass of 1 electron = 9.11 × 10-31 kg ∴mass of 1 mole of electrons = (6.022 × 1023) × (9.11 × 10-31) = 5.486 × 10-7 kg
Charge on one electron = 1.602 × 10-19 coulomb. ∴ Charge on 1 mole of electrons = (1.602 × 10-19 ) × (6.022 × 1023 ) = 9.65 × 104 coulomb. NCERT Solutions of Atomic Structure.
2. (i) Calculate the total number of electrons present in one mole of methane.
Ans:- 1 molecule of CH4 contains electrons = 6+ 4 = 10 ∴ 1 mole, i.e., 6.022 × 1023 molecules will contain electrons = 6.022 × 1024
(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).
Ans:- (a) 1 g-atom of 14C = 14 g = 6.022 × 1023 atoms = (6.022 × 1023 ) × 8 neutrons (as each 14C has 14 – 6 = 8 neutrons)
Thus 14 g or 14000 mg have 8 × 6.022 × 1023 neutrons. ∴ 7 mg will have neutrons = 8 × 6.022 × 1023 × 7/14000 = 2.4088 × 1021
(b) Mass of 1 neutron = 1.675 × 10-27 kg ∴ Mass of 2.4088 × 1021 neutrons = (2.4088 × 1021 ) × (1.675 × 10-27 kg ) = 4.0347 × 10-6 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?
(a) 1 mol of NH3 = 17 g NH3 = 6.022 × 1023 molecules of NH3 = ( 6.022 × 1023) × (7 + 3) protons = 6.022 × 1024 protons ∴ 34 mg, i.e., 0.034 g NH3 = 6.022 × 1024 /17 × 0.034 = 1.2044 × 1022 protons
(b) Mass of 1 proton = 1.6726 × 10-27 kg ∴ Mass of 1.2044 × 1022 protons = ( 1.6726 × 10-27 ) × ( 1.2044 × 1022) kg = 2.0145 × 10-5 kg.
3. How many neutrons and protons are there in the following nuclei? NCERT Solutions of Atomic Structure.
136C, 168O , 2412Mg ,5626Fe, 8838Sr
Ans:- In 136C, Z = 6, A = 13, As Z = p = 6 and n = A – Z = 13 – 6 = 7
In 168O, Z = 8, A = 16, As Z = p = 8 and n = A – Z = 16 – 8 = 8
In 2412Mg, Z = 12, A = 24, As Z = p = 12 and n = A – Z = 24 – 12 = 12
In 5626Fe, Z = 26, A = 56, As Z = p = 26 and n = A – Z = 56 – 26 = 30
In 8838Sr, Z = 38, A = 88, As Z = p = 38 and n = A – Z = 88 – 38 = 50.
4. 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 NCERT Solutions of Atomic Structure.
Ans:- (i) 3517Cl, (ii) 23392U (iii) 94Be
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.
Ans:- λ = 580 nm = 580 × 10-9 m
Frequency, ν = c/λ = 3.0 × 108 ms-1 / 580 × 10-9 m = 5.17 × 1014 s-1
Wave number, () = 1/λ = 1/ 580 × 10-9 m = 1.72 × 106 ms-1
6. Find energy of each of the photons which
(i) correspond to light of frequency 3× 1015 Hz. NCERT Solutions of Atomic Structure.
Ans:- ν = 3 × 1015 Hz and E = hv = (6.626 × 10-34)Js × (3 × 1015 s-1) = 1.988 × 10-18 J
(ii) have wavelength of 0.50 Å.
Ans:- λ = 0.50 × 10-10 m and E = hc/λ = (6.626 × 10-34) × (3.0 × 108 ms-1)/ 0.50 × 10-10 m = 3.98 × 10-15 J
7. Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10–10 s.
