NCERT Solutions for Class 11 Chemistry Chapter 2 Structure of Atom includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Chemistry exams must go through NCERT Solutions for Class 11 Chemisrty Chapter 2 Structure of Atom. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.
Chemistry Part 1 Class 11 will introduce you to basic concepts of subjects, about atoms, chemical bonding and other topics. There are 7 chapters through which learners can study in an organized manner and improve their studying habits.
Class 11th Chapter -2 Structure of Atom NCERT CHEMISRTY SOLUTION |
Page No: 65
Exercises
2.1. (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Answer
(i) Mass of one electron = 9.11×10-31kg
∴ 1g = 10-3 kg = (1/9.11×10-31)×10-3 electrons = 1.098×1027
(ii) Mass of one electron = 9.11×10-31kgExercises
2.1. (i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Answer
(i) Mass of one electron = 9.11×10-31kg
∴ 1g = 10-3 kg = (1/9.11×10-31)×10-3 electrons = 1.098×1027
∴ Mass of 1 mole of electrons = (9.11×10-31)×(6.022×1023) = 5.48×10-7 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 coulombs.
2.2. (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 = 1.675×10-27kg).
(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 changed ?
Answer
(i) Electrons present in 1 molecule of methane (CH4) = 6+4 = 10
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 coulombs.
2.2. (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 = 1.675×10-27kg).
(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 changed ?
Answer
(i) Electrons present in 1 molecule of methane (CH4) = 6+4 = 10
∴ Electrons in mol i.e. 6.022×1023 molecules = 6.022×1024
(ii) (a) Number of atoms in 14C in 1 mole= 6.022×1023 atoms1 atom of 14C contains = 14-6 = 8 neutrons.
∴ The number of neutrons in 14 g of 14C = 6.022×1023×8 neutrons
Number of neutrons in 7 mg = (6.022×1023×8×7)/14000 = 2.4088×1021 neutrons
Number of neutrons in 7 mg = (6.022×1023×8×7)/14000 = 2.4088×1021 neutrons
(b) Mass of one neutron = 1.674×10-27kg
Mass of total neutrons in 7 g of 14C = (2.4088×1021) (1.675 ×10-27 kg) = 4.035 ×10-6 kg
(iii) (a) 1 mol of NH3 = 17g NH3 = 6.022×1023 molecules of NH3
Mass of total neutrons in 7 g of 14C = (2.4088×1021) (1.675 ×10-27 kg) = 4.035 ×10-6 kg
(iii) (a) 1 mol of NH3 = 17g NH3 = 6.022×1023 molecules of NH3
1 atom of NH3 contains = 7+3 = 10 protons
∴ The number of protons in 1 mol of NH3 = 6.022×1024 protons.
Number of protons in 34 mg of NH3 = (6.022×1024×34)/17×1000 = 1.2044×1022 protons.
(b) Mass of one proton = 1.6726×10-27kg
∴ Mass of 1.2044×1022 protons = (1.6726×10-27) ×(1.2044×1022) kg = 2.0145×10-5 kg.
No, there will be no effect of temperature and pressure.
2.3. How many neutrons and protons are there in the following nuclei ?
13C6, 16O8, 24Mg12, 56Fe26, 88CSr38
Answer
Nucleus Z A Protons(Z) Neutrons(A-Z)13C6, 16O8, 24Mg12, 56Fe26, 88CSr38
Answer
13C6 6 13 6 13-6=7
16O8 8 16 8 16-8=8
24Mg12 12 24 12 24-12=12
56Fe26 26 56 26 56-26=30
88CSr38 38 88 38 88-38=50
2.4. 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.
Answer
(i) 35Cl17
(ii) 233U92
(iii) 9Be4 Answer
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 (ṽ) = 1/λ = 1/580×10-9 m = 1.72×106 m-1(i) correspond to light of frequency 3×1015 Hz.
(ii) have wavelength of 0.50 Å.
Answer
(i) ν = 3×1015 Hz
E = hν = (6.626×10-34 Js)×(3×1015 s-1) = 1.988×10-18 J
(ii) λ = 0.50×10-10 m
E = hν = hc/λ = (6.626×10-34 Js)×(3×1015 s-1)/0.50×10-10 m = 3.98×10-15 J
2.7. Calculate the wavelength, frequency and wave number of a light wave whose period is
2.0×10-10 s.
Answer
Frequency (ν) = 1/Period = 1/2.0×10-10 s = 5×109 s-1.
Wavelength (λ) = c/ν = 3.0×108 ms-1/5×109 s-1 = 6.0×102 m
Wave number (ṽ) = 1/λ = 1/6.0×102 m = 16.66 m-1
Answer
λ = 4000 pm = 4000×10-12 m = 4×10-9 m
E = Nhν = Nhc/λ
∴ N = E×λ/h×c = (1J×4×10-9 m)/(6.626×10-34 Js×3.0×108 ms-1) = 2.012×1016 photons.
