Question1
(i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Solution:
(i) 9.11 Ã— 10^{28} g is the mass of 1 electron
No. of electrons 1 g of mass = = 0.1099 Ã— 10^{28 }
= 1.099 Ã— 10^{27}
(ii) Mass of 1 electron = 9.1 Ã— 10^{31} kg
Mass of 6.023 Ã— 10^{23} electrons
= 6.023 Ã— 10^{23} Ã— 1.6 Ã— 10^{31} g
= 5.48 Ã— 10^{7} kg
Charge of one electron = 1.6 Ã— 10^{19} coulomb
Charge of 6.023 Ã— 10^{23} electrons
= 6.023 Ã— 10^{23} Ã— 1.6 Ã— 10^{19 }= 9.65 Ã— 10^{4}C.
(ii) Calculate the mass and charge of one mole of electrons.
Solution:
(i) 9.11 Ã— 10^{28} g is the mass of 1 electron
No. of electrons 1 g of mass = = 0.1099 Ã— 10^{28 }
= 1.099 Ã— 10^{27}
(ii) Mass of 1 electron = 9.1 Ã— 10^{31} kg
Mass of 6.023 Ã— 10^{23} electrons
= 6.023 Ã— 10^{23} Ã— 1.6 Ã— 10^{31} g
= 5.48 Ã— 10^{7} kg
Charge of one electron = 1.6 Ã— 10^{19} coulomb
Charge of 6.023 Ã— 10^{23} electrons
= 6.023 Ã— 10^{23} Ã— 1.6 Ã— 10^{19 }= 9.65 Ã— 10^{4}C.
Question2
(i) Calculate the number of electrons present in 1 mol of methane.
(ii) Find a) the total number and (b) the total mass of neutrons in 7 mg of ^{14}C.
(Assume that mass of a neutron = 1.675 Ã— 10^{27} kg)
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH_{3} at S.T.P. Will the answer change if the temperature and pressure are changed ?
Solution:
(i) 1 molecule of methane contains = 6 + 4 = 10 electrons
6.022 Ã— 10^{23} molecules (1 mole) of methane contain electrons
= 6.022 Ã— 10^{23} Ã— 10
= 6.022 Ã— 10^{24}
(ii) Atomic weight mass of C^{14} = 14 g
14 g of carbon has 6.022 Ã— 10^{23} neutrons
No of neutrons in
7 Ã— 10^{3} g of carbon
= = 3.011 Ã— 10^{20 }Mass of 1 neutron = 1.675 Ã— 10^{27} kg
^{ }Mass of 3.011 Ã— 10^{20} neutrons
^{ }= 1.675 Ã— 10^{27} Ã— 3.011 Ã— 10^{20} kg
^{ }= 5.04 Ã— 10^{7} kg
(iii) 17 g of NH_{3} has 6.022 Ã— 10^{23} protons
^{ }No of protons in 34 Ã— 10^{3} g of NH_{3} = protons
^{ }= 12.044 Ã— 10^{20 }= 1.2044 Ã— 10^{21 }Mass of 1 proton = 1.67 Ã— 10^{27} kg
Mass of 1.2044 Ã— 10^{21 }= 1.2044 Ã— 10^{21} Ã—1.67 Ã— 10^{27 }= 2.015 Ã— 10^{6}
No. Answer will not change with temperature and pressure.
(ii) Find a) the total number and (b) the total mass of neutrons in 7 mg of ^{14}C.
(Assume that mass of a neutron = 1.675 Ã— 10^{27} kg)
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH_{3} at S.T.P. Will the answer change if the temperature and pressure are changed ?
Solution:
(i) 1 molecule of methane contains = 6 + 4 = 10 electrons
6.022 Ã— 10^{23} molecules (1 mole) of methane contain electrons
= 6.022 Ã— 10^{23} Ã— 10
= 6.022 Ã— 10^{24}
(ii) Atomic weight mass of C^{14} = 14 g
14 g of carbon has 6.022 Ã— 10^{23} neutrons
No of neutrons in
7 Ã— 10^{3} g of carbon
= = 3.011 Ã— 10^{20 }Mass of 1 neutron = 1.675 Ã— 10^{27} kg
^{ }Mass of 3.011 Ã— 10^{20} neutrons
^{ }= 1.675 Ã— 10^{27} Ã— 3.011 Ã— 10^{20} kg
^{ }= 5.04 Ã— 10^{7} kg
(iii) 17 g of NH_{3} has 6.022 Ã— 10^{23} protons
^{ }No of protons in 34 Ã— 10^{3} g of NH_{3} = protons
^{ }= 12.044 Ã— 10^{20 }= 1.2044 Ã— 10^{21 }Mass of 1 proton = 1.67 Ã— 10^{27} kg
Mass of 1.2044 Ã— 10^{21 }= 1.2044 Ã— 10^{21} Ã—1.67 Ã— 10^{27 }= 2.015 Ã— 10^{6}
No. Answer will not change with temperature and pressure.
Question3
How many protons and neutrons are there in the following nuclei:
Solution:
C : Protons = 6, Neutrons = 7
O : P = 8, N = 8
Mg : P = 12, N = 12.
Solution:
C : Protons = 6, Neutrons = 7
O : P = 8, N = 8
Mg : P = 12, N = 12.
Question4
Yellow light emitted from a Sodium lamp has wavelength (Î» ) of 580 nm. Calculate frequency (v) and wave number () of yellow light.
Solution:
Î» = 580 mm
= 580 Ã— 10^{9} m
c = Î» v
where C = vel of light (3.0 Ã— 10^{8}ms^{1})
Frequency (