Ans:- Frequency (v) = 1/ period = 1/ 2.0 × 10-10 s = 5 × 109 s-1
Wavelength (λ) = c/v = 3.0 × 108 ms-1 / 5 × 109 s-1 = 6.02 × 10-2 m
Wave number, () = 1 / λ = 1 / 6.02 × 10-2 m = 16.66 m-1
8. What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Ans:- λ = 4000 pm = 4000 × 10-12 m = 4 × 10-9 m and E = Nhv = Nhc/λ ∴ N = E × λ / h × c
∴ N = (1J) × (4 × 10-9 m ) /(6.626 × 10-34 Js) × (3.0 × 108 ms-1) = 2.012 × 1016 photons. NCERT Solutions of Atomic Structure.
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).
Ans:- (i) the energy of the photon (E) = hv = hc/λ = (6.626 × 10-34 Js) × (3.0 × 108 ms-1) / 4 × 10-7 m = 4.97 × 10-19 J = 4.97 × 10-19 / 1.602 × 10-19 eV = 3.10 eV.
(ii) the kinetic energy of the emission (1/2mv2) = hv – hvo = 3.10 – 2.13 = 0.97 eV.
(iii) 1/2mv2 = 0.97 eV = 0.97 × 1.6020 × 10–19 J or 1/2 × (9.11 × 10-31 kg) × V2 = 0.97 eV = 0.97 × 1.6020 × 10–19 J or v2 = 0.341 × 1012 = 34.1 × 1010 ∴ v = 5.24 × 104 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.
Ans:- E = Nhc/λ = (6.02 × 1023 mol-1) × (6.626 × 10-34 Js) × (3.0 × 108 ms-1) / 242 × 10-9 m = 4.945 × 105 J mol-1 = 494.5 KJmol-1.
11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second. NCERT Solutions of Atomic Structure.
Ans:- Energy emitted by the bulb = 25 watt = 25 Js–1 (∴1 watt = 1 Js–1)
Energy of 1 photon (E) = hv = hc / λ here, λ = 0.57 μm = 0.57 × 10–6 m, h = 6.626 × 10-34 Js and c = 3.0 × 108 ms-1
Now putting the value of λ, h and c. E = (6.626 × 10-34 Js) × (3.0 × 108 ms-1) /0.57 × 10–6 m = 3.48 × 10–19 J
∴ No. of photons emitted per second = 25 Js–1 / 3.48 × 10–19 J = 7.18 × 1019 photons.
12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Calculate threshold frequency 
and work function (W0) of the metal.
Ans:- Threshold wavelength (λo) = 6800 Ao = 6800 × 10–10 m. As c = vλ ∴ λo = 3.0 × 108 ms-1 / 6800 × 10–10 m = 4.41 × 1014 s–1
Work function (W0) = hvo = (6.626 × 10-34 Js) × (4.41 × 1014 s–1 ) = 2.92 × 10–19 J
13. 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:- Wave number, () = R (1/n12 – 1/n22) = 109677 (1/22 – 1/42) cm–1 = 20564.4 cm–1
∴ λ = 1/ = 1/ 20564.4 cm–1 = 486 × 10–7 cm = 486 × 10–9 m = 486 nm
14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your Sol. with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit). NCERT Solutions of Atomic Structure.
Ans:- En = – 21.8 × 10–19/n2 J atom –1 For ionization from 5th orbit, n1 = 5, n2 = ∞ ∴ ΔE = E2 – E1 = – 21.8 × 10–19(1/n22 – 1/n12) = 21.8 × 10–19(1/n12 – 1/n22) = 21.8 × 10–19 (1/52 – 1/∞) = 8.72 × 10–20 J
For ionization from 1st orbit, n1 = 1, n2 = ∞ hence, ΔE = 21.8 × 10–19 (1/12 – 1/∞) = 21.8 × 10–19 J
ΔE’/ΔE = 21.8 × 10–19 J / 8.72 × 10–20 J = 25 Thus, the energy required to remove electron from 1st orbit is 25 times than that required to electron from 5th orbit.