2.9. 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).
Answer
(i) Energy of the photon (E) = hν = 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
(ii) Kinetic energy of emission (1/2 mv2) = hν- hνo = 3.10-2.13 = 0.97 eV
(iii) 1/2 mv2 = 0.97 eV = 0.97×1.602×10-19 J
⇒ 1/2×(9.11×10-31 kg)×v2 = 0.97×1.602×10-19 J
⇒ v2 = 0.341×1012 = 34.1×1010
⇒ v = 5.84×105 ms-1
2.10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol-1.
Answer
∴ N = E×λ/h×c = (1J×4×10-9 m)/(6.626×10-34 Js×3.0×108 ms-1) = 2.012×1016 photons.
2.9. 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).
Answer
(i) Energy of the photon (E) = hν = 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
(ii) Kinetic energy of emission (1/2 mv2) = hν- hνo = 3.10-2.13 = 0.97 eV
(iii) 1/2 mv2 = 0.97 eV = 0.97×1.602×10-19 J
⇒ 1/2×(9.11×10-31 kg)×v2 = 0.97×1.602×10-19 J
⇒ v2 = 0.341×1012 = 34.1×1010
⇒ v = 5.84×105 ms-1
2.10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol-1.
Answer
E = Nhν = Nhc/λ = (6.022×1023 mol-1)×(6.626×10-34 Js×3.0×108 ms-1)/242×10-9 m
= 4.945×105 Jmol-1 = 494.5 kJmol-1
Answer
Energy emitted by the bulb = 25 watt = 25 Js-1
Energy of one photon (E) = hν = hc/λ
λ = 0.57µm = 0.57×10-6 m
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 sec = 25 Js-1/3.48×10-19 J = 7.18×1019
2.12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (νo) and work function (wo) of the metal.
Answer
c = νλ
∴ νo = c/λo = 3.0×108 ms-1/6800×10-10 m = 4.14×1014 s-1
Work function (wo) = hνo = 6.626×10-34 Js×4.14×1014 s-1 = 2.92×10-19 J
2.13. 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?
Answer
ṽ = R(1/n12 – 1/n22) = 109677(1/22 – 1/42) cm-1 = 20564.4 cm-1
λ = 1/ν = 1/20564.4 = 486×10-7 cm = 486×10-9 m = 486 nm
Page No: 66
2.14. 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).
Answer
En = -21.8×10-19/n2 Jatom-1
For ionization from 5th orbit, n1 = 5, n2 = ∞
λ = 1/ν = 1/20564.4 = 486×10-7 cm = 486×10-9 m = 486 nm
Page No: 66
2.14. 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).
Answer
En = -21.8×10-19/n2 Jatom-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 = ∞
∴ ΔE’ = 21.8×10-19×(1/12 – 1/∞) = 21.8×10-19 J
ΔE’/ΔE = 21.8×10-19/8.72×10-20 = 25
Hence, 25 times less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.2.15. What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
Answer
The number of spectral lines produced when an electron in the nthlevel drops down to the ground state is given by n(n-1)/2.
Answer
The number of spectral lines produced when an electron in the nthlevel drops down to the ground state is given by n(n-1)/2.
Given, n=6
∴ Number of spectral lines = 6×5/2 = 15
also given by, ∑(n2 – n1) = ∑(6-1) = ∑5 = 5+4+3+2+1 = 15
2.16. (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.
Answer
(i) En = -21.8×10-19/n2 J
∴ E5 = -21.8×10-18/52 J = 8.72×10-20 J
(ii) For H atom, rn = 0.529×n2 Å
∴ r5 = 0.529×52 = 13.225 Å = 1.3225 nm
2.17. Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.
Answer
For the Balmer series, n1 = 2. Hence, ṽ = R(1/22 – 1/n22)
ṽ = 1/λ (inversely proportional)
For λ to be maximum, ṽ should be minimum. This can be happened when n2 is minimum i.e. n2 = 3. Hence, ṽ = (1.097×107 m-1) (1/22 – 1/32) = 1.097×107×5/36 m-1 = 1.523×106 m-1
Answer
1erg = 10-7 J
As ground state electronic energy is –2.18×10-11 ergs, this means that En = -21.8×10-11/n2 ergs.
ΔE = E5 – E1 = 2.18×10-11 (1/12 – 1/52) = 2.18×10-11(24/25) = 2.09×10-11 ergs = 2.09×10-18 J
When electron returns to ground state (n=1), energy emitted = 2.09×10-11 ergs.
As, E = hν = hc/λ
⇒ λ = hc/E = (6.626×10-27 erg sec) (3.0×1010 cm s-1)/2.09×10-11 ergs
= 9.51×10-6 cm = 951×10-8 cm = 951 Å
2.19. 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?
