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1. Stoichiometry 1
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Lecture1.1
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Lecture1.3
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2. Stoichiometry 2
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Lecture2.1
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Lecture2.3
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3. Thermodynamics & Thermochemistry
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Lecture3.1
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4. Atomic Structure
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5. Chemical equilibrium
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6. Ionic Equilibrium
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Lecture6.1
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Lecture6.17
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7. Introduction & Development of Org. Chemistry
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Lecture7.1
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Lecture7.2
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Lecture7.3
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8. Nomenclature of Org. Compounds
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Lecture8.1
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9. GOC 1- Hybridisation, Resonance, Aromaticity
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Lecture9.1
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10. GOC 2 - Substituent effect
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Lecture10.1
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11. GOC 2 - Reactive Intermediate
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Lecture11.1
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12. GOC 2 - Acid, base, Electrophile, Nucleophile
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Lecture12.1
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13. Isomerism
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14. Reaction Mechanism
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15. Alkane
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Lecture15.1
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Lecture15.3
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16. Alkene
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Lecture16.1
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17. Alkyl Halides
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Lecture17.1
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18. Chemical Bonding
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Lecture18.30
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19. Periodic Table
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Lecture19.1
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Lecture19.6
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Lecture19.7
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20. Metallurgy
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Lecture20.1
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21. Hydrogen and its Compounds
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22. S block metals
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Lecture22.1
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23. p block elements
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Lecture23.4
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Lecture23.5
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Lecture23.6
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Lecture23.7
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Lecture23.8
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NCERT Solutions – Atomic Structure
(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 = 9.11×10-31kg
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 NH3Â at STP.
Will the answer change if the temperature and pressure are changed ?
Answer
(i) Electrons present in 1 molecule of methane (CH4) =Â 6+4 = 10
1 atom of 14C contains = 14-6 = 8 neutrons.
Number of neutrons in 7 mg = (6.022×1023×8×7)/14000 = 2.4088×1021 neutrons
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 NH3Â
13C6, 16O8, 24Mg12, 56Fe26, 88CSr38
Answer
Nucleus     Z      A     Protons(Z)     Neutrons(A-Z)
13C6 Â Â Â Â Â Â Â Â Â 6 Â Â Â Â Â 13 Â Â Â Â Â Â 6 Â Â Â Â Â Â Â Â Â Â Â 13-6=7
16O8 Â Â Â Â Â Â Â Â Â 8 Â Â Â Â Â 16 Â Â Â Â Â Â 8 Â Â Â Â Â Â Â Â Â Â Â 16-8=8
24Mg12 Â Â Â Â Â Â 12 Â Â Â Â Â 24 Â Â Â Â Â Â 12 Â Â Â Â Â Â Â Â Â Â 24-12=12
56Fe26 Â Â Â Â Â Â Â 26 Â Â Â Â 56 Â Â Â Â Â Â 26 Â Â Â Â Â Â Â Â Â Â 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.
(iii)Â 9Be4Â
2.5. Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (ν) and wave number (ṽ) of the yellow light.
wave number (ṽ) = 1/λ = 1/580×10-9 m = 1.72×106 m-1
2.6. Find energy of each of the photons which
(i) correspond to light of frequency 3×1015 Hz.
(ii) have wavelength of 0.50 Ã….
2.8. What is the number of photons of light with a wavelength of 4000 pm that provide 1J of energy?
∴ 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
2.11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second.
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.
λ = 1/ν = 1/20564.4 = 486×10-7 cm =  486×10-9 m = 486 nm
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-1Â
For ionization from 5th orbit, n1 = 5, n2 = ∞
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.
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.
For the Balmer series, n1 = 2. Hence, ṽ = R(1/22 – 1/n22)
2.18. 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.
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
K.E. = 1/2 mv2
∴ v = √2 K.E./m
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
Isoelectronic are the species having same number of electrons.
A positive charge means the shortage of an electron.
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)Â Na+Â andMg2+Â (10 electrons each)
2) K+, Ca2+, S2- and Ar (18 electrons each)
∴ electronic configuration of H– = 1s2
(b) 11Na = 1s22s22p63s1 . A positive charge means the shortage of an electron.
(d)Â 9F =Â 1s22s22p5
(ii) (a)Â 3s1
Completing 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
electronic configuration =Â 1s22s1
∴ Atomic number of the element = 2+1 = 3
Hence, the element with the electronic configuration [He]2s1 is lithium (Li).
(b)Â [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)Â [Ar] 4s23d1
electronic configuration =Â 1s22s22p63s23p64s23d1
Hence, the element with the electronic configuration [Ar] 4s23d1Â 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 l = 4, minimum value of n = 5
For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (l) = 2
Magnetic quantum number (ml) = -2,-1,0,1,2
∴ 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
2.27. Give the number of electrons in the species H2+ , H2 and O2+
H2+Â = 2-1 = 1 electron
H2 =Â 1HÂ +Â 1H = 2 electrons
O2+Â = 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
m2Â = -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
(f) Possible
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
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, á¹½ = (2Ï€2mZ2e4/ch3)×(1/n12 – 1/n22) = RZ2(1/n12 – 1/n22)
∴ For He+ spectrum, Balmer transition, n=4 to n=2
á¹½ = 1/λ = RZ2(1/22 – 1/42)  = R×4×3/16 = 3R/4
For hydrogen spectrum ,
á¹½ = 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–Â
The ionization energy for the H atom in the ground state is 2.18×10-18 J atom-1.
Answer
Diameter of a carbon atom = 0.15 nm = 0.15×10-9 m = 1.5×10-10 m
2.36. 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.
2.38. A certain particle carries 2.5×10-16 C of static electric charge. Calculate the number of electrons present in it.
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 =Â 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/Â 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.
Then, number of neutrons = x+(30.4 x/Â 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.
The increasing order of frequency is as follows:
Radiation from FM radio < amber light < radiation from microwave oven < X- rays < cosmic rays
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-1
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.
Energy = Nhν = (2.5×1015)×(6.626×10-34 Js)×(0.5×109 s-1) = 8.28×10-10 J
∴ 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
2.52. Following results are observed when sodium metal is irradiated with different wavelengths.     Calculate (a) threshold wavelength and, (b) Planck’s constant.
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/ZÂ Ã… = 52.9n2/Z pm
r1Â = 1.3225 nm = 1322.5 pm = 52.9n12
r2Â = 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 n2Â = 2, n1Â = 5. Thus the transition is from 5th orbit to 2nd orbit. It belongs to Balmer series.
á¹½ = 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.
Δx = 0.002 nm = 2×10-12 m
(b)Â 14Si = 1s2Â 2s2Â 2p6Â 3s2Â 3 px1Â py1. 2Â 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 = n2Â = 42Â = 16
Each orbitals has one electron with ms = -1/2. Hence, there will be 16 electrons with ms = -1/2.