15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
Ans:- No. of lines produced when electron from nth shell drops to ground state = n(n – 1)/2. Here n = 6 ∴ No. of lines produced = 6(6 – 1)/2 = 15
16. (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:- (i) En = –2.18 × 10–18/n2 J ∴ E5 = –2.18 × 10–18/52 J = – 8.72 × 10–20 J
(ii) For H- atom, rn = 0.529 × n2 Ao ∴ r5 = 0.529 × 52 Ao = 13.225 Ao = 1.3225 nm.
17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen. NCERT Solutions of Atomic Structure.
Ans:- For Balmer series, n1 = 2, Hence, Wave number, () = R (1/22 – 1/n22) Now = 1/λ, For λ to be longest, should be minimum. This can be so only when n2 is minimum, i.e., n2 = 3, Hence, Wave number, () = (1.097 × 107 m–1) × (1/22 – 1/322) = 1.097 × 107 × 5/36 m–1 = 1.523 × 106 m–1
18. 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:- As ground state electronic energy is –2.18×10-11 ergs, this means that En = –2.18×10-11 / n2 erg and ΔE = E5 – E1 = 2.18×10-11 (1/12 – 1/52) = 2.18×10-11 (24/25) = 2.09 ×10-11 erg = 2.09 ×10-18 J (∴1 erg = 10–7 J)
When electron returns to ground state (i.e., to n = 1), energy emitted = 2.09 ×10-11 erg. As E = hc/λ or λ = hc/E = (6.626 × 10-27 erg sec) × (3.0 × 1010 cms-1)/2.09 ×10-11 erg = 9.51 × 10–6 cm = 951 Ao
19. The electron energy in hydrogen atom is given by E n = (–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:- ΔE = E∞ – E2 = 0 – (–2.18 × 10–18/22 J atom–1 ) = 5.45 × 10–19 J atom–1
Now , ΔE = hc/λ or λ = hc/ΔE = (6.626 × 10-34 Js) × (3.0 × 108 ms-1) / 5.45 × 10–19 J = 3.647 × 10–7 m = 3.647 × 10–7 cm.
20. Calculate the wavelength of an electron moving with a velocity of 2.05×107 ms-1. NCERT Solutions of Atomic Structure.
Ans:- By de Broglie equation, λ = h/ mv = 6.626 × 10-34 Js/ (9.11 × 10–31 Kg)(2.05×107 ms-1 ) = 3.55 × 10–11 m.
21. The mass of an electron is 9.11×10-31 kg. If its K.E. is 3.0×10-25 J, calculate its wavelength.
Ans:- K.E. = 1/2 mv2 and v = √2K.E./m = √2 × 3.0×10-25 J/√9.11 × 10–31 Kg = 812 ms–1
By de Broglie equation, λ = h/ mv = 6.626 × 10-34 Js/ (9.11 × 10–31 Kg)(812 ms–1 ) = 8.967 × 10–7 m = 8967 Ao
22. Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na+ , K+ , Mg2+ , Ca2+ , S2- , Ar.
Ans:- No. of electrons are: Na+ = 11 – 1 = 10, K+ = 19 – 1 = 18, Mg2+ = 12 – 2 = 10, Ca2+ = 20 – 2 = 18, S2– = 16 + 2 = 18, Ar = 18. Hence, isoelectronic species are Na+ and Mg2+ ; K+, Ca2+, S2– and Ar.
23. (i) Write the electronic configurations of the following ions: (a) H– (b) Na+ (c) O2- (d) F–
Ans:- (a) H+ = 1s0 (b) Na+ = 1s2 2s2 2p6 (c) O2– = 1s2 2s2 2p6 (d) F– = 1s2 2s2 2p6
(ii) What are the atomic numbers of elements whose outermost electrons are represented by (a) 3s1 (b) 2p3 and (c) 3p5 ? NCERT Solutions of Atomic Structure.