Answer
ΔE = E∞– E2 = 0 – (–2.18×10-18 J atom-1/22) = 5.45×10-19 J atom-1
ΔE = hν = hc/λ
⇒ λ = hc/ΔE = (6.626×10-34 Js)×(3.0×108 ms-1)/5.45×10-19 J = 3.674×10-7 m = 3.674×10-5 cm
2.20. Calculate the wavelength of an electron moving with a velocity of 2.05×107 ms-1.
Answer
ΔE = hν = hc/λ
⇒ λ = hc/ΔE = (6.626×10-34 Js)×(3.0×108 ms-1)/5.45×10-19 J = 3.674×10-7 m = 3.674×10-5 cm
2.20. Calculate the wavelength of an electron moving with a velocity of 2.05×107 ms-1.
Answer
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
2.21. The mass of an electron is 9.11×10-31 kg. If its K.E. is 3.0×10-25 J, calculate its wavelength.
Answer
K.E. = 1/2 mv2
= 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 m2.22. Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na+ , K+ , Mg2+ , Ca2+ , S2- , Ar.
Answer
Notes:
Isoelectronic are the species having same number of electrons.A positive charge means the shortage of an electron.
A negative charge means gain of electron.
Number of electrons in Na+ = 11-1 = 10
Number of electrons in K+ = 19-1 = 18
Number of electrons in Mg2+ = 12-2 = 10
Number of electrons in Ca2+ = 20-2 = 18
Number of electrons in S2- = 16+2 = 18
Number of electrons in Ar = 18
Hence, the following are isoelectronic species:
1) Na+ andMg2+ (10 electrons each)
2) K+, Ca2+, S2- and Ar (18 electrons each)
Number of electrons in K+ = 19-1 = 18
Number of electrons in Mg2+ = 12-2 = 10
Number of electrons in Ca2+ = 20-2 = 18
Number of electrons in S2- = 16+2 = 18
Number of electrons in Ar = 18
Hence, the following are isoelectronic species:
1) Na+ andMg2+ (10 electrons each)
2) K+, Ca2+, S2- and Ar (18 electrons each)
2.23. (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 .
Answer
(i) (a) 1H = 1s1 . A negative charge means gain of electron.
∴ electronic configuration of H– = 1s2
(b) 11Na = 1s22s22p63s1 . A positive charge means the shortage of an electron.
∴ electronic configuration of H– = 1s2
(b) 11Na = 1s22s22p63s1 . A positive charge means the shortage of an electron.
∴ electronic configuration of Na+ = 1s22s22p6
(c) 8O = 1s22s22p4
∴ electronic configuration of O2- = 1s22s22p6
(d) 9F = 1s22s22p5∴ electronic configuration of F– = 1s22s22p6
(ii) (a) 3s1Completing the electron configuration of the element as 1s22s22p63s1
∴ Number of electrons present in the atom of the element = 2+2+6+1 = 11
∴ Atomic number of the element = 11
(b) 2p3
Completing the electron configuration of the element as 1s22s22p3
∴ Number of electrons present in the atom of the element = 2+2+3 = 7
∴ Atomic number of the element = 7
(c) 3p5
Completing the electron configuration of the element as 1s22s22p63s23p5
∴ Number of electrons present in the atom of the element = 2+2+6+2+5 = 17
∴ Atomic number of the element = 9
(iii) (a) [He]2s1
electronic configuration = 1s22s1
∴ Atomic number of the element = 2+1 = 3
Hence, the element with the electronic configuration [He]2s1 is lithium (Li).
(b) [Ne]3s23p3
electronic configuration = 1s22s22p63s23p3
∴ Atomic number of the element = 2+2+6+2+3 = 15
Hence, the element with the electronic configuration [Ne]3s23p3 is phosphorus (P).
(c) [Ar] 4s23d1
electronic configuration = 1s22s22p63s23p64s23d1
electronic configuration = 1s22s1
∴ Atomic number of the element = 2+1 = 3
Hence, the element with the electronic configuration [He]2s1 is lithium (Li).
(b) [Ne]3s23p3
electronic configuration = 1s22s22p63s23p3
∴ Atomic number of the element = 2+2+6+2+3 = 15
Hence, the element with the electronic configuration [Ne]3s23p3 is phosphorus (P).
(c) [Ar] 4s23d1
electronic configuration = 1s22s22p63s23p64s23d1
∴ Atomic number of the element = 2+2+6+2+6+2+1 = 21
Hence, the element with the electronic configuration [Ar] 4s23d1 is scandium (Sc).
2.24. What is the lowest value of n that allows g orbitals to exist?
Answer
For g-orbitals, l = 4.
Hence, the element with the electronic configuration [Ar] 4s23d1 is scandium (Sc).
2.24. What is the lowest value of n that allows g orbitals to exist?
Answer
For g-orbitals, l = 4.
For any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n – 1).
∴ For l = 4, minimum value of n = 5
∴ For l = 4, minimum value of n = 5
2.25. An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
Answer
For the 3d orbital:Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (ml) = -2,-1,0,1,2
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.
Answer
(i) For neutral atom, number of protons = number of electrons.
∴ Number of protons in the atom of the given element = 29 = Atomic number
(ii) The electronic configuration of the atom with Z=29 is 1s22s22p63s23p64s13d10
∴ Number of protons in the atom of the given element = 29 = Atomic number
(ii) The electronic configuration of the atom with Z=29 is 1s22s22p63s23p64s13d10
Answer
H2+ = 2-1 = 1 electron
H2 = 1H + 1H = 2 electrons
O2+ = 16-1 = 15 electrons
2.28. (i) An atomic orbital has n = 3. What are the possible values of l and ml ?
(ii) List the quantum numbers (ml and l ) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible?
1p, 2s, 2p and 3f
Answer
(i) For a given value of n, l can have values from 0 to (n-1).
∴ For n = 3 , l = 0, 1, 2
For a given value of l, ml can have (2l+1) values.
When l = 0, m = 0
l = 1, m = – 1, 0, 1
l = 2, m = – 2, – 1, 0, 1, 2
l = 3, m = -3, -2, -1, 0, 1, 2, 3
(ii) For 3d orbital, n = 3, l = 2.
∴ For l = 2
m2 = -2, -1, 0, 1, 2
(iii) 1p is not possible because when n = 1, l = 0. (for p, l = 1)
2s is possible because when n=2, l = 0,1 (for s, l=0)
2p is possible because when n=2, l = 0,1 (for p, l=1)
3f is not possible because when n=3, l = 0, 1 , 2 (for f, l=3)
2.29. 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.
Answer
(a) 1s
(b) 3p
(c) 4d
(d) 4f
2.30. 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
Answer
(a) Not possible because n≠0
(b) Possible
(c) not possible because when n=1, l≠1
(d) Possible
(e) Not possible because when n=3, l≠3
(f) Possible
Page No: 67
2.31. How many electrons in an atom may have the following quantum numbers?
(a) n = 4, ms = –½ (b) n = 3, l = 0
Answer
(i) The total number of electrons in n is given by 2n2
n=4, Number of electrons = 2×42 = 32
H2+ = 2-1 = 1 electron
H2 = 1H + 1H = 2 electrons
O2+ = 16-1 = 15 electrons
2.28. (i) An atomic orbital has n = 3. What are the possible values of l and ml ?
(ii) List the quantum numbers (ml and l ) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible?
1p, 2s, 2p and 3f
Answer
(i) For a given value of n, l can have values from 0 to (n-1).
∴ For n = 3 , l = 0, 1, 2
For a given value of l, ml can have (2l+1) values.
When l = 0, m = 0
l = 1, m = – 1, 0, 1
l = 2, m = – 2, – 1, 0, 1, 2
l = 3, m = -3, -2, -1, 0, 1, 2, 3
(ii) For 3d orbital, n = 3, l = 2.
∴ For l = 2
m2 = -2, -1, 0, 1, 2
(iii) 1p is not possible because when n = 1, l = 0. (for p, l = 1)
2s is possible because when n=2, l = 0,1 (for s, l=0)
2p is possible because when n=2, l = 0,1 (for p, l=1)
3f is not possible because when n=3, l = 0, 1 , 2 (for f, l=3)
2.29. 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.
Answer
(a) 1s
(b) 3p
(c) 4d
(d) 4f
2.30. 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
Answer
(a) Not possible because n≠0
(b) Possible
(c) not possible because when n=1, l≠1
(d) Possible
(e) Not possible because when n=3, l≠3
(f) Possible
Page No: 67
2.31. How many electrons in an atom may have the following quantum numbers?
(a) n = 4, ms = –½ (b) n = 3, l = 0
Answer
(i) The total number of electrons in n is given by 2n2
n=4, Number of electrons = 2×42 = 32
Half of 32 electrons will have spin quantum number ms = –½ i.e. 16 electrons
(ii) n=3 and l=0 means it is 3s orbital which can have only 2 electrons.
2.32. 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.
Answer
According to Bohr postulate of angular momentum,
mvr = nh/2π ⇒ 2πr = nh/mv …(i)
According to de Broglie equation, λ=h/mv …(ii)
Substituting the value of eqn (ii) in eqn (i) we get,
2πr = nλ
Thus, circumference of the Bohr orbit for the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.
2.33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?
Answer
For H-like particles, ṽ = (2π2mZ2e4/ch3)×(1/n12 – 1/n22) = RZ2(1/n12 – 1/n22)
∴ For He+ spectrum, Balmer transition, n=4 to n=2
ṽ = 1/λ = RZ2(1/22 – 1/42) = R×4×3/16 = 3R/4
For hydrogen spectrum ,
ṽ = 1/λ = R(1/n12 – 1/n22) = 3R/4 ⇒ (1/n12 – 1/n22) = 3/4
which can be true only for n1=1 and n2=2 i.e. transition from n=2 to n=1.