Ans:- (a) 1s2 2s2 2p6 3s1 (Z = 11) (b) 1s2 2s2 2p3 (Z = 7) (c) 1s2 2s2 2p6 3s2 3p6 4s2 3d6 (Z = 26)
(iii) Which atoms are indicated by the following configurations ? (a) [He]2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d .
Ans:- (a) [He]2s1 = Li (Z = 3) (b) [Ne] 3s2 3p3 = P (Z = 15) (c) [Ar] 4s2 3d1 = Sc (Z = 21)
24. What is the lowest value of n that allows g orbitals to exist?
Ans:- For g- subshell, l = 4, As l = 0 to n–1, to have l = 4, minimum value of n = 5
25. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
Ans:- For 3d orbitals, n = 3, l = 2 and for l = 2 ml = –2, –1, 0, +1, +2.
26. 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:-(i) For neutral atom, no. of protons = no. of electrons = 29. Thus, atomic number of the element = 29. Hence, (ii)Electronic Configuration of the element with Z = 29 will be 1s2 2s2 2p6 3s2 3p6 3d10 4s1 and element is Cu (Z = 29).
27. Give the number of electrons in the species H2+ , H2 and O2+
Ans:- No. of electrons in H2+ ion = 2 – 1 = 1. No. of electrons in H2 = 1 + 1 = 2. No. of electrons in O2+ = 16 – 1 = 15. NCERT Solutions of Atomic Structure.
28. (i) An atomic orbital has n = 3. What are the possible values of l and ml ?
Ans:- when n = 3, l = 0, 1, 2, when l = 0, ml = 0, when l = 1, ml = –1, 0, +1, when l = 2, ml = –2, –1, 0, + 1, + 2
(ii) List the quantum numbers (ml and l ) of electrons for 3d orbital.
Ans:- For 3d-orbital, n = 3, l =2, For l = 2, ml = –2, –1, 0, + 1, + 2
(iii) Which of the following orbitals are possible?
1p, 2s, 2p and 3f
Ans:- 1p is not possible because, when n = 1, l = 0 only (for p, l = 1), 2s is possible because when n = 2, l = 0, 1 (for s, l = 0). 3p is possible because when n = 3, l = 0, 1, 2 (for p, l = 1). 3f is not possible because, when n = 3, l = 0, 1, 2 only (for f, l = 3).
29. 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:- (a) 1s (b) 3p (c) 4d (d) 4f
30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.
(a) n = 0, l = 0, ml = 0, ms = +½
(b) n = 1, l = 0, ml = 0, ms = –½
(c) n = 1, l = 1, ml = 0, ms = +½
(d) n = 2, l = 1, ml = 0, ms = –½
(e) n = 3, l = 3, ml = –3, ms = +½
(f) n = 3, l = 1, ml = 0, ms = +½
Ans:- (a) Not possible because n ≠ 0 (b) possible (c) Not possible because n = 1, l ≠ 1 (d) possible (e) Not possible because when n = 3, l ≠ 3 (f) possible
31. How many electrons in an atom may have the following quantum numbers?
(a) n = 4, ms = –½ (b) n = 3, l = 0 NCERT Solutions of Atomic Structure.
Ans:- (a) Total electrons in n = 4 are 2n2, i.e., 2 × 42 = 32. Half of them, i.e., 16 electrons have ms = – 1/2
(b) n = 3, l = 0 means 3s orbital which can have 2 electrons.
32. 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:- According to Bohr’s postulate of angular momentum, mvr = nh/2π or 2πr = nh/mv, ————–(1) again According to de Broglie equation, λ = h/mv. Substituting this value in eqation (1), we get 2π r = nλ which implies that the circumference (2π r) of the Bohr’s orbit for hydrogen atom is an integral multiple of de Broglie wavelength.