2.34. 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.
According to de Broglie equation, λ=h/mv …(ii)
Substituting the value of eqn (ii) in eqn (i) we get,
2πr = nλ
Thus, circumference of the Bohr orbit for the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.
2.33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?
Answer
For H-like particles, ṽ = (2π2mZ2e4/ch3)×(1/n12 – 1/n22) = RZ2(1/n12 – 1/n22)
∴ For He+ spectrum, Balmer transition, n=4 to n=2
ṽ = 1/λ = RZ2(1/22 – 1/42) = R×4×3/16 = 3R/4
For hydrogen spectrum ,
ṽ = 1/λ = R(1/n12 – 1/n22) = 3R/4 ⇒ (1/n12 – 1/n22) = 3/4
which can be true only for n1=1 and n2=2 i.e. transition from n=2 to n=1.
2.34. 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.
For H-like particles, En = – (2π2mZ2e4/n2h2)
For H-atom, I.E. = E-E1 = 0 – (-2π2m×22×e4/12×h2)
= (4×2π2me4/h2) = 4×2.18×10-18 J = 8.72×10-18 J
2.35. 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.
Answer
Diameter of a carbon atom = 0.15 nm = 0.15×10-9 m = 1.5×10-10 m
Length along which atoms are to be placed = 20 cm = 2×10-1 m
∴ No. of C-atoms which can be placed along the line = 2×10-1 m/1.5×10-10 m = 1.33×109
Answer
Total length = 2.4 cm
Total number of atoms along the length = 2×108
∴ Diameter of each atom = 2.4 cm/2×108 = 1.2×10-8 cm
∴ Radius of the atom = Diameter/2 = 1.2×10-8 cm/2 = 0.6×10-8 cm
2.37. 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.
Answer
(a) Radius of zinc atom = 2.6Å/2 = 1.3Å = 1.3×10-10 m = 130×10-12 pm
(b) Given length = 1.6 cm = 1.6×10-2 m
Diameter of one atom = 2.6 Å = 2.6×10-10 m
∴ No. of atoms present along the length = 1.6×10-2/2.6×10-10 = 6.154×107
Answer
Charge on one electron = 1.602×10-19 C
∴ Number of electrons carrying 2.5×10-16 C charge = 2.5×10-16/1.602×10-19 =1560
2.39. 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 –1.282×10-18 C, calculate the number of electrons present on it.
Answer
As in the above question,
Number of electrons present in oil drop = –1.282×10-18/1.602×10-19 = 8
2.40. 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 ?
Answer
Heavy atoms have nucleus carrying large amount of positive charge. Therefor, some α-particles will easily deflected back. Also a number of α-particles are deflected through small angles because of large positive charge.
If light atoms are used, their nuclei will have small positive charge, hence the number of α-particles getting deflected even through small angles will be negligible.
2.41. Symbols 79Br35 and 79Br can be written, whereas symbols 35Br79 and 35Br are not acceptable. Answer briefly.
Answer
35Br79 is not acceptable because atomic number should be written as subscript, while mass number should be written as superscript. 35Br is not acceptable because atomic number of an element is fixed. However, mass number is not fixed as it depends upon the isotopes taken. Hence, it is essential to indicate mass number.
2.42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Answer
Mass number = protons+neutrons = p+n = 81 (given)
Let p be x, then neutrons = x + (31.7/100)x = 1.317 x
∴ x + 1.317 x = 81
⇒ 2.317 x = 81
⇒ x = 81/2.317 = 35
Thus, protons = 35 = atomic number.
The symbol of the element is 81Br35 or 8135Br
2.43. 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.
Answer
Let the number of electrons in the ion = x
Then, number of neutrons = x+(11.1 x/ 100) = 1.111 x
Number of electrons in the neutral atom = x-1 (ion possesses one unit of negative charge)
∴ Number of protons = x-1
Mass number = No. of protons + No. of neutrons
∴ 1.111 x + x – 1 = 37
⇒ 2.111x = 38
⇒ x = 18
∴ No. of protons = Atomic no. = x-1 = 18-1 = 17
The symbol of the ion is 3717Cl-1
2.44. 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.
Answer
35Br79 is not acceptable because atomic number should be written as subscript, while mass number should be written as superscript. 35Br is not acceptable because atomic number of an element is fixed. However, mass number is not fixed as it depends upon the isotopes taken. Hence, it is essential to indicate mass number.
2.42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Answer
Mass number = protons+neutrons = p+n = 81 (given)
Let p be x, then neutrons = x + (31.7/100)x = 1.317 x
∴ x + 1.317 x = 81
⇒ 2.317 x = 81
⇒ x = 81/2.317 = 35
Thus, protons = 35 = atomic number.
The symbol of the element is 81Br35 or 8135Br
2.43. 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.