33. 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 H-like particles in general, Wave number () = 2π2mZ2e4/ch2 (1/n12 – 1/n22) = RZ2 (1/n12 – 1/n22)
∴ For He+ spectrum, for Balmer transition, n = 4 to n = 2, = 1/λ = RZ2 (1/22 – 1/42) = R × 4 × 3/16 = 3R/4
For hydrogen spectrum, = 1/λ= R × 1× (1/n12 – 1/n22) = 3/4 R or (1/n12 – 1/n22) = 3/4 which can be so for n1 = 1 and n2 = 2, i.e., the transition is from n = 2 to n = 1
34. Calculate the energy required for the process
He+(g) → He2+ (g) + e–
The ionization energy for the H atom in the ground state is 2.18×10-18 J atom-1. NCERT Solutions of Atomic Structure.
Ans:- For H- like particles, En = – 2π2mZ2e4/ n2h2 For H-atom, I.E. = E∞ – E1 = 0 – (– 2π2me4/ 12 × h2 ) = 2π2me4/ h2 = 2.18 × 10–18 J atom-1 (given) , For the given process, Energy required = E∞ – E1 = 0 – ( 2π2m22e4/ 12 × h2 ) = 4 × 2π2me4/ h2 = 4 × 2.18 × 10–18 J atom-1 = 8.72 × 10–18 J.
35. 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 atoms = 0.15 pm = 1.5 × 10 –10 m. Length along which atoms are to be placed = 20 cm = 2 × 10–1 m. ∴ No. of carbon atoms which can be placed along the length = 2 × 10–1 m / 1.5 × 10 –10 m = 1.33 × 109
36. 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. NCERT Solutions of Atomic Structure.
Ans:- Total length = 3.0 cm. Total number of atoms along the length = 2 × 108 ∴ Diameter of each atom = 3.0 cm /2 × 108 = 1.5 × 10–8 cm
∴ Radius of the atom = 1.5 × 10–8 cm /2 = 0.075 × 10–7 cm = 0.075 × 10–9 m = 0.075 nm.
37. 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 Ao / 2 = 1.3 Ao = 1.3 × 10–10 m = 130 pm.
(b) Given length = 1.6 cm = 1.6 × 10–2 m. Diameter of one atom = 2.6 Ao = 2.6 × 10–10 m ∴ No. of atoms present along the length = 1.6 × 10–2 m /2.6 × 10–10 m = 6.154 × 107
38. A certain particle carries 2.5 × 10-16 C of static electric charge. Calculate the number of electrons present in it.
Ans:- Charge carried by one electron = 1.602 × 10–19 C. ∴ Electrons present in particles carrying 2.5 × 10-16 C charge = 2.5 × 10-16 C / 1.602 × 10–19 C = 1560
39. In Millikan’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 –1.282×10-18 C, calculate the number of electrons present on it.
Ans:- Charge carried by one electron = 1.602 × 10–19 C. ∴ No. of electrons present on the oil drop = –1.282×10-18 C / –1.602 × 10–19 C = 8
40. 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 ? NCERT Solutions of Atomic Structure.
Ans:- Heavy atoms have a heavy nucleus carrying a large amount of positive charge. Hence, some α- particles are easily deflected back on hitting the nucleus. Also a number of α- particles are deflected through small angles because of large positive charge on the nucleus. If light atoms are used, their nuclei will be light and moreover, they will have small positive charge on the nucleus. Hence, the number of particles deflected back and those deflected through some angle will be negligible.
41. Symbols 79Br35 and 79Br can be written, whereas symbols 35Br79 and 35Br are not acceptable . Answer briefly.
Ans:- Atomic number of an element is fixed. However, mass number is not fixed as it depends on the isotope taken. Hence, it is essential to indicate the mass number.
42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol. NCERT Solutions of Atomic Structure.
Ans:- Mass number = 81, that means p + n = 81. If protons = x, then neutrons = x + 31.7 /100 × x = 1.317 x ∴ x + 1.317 x = 81 or 2.317 x = 81 or x = 81/2.317 = 35. Thus, protons = 35 and Z = 35. Hence, the formula is 8135Br .
43. 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:- Suppose the number of electrons in the ion = x. Then number of neutrons = x + 11.1/100 × x = 1.111 x. No. of electrons in the neutral atom = x – 1 ∴ No. of protons = x– 1. As mass number = p + n ∴ 37 = 1.111 x + (x – 1) or 2.111 x = 38 or x = 18 ∴ No. of protons = x – 1 = 18 – 1 = 17. Hence, the symbol of the ion will be 3717Cl-1
44. 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:- Suppose the number of electrons in the ion M3+ = x and number of neutrons = x + 30.4/100 × x = 1.304 x . Number of electrons in the neutral atom = x + 3 ∴ No. of protons = x + 3. As Mass no. = p + n. Or 56 = x + 3 + 1.304 x or 2.304 x = 53 or x = 23 ∴ No. of protons = Atomic number = x +3 = 23 + 3 = 26. Hence, the symbol of the ion will be 5626Fe3+
45. 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 < amber colour < microwave < FM.
46. 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. NCERT Solutions of Atomic Structure.
Ans:- E = Nhv = Nhc/λ = (5.6×1024 ).(6.626 × 10-34 Js) × (3.0 × 108 ms-1)/ (337.1 × 10–9 m) = 3.3 × 106 J
47. 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 = 616 nm = 616 × 10–9 m. (a) Frequency, v = c/λ = 3.0 × 108 ms-1 / 616 × 10–9 m = 4.87 × 1014 s–1
(b) Velocity of the radiation = 3.0 × 108 ms-1 ∴ Distance travelled in 30 a = 30 × 3.0 × 108 m = 9.0 × 109 m
(c) E = hc / λ = (6.626 × 10-34 Js) × (3.0 × 108 ms-1 )/616 × 10–9 m = 32.27 × 10–20 J
(d) No. of photons in 2 J of energy = 2/ 32.27 × 10–20 = 6.2 × 1018
48. 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 = hc/λ = (6.626 × 10-34 Js) × (3.0 × 108 ms-1 )/(600 × 10–9 m) = 3.313 × 10–19 J. Total energy received = 3.15 × 10–18 J ∴ No of photons received = 3.15 × 10–18 J / 3.313 × 10–19 J = 9.51 ≅ 10
49. 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. NCERT Solutions of Atomic Structure.
Ans:- Frequency = 1/ period = 1 / 2 × 10–9 s = 0.5 × 109 s–1 ∴ Energy = N hv = (2.5×1015 ).(6.626 × 10-34 Js) (0.5 × 109 s–1 ) = 8.28 × 10–10 J.
50. 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 = 589 × 10–9 m ∴ v1 = c/λ1 = 3.0 × 108 ms-1 /589 × 10–9 m = 5.093 × 1014 s-1
λ2 = 589.6 nm = 589.6 × 10–9 m ∴ v2 = c/λ2 = 3.0 × 108 ms-1 /589.6 × 10–9 m = 5.088 × 1014 s-1
ΔE = E2 – E1 = h (v2 – v1) = (6.626 × 10-34 Js)(5.093 – 5.088) × 1014 s-1 = 3.31 × 10–22 J
51. The work function for cesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the cesium element is irradiated with a wavelength 500nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Ans:- (a) Work function (Wo) = hvo ∴ vo = Wo / h = 1.9 × 1.602 × 10–19 J/ 6.626 × 10-34 Js = 4.59 × 1014 s–1
(b) λo = c / vo = 3.0 × 108 ms-1 / 4.59 × 1014 s–1 = 6.54 × 10–7 m = 654 m
(c) K.E. of ejected electron = h (v – vo) = hc (1/λ – 1/λo) = (6.626 × 10-34 Js) × (3.0 × 108 ms-1 ) (1/500 ×10–9 m — 1/654 × 10–9 m ) = 6.626 × 3.0 × 10–26 / 10–9 (154/500 × 654) J = 9.36 × 10–20 J
K.E. = 1/2 mv2 = 9.36 × 10–20 J or kg m2s–2 ∴ 1/2 × ( 9.11 × 10–31 kg) v2 = 9.36 × 10–20 kg m2s–2 or v2 = 2.055 × 1011 m2s–2 = 20.55 × 1010 m2 s–2 or v = 4.53 × 105 ms-1
52. Following results are observed when sodium metal is irradiated with different wavelengths. NCERT Solutions of Atomic Structure.