Answer
Let the number of electrons in the ion = x
Then, number of neutrons = x+(11.1 x/ 100) = 1.111 x
Number of electrons in the neutral atom = x-1 (ion possesses one unit of negative charge)
∴ Number of protons = x-1
Mass number = No. of protons + No. of neutrons
∴ 1.111 x + x – 1 = 37
⇒ 2.111x = 38
⇒ x = 18
∴ No. of protons = Atomic no. = x-1 = 18-1 = 17
The symbol of the ion is 3717Cl-1
2.44. 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.
Answer
Let the number of electrons in the ion = x
Then, number of neutrons = x+(30.4 x/ 100) = 1.304 x
Number of electrons in the neutral atom = x+3 (ion possesses 3 units of positive charge)
∴ Number of protons = x+3
Mass number = No. of protons + No. of neutrons
∴ 1.304 x + x +3 = 56
⇒ 2.304x = 53
⇒ x = 23
∴ No. of protons = Atomic no. = x+3 = 23+3 = 26
The symbol of the ion is 5626Fe+3
2.45. 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.
Then, number of neutrons = x+(30.4 x/ 100) = 1.304 x
Number of electrons in the neutral atom = x+3 (ion possesses 3 units of positive charge)
∴ Number of protons = x+3
Mass number = No. of protons + No. of neutrons
∴ 1.304 x + x +3 = 56
⇒ 2.304x = 53
⇒ x = 23
∴ No. of protons = Atomic no. = x+3 = 23+3 = 26
The symbol of the ion is 5626Fe+3
2.45. 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.
Answer
The increasing order of frequency is as follows:
Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays2.46. 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.
Answer
E = Nhν = Nhc/λ = (5.6×1024)×(6.626×10-34 Js×3.0×108 ms-1)/337.1×10-9 m = 3.3×106 J
2.47. 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.
Answer
λ = 616 nm = 616×10-9 m
(a) Frequency, ν = 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-1Answer
E = Nhν = Nhc/λ = (5.6×1024)×(6.626×10-34 Js×3.0×108 ms-1)/337.1×10-9 m = 3.3×106 J
2.47. 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.
Answer
λ = 616 nm = 616×10-9 m
(a) Frequency, ν = c/λ = 3.0×108 ms-1/616×10-9 m = 4.87×1014 s-1
∴ Distance travelled in 30 s = 30×3×108 m = 9.0×109 m
(c) E = hν = hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/616×10-9 m = 32.27×10-20 J
(d) No. of quanta in 2J of energy = 2J/32.27×10-20 J = 6.2×1018
2.48. 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.
Answer
Energy of one photon = hν = 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 (approx 10)
Page No: 68
2.49. 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.
Answer
Frequency = 1/2×10-19 s = 0.5×109 s-1
Energy = Nhν = (2.5×1015)×(6.626×10-34 Js)×(0.5×109 s-1) = 8.28×10-10 J2.50. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calcualte the frequency of each transition and energy difference between two excited states.
Answer
λ1 = 589 nm = 589×10-9 m
∴ ν1 = 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
∴ ν2 = c/λ2 = 3.0×108 ms-1/589.6×10-9 m = 5.088×1014 s-1
ΔE = E2 – E1 = h(ν2 – ν1) = (6.626×10-34 Js)×(5.093-5.088)×1014 s-1 = 3.31×10-22 J
2.51. 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 500nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Answer
(a) Work function (W0) = hν0
∴ ν0 = W0/h = 1.9×1.602×10-19 J/6.626×10-34 Js = 4.59×1014 s-1 (1eV = 1.602×10-19 J)
(b) λ0 = c/ν0 = 3.0×108 ms-1/4.59×1014 s-1 = 6.54×10-7 m = 654×10-9 m = 654 nm
(c) K.E. of ejected electron = h(ν – ν0) = hc (1/λ – 1/λ0 )
= (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
∴ 1/2×(9.11×10-31 kg) v2 = 9.36×10-20 kgm2s-2
⇒ v2 = 2.055×1011 m2s-2 = 20.55×1010 m2s-2
∴ ν1 = 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
∴ ν2 = c/λ2 = 3.0×108 ms-1/589.6×10-9 m = 5.088×1014 s-1
ΔE = E2 – E1 = h(ν2 – ν1) = (6.626×10-34 Js)×(5.093-5.088)×1014 s-1 = 3.31×10-22 J
2.51. 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 500nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Answer
(a) Work function (W0) = hν0
∴ ν0 = W0/h = 1.9×1.602×10-19 J/6.626×10-34 Js = 4.59×1014 s-1 (1eV = 1.602×10-19 J)
(b) λ0 = c/ν0 = 3.0×108 ms-1/4.59×1014 s-1 = 6.54×10-7 m = 654×10-9 m = 654 nm
(c) K.E. of ejected electron = h(ν – ν0) = hc (1/λ – 1/λ0 )
= (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
∴ 1/2×(9.11×10-31 kg) v2 = 9.36×10-20 kgm2s-2
⇒ v2 = 2.055×1011 m2s-2 = 20.55×1010 m2s-2
⇒ v = 4.53×105 ms-1
λ (nm) 500 450 400
v×10-5(cm s-1) 2.55 4.35 5.35
Answer
Let the threshold wavelength to be λ0 nm = λ0×10-9 m .