Ans:- Soppose threshold wavelength = λo nm = λo × 10–9 m. Then h (v – vo) = 1/ mv2 or hc (1/λ – 1/λo) = 1/ mv2
Substituting the given result of the experiments, we get hc/ 10–9 (1/500 – 1/λo ) = 1/2 m (2.55 × 106)2 ———-(i), hc/ 10–9 (1/450 – 1/λo ) = 1/2 m (4.35 × 106)2 ————(ii), hc/ 10–9 (1/400 – 1/λo ) = 1/2 m (5.20 × 106)2 ———————-(iii) Dividing equation (ii) by equation (i), we get λo – 450 /450 λo × 500 λo / λo – 500 = (4.35/ 2.55)2 or λo – 450 /λo – 500 = 450/500 (4.35/ 2.55)2 = 2.619 or λo – 450 = 2.619λo – 1309.5 or 1.619λo = 859.5 ∴ λo = 531 nm
Substituting this value in eqation (iii), we get( h × (3 × 108) / 10–9) (1/400 – 1/531) = 1/2 (9.11 × 10–31) (5.20 × 106)2 ∴ h = 6.66 × 10–34 Js.
53. 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:- Energy of the incident radiation = work function × kinetic energy of phenomenon. Energy of incident radiation (E) = hv = hc /λ = (6.626 × 10-34 Js) × (3.0 × 108 ms-1 )/ (256.7 × 10–9 m) = 7.74 × 10–19 J = 4.83 eV. The potential applied gives the kinetic energy to the electron. Hence, kinetic energy of the electron = 0.35 eV. ∴ work function = 4.83 eV – 0.35 eV = 4.48 eV
54. 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. NCERT Solutions of Atomic Structure.
Ans:- Energy of the incident photon = hc/λ = (6.626 × 10-34 Js) × (3.0 × 108 ms-1 )/ (150 × 10–12 m) = 13.25 × 10–16 J.
Energy of the electron ejected = 1/2 mv2 = 1/2 ( 9.11 × 10–31 kg)(1.5 × 107 ms-1)2 = 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 × 10–19 eV = 7.63 × 103 eV
55. 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 = (3.0 × 108 ms-1 )/1285 × 10–9 m = 3.29 × 1015 (1/32 – 1/n2) or 1/n2 = 1/9 – (3.0 × 108/1285 × 10–9 m)(1/3.29 × 1015) = 0.111 – 0.071 = 0.04 = 1/25 or n2 = 25 ∴ n = 5
56. 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:- Radius of nth orbit of H- like particles = 0.529 n2/Z Ao = 52.9 n2/Z pm r1 = 1.3225 nm = 1322.5 pm = 52.9 n12 and r2 = 211.6 pm = 52.9 n22/Z
∴ r1/r2 = 1322.5/211.6 = n12 / n22 or n12 / n22 = 6.25 or n1/n2 = 2.5 ∴ If n2 = 2, n1 = 5, Thus , the transition is from 5th orbit to 2nd orbit. It belongs to Balmer series. And = 1.097 × 107 m–1(1/22 – 1/52) = 1.097 × 21/100 × 107 m–1 or λ = 1/ = 100 / 1.097 × 21× 107 m = 434 nm. It lies in the visible region.
57. 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. NCERT Solutions of Atomic Structure.