Following equation holds true for photoelectric emission in given case:
K.E. = 1/2 mv2 = h(ν – ν0)
⇒1/2 mv2 = hν – hν0
⇒ hν0 = hν – 1/2 mv2
⇒ hc/λ0 = hc/λ – 1/2 mv2
(a) Substituting the value of λ and v from the above given data, we get three values of λ0 as,
λ0(1) = 541 nm
λ0(2) = 546 nm
λ0(3) = 542 nm
Threshold frequency = λav = {λ0(1)+λ0(2)+λ0(3)}/3 = (541+546+542)/3 = 543 (approx 540)
(b) Part of this question can’t be solved due to incorrect value of v i.e 5.35.
Students can assume this value as 5.20 if they want to solve this question.
2.53. 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.
Answer
Energy of the incident radiation = Work function + Kinetic energy of photoelectron
E = 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 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
2.54. 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.
Answer
Energy of the incident photon= hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/(150×10-12m) = 13.25×10-16 J
Energy of the electron ejected = 1/2 mv2 = 1/2×(9.11×10-31kg)×(1.5×107ms-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 = 12.225×10-16/1.602×10-19 eV = 7.63×103 eV
2.55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted 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.
Answer
ν = c/λ = 3.0×108 ms-1/1285×10-9 m = 3.29×1015 (1/32 – 1/n2)
⇒ 1/n2 = 1/9 – (3.0×108 ms-1/1285×10-9 m)×(1/3.29×1015) = 0.111-0.071 = 0.04 = 1/25
⇒ n2 = 25
⇒ n = 5
The radiation corresponding to 1285 nm lies in the infrared region.
2.56. 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.
Answer
Radius of nth orbit of H-like particles = 0.529n2/Z Å = 52.9n2/Z pm
r1 = 1.3225 nm = 1322.5 pm = 52.9n12
r2 = 211.6 pm = 52.9n22/Z
∴ r1/r2 = 1322.5 pm/211.6 pm = n12/n22
⇒ n12/n22 = 6.25
⇒ n1/n2 = 2.5
If n2 = 2, n1 = 5. Thus the transition is from 5th orbit to 2nd orbit. It belongs to Balmer series.
ṽ = 1.097×107 m-1 (1/22 – 1/52) = 1.097×107×21/100 m-1
λ = 1/ṽ = 100/(1.097×21×107) m = 434×10-9 m = 434 nm
It lies in visible range.
2.57. 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.
Answer
λ = h/mv = 6.626×10-34 kgm2s-1/(9.11×10-31 kg) (1.6×106 ms-1) = 4.55×10-10 m = 455 pm
2.58. 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.
Answer
Mass of neutron = 1.675×10-27 kg
λ = h/mv
⇒ v = h/mλ = 6.626×10-34 kgm2s-1/(1.675×10-27 kg) (800×10-12 m) = 4.94×104 ms-1
2.59. 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.
K.E. = 1/2 mv2 = h(ν – ν0)
⇒1/2 mv2 = hν – hν0
⇒ hν0 = hν – 1/2 mv2
⇒ hc/λ0 = hc/λ – 1/2 mv2
λ0(1) = 541 nm
λ0(2) = 546 nm
λ0(3) = 542 nm
Threshold frequency = λav = {λ0(1)+λ0(2)+λ0(3)}/3 = (541+546+542)/3 = 543 (approx 540)
(b) Part of this question can’t be solved due to incorrect value of v i.e 5.35.
Students can assume this value as 5.20 if they want to solve this question.
2.53. 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.
Answer
Energy of the incident radiation = Work function + Kinetic energy of photoelectron
E = 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 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
2.54. 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.
Answer
Energy of the incident photon= hc/λ = (6.626×10-34 Js×3.0×108 ms-1)/(150×10-12m) = 13.25×10-16 J
Energy of the electron ejected = 1/2 mv2 = 1/2×(9.11×10-31kg)×(1.5×107ms-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 = 12.225×10-16/1.602×10-19 eV = 7.63×103 eV
2.55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represeted 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.
Answer
ν = c/λ = 3.0×108 ms-1/1285×10-9 m = 3.29×1015 (1/32 – 1/n2)
⇒ 1/n2 = 1/9 – (3.0×108 ms-1/1285×10-9 m)×(1/3.29×1015) = 0.111-0.071 = 0.04 = 1/25
⇒ n2 = 25
⇒ n = 5
The radiation corresponding to 1285 nm lies in the infrared region.
2.56. 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.