Ans:- λ = h / mv = 6.626 × 10–34 kg m2 s–1 / (9.11 × 10–31 kg) (1.6 × 106 ms-1) = 4.55 × 10–10 m = 455 pm.
58. 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:- λ = h / mv or v = h / mλ = 6.626 × 10–34 kg m2 s–1 / (1.675 × 10–27 kg)(800 × 10–12 m) = 4.94 × 104 ms-1
59. 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:- λ = h / mv = 6.626 × 10–34 kg m2 s–1 / (9.11 × 10–31 kg)(2.19 × 106 ms-1) = 3.32 × 10–10 m = 332 pm
60. 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 = 6.626 × 10–34 kg m2 s–1 / (0.1 kg)(4.37 × 105 ms-1) = 1.516 × 10–28 m
61. 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/(4π×0.05 nm), is there any problem in defining this value. NCERT Solutions of Atomic Structure.
Ans:- Δx = 0.002 nm = 2 × 10–12 m and Δx × Δp = h / 4π ∴ Δp = h / 4π.Δx = 6.626 × 10–34 kg m2 s–1 / 4 × 3.14 × (2 × 10–12 m) = 2.638 × 10–23 kg ms-1
Actual momentum = h / 4π × 0.05 nm = 6.626 × 10–34 kg m2 s–1 / 4 × 3.14 × 5 × 10–11 m = 1.055 × 10–24 kg ms-1. It cannot be defined as the actual magnitude of the momentum is smaller than the uncertainty.
62. 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 , m s = –1/2
2. n = 3, l = 2, ml = 1 , m s = +1/2
3. n = 4, l = 1, ml = 0 , m s = +1/2
4. n = 3, l = 2, ml = –2 , m s = –1/2
5. n = 3, l = 1, ml = –1 , m s = +1/2
6. n = 4, l = 1, ml = 0 , m s = +1/2
Ans:- The orbitals occupied by the electrons are (i) 4d (ii) 3d (iii) 4p (iv) 3d (v) 3p (vi) 4p ; Their energy will be in the order : (v) < (ii) = (iv) < (vi) = (iii) < (i)
63. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron experiences the lowest effective nuclear charge ? NCERT Solutions of Atomic Structure.
Ans:- 4p electrons, being farthest from the nucleus, experience the lowest effective nuclear charge.
64. 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:- (i) 2s is closer to the nucleus than 3s. Hence, 2s will experience larger effective nuclear charge. For the same reason in (ii) 4d and in (iii) 3p.
65. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?
Ans:- Silicon has greater nuclear charge (+14) than aluminium (+13). Hence, the unpaired 3p electrons in case of silicon will experience more effective nuclear charge from the nucleus.
66. Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Ans:- (a) 15P = [Ne] 3s2 3px1 3py1 3pz1, No. of unpaired electrons = 3 (b) 14 Si = [Ne] 3s2 3px2 3py1 3pz1, No. of unpaired electrons = 2. (c) 24Cr = [Ar] 3d5 4s1 , No. of unpaired electrons = 6. (d) 26Fe = [Ar] 3d6 4s2 , No. of unpaired electrons = 4 in 3d. (e) 36Kr = It is a noble gas. All orbitals are completely filled. Unpaired electrons = 0
67. (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, l = 0, 1, 2, 3 (4 subshells as s, p, d and f) (b) No. of orbitals in 4th shell = n2 = 42 = 16. Each orbital has one electron with ms = –1/2, Hence, there will be 16 electrons with ms = –1/2.
Conclusion of NCERT Solutions of Atomic Structure
Atomic structure is the second chapter of chemistry in class 11th. There are 67 questions in the exercise of this chapter. All the questions have been solved by the easiest method in proper manner. Required formulas have been applied at proper steps. We hope that students will take benefits from this NCERT Solutions of Atomic Structure very much. At final you all are requested to share this article among your friends and favourites. Thanks a lot for this kind attention.