Answer
Radius of nth orbit of H-like particles = 0.529n2/Z Å = 52.9n2/Z pm
r1 = 1.3225 nm = 1322.5 pm = 52.9n12
r2 = 211.6 pm = 52.9n22/Z
∴ r1/r2 = 1322.5 pm/211.6 pm = n12/n22
⇒ n12/n22 = 6.25
⇒ n1/n2 = 2.5
If n2 = 2, n1 = 5. Thus the transition is from 5th orbit to 2nd orbit. It belongs to Balmer series.
ṽ = 1.097×107 m-1 (1/22 – 1/52) = 1.097×107×21/100 m-1
λ = 1/ṽ = 100/(1.097×21×107) m = 434×10-9 m = 434 nm
It lies in visible range.
2.57. 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.
Answer
λ = h/mv = 6.626×10-34 kgm2s-1/(9.11×10-31 kg) (1.6×106 ms-1) = 4.55×10-10 m = 455 pm
2.58. 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.
Answer
Mass of neutron = 1.675×10-27 kg
λ = h/mv
⇒ v = h/mλ = 6.626×10-34 kgm2s-1/(1.675×10-27 kg) (800×10-12 m) = 4.94×104 ms-1
2.59. 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.
Answer
λ = h/mv = 6.626×10-34 kgm2s-1/(9.11×10-31 kg) (2.19×106 ms-1) = 3.32×10-10 m =332 pm
2.60. 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, calcualte the wavelength associated with this velocity.
Answer
λ = h/mv = 6.626×10-34 kgm2s-1/(0.1 kg) (4.37×105 ms-1) = 1.516×10-28 m
2.61. 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.
Answer
Δx = 0.002 nm = 2×10-12 mΔx × Δp = h/4π
∴ Δp = h/4πΔx = 6.626×10-34 kgm2s-1/(4×3.14×2×10-12 m) = 2.638×10-23 kgms-1
Actual momentum = h/(4π×0.05 nm) = h/(4π×5×10-11 m)
= 6.626×10-34 kgm2s-1/(4×3.14×5×10-11 m) = 1.055×10-24 kgms-1
It cannot be defined because the actual magnitude of the momentum is smaller than the uncertainty.
2.62. 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
Answer
The orbitals occupied by the electrons are:
(1) 4d
(2) 3d
(3) 4p
(4) 3d
(5) 3p
(6) 4p
Same orbitals will have same energy and higher the value of (n+l) higher is the energy,
Their energies will be in order: (5)<(2)=(4)<(6)=(3)<(1)
Page No: 69
2.63. 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 ?
Answer
4p electrons, being farthest from the nucleus experience the lowest effective nuclear charge.
2.64. 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.
Answer
(i) 2s is closer to the nucleus than 3s. Hence, 2s will experience larger effective nuclear charge.
(ii) 4d (Reason being the same as above)
(iii) 3p (Reason being the same as above)
2.65. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?
Answer
Silicon has greater nuclear charge (+14) than aluminium (+13). Hence, the unpaired 3p electron in case of silicon will experience more effective nuclear charge.
2.66. Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Answer
(a) 15P = 1s2 2s2 2p6 3s2 3 px1 py1 pz1. 3 unpaired electrons. (in 3p)
(b) 14Si = 1s2 2s2 2p6 3s2 3 px1 py1. 2 unpaired electrons.(in 3p)
(c) 24Cr = 1s2 2s2 2p6 3s2 3p6 3d5 4s1. 6 unpaired electrons. (5 in 3d and 1 in 4s)
(d) 26Fe = 1s2 2s2 2p6 3s2 3p6 3d6 4s2. 4 unpaired electrons. (in 3d)
(e) 36Kr = It is a Noble gas. All orbitals are filled. No unpaired electrons.
2.67. (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 ?
Answer
(a) n=4, l = 0, 1, 2, 3. 4 sub-shells are associated with n = 4
(b) No. of orbitals in the shells = n2 = 42 = 16
Each orbitals has one electron with ms = -1/2. Hence, there will be 16 electrons with ms = -1/2.
(b) 14Si = 1s2 2s2 2p6 3s2 3 px1 py1. 2 unpaired electrons.(in 3p)
(c) 24Cr = 1s2 2s2 2p6 3s2 3p6 3d5 4s1. 6 unpaired electrons. (5 in 3d and 1 in 4s)
(d) 26Fe = 1s2 2s2 2p6 3s2 3p6 3d6 4s2. 4 unpaired electrons. (in 3d)
(e) 36Kr = It is a Noble gas. All orbitals are filled. No unpaired electrons.
2.67. (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 ?
Answer
(a) n=4, l = 0, 1, 2, 3. 4 sub-shells are associated with n = 4
(b) No. of orbitals in the shells = n2 = 42 = 16
Each orbitals has one electron with ms = -1/2. Hence, there will be 16 electrons with ms = -1/2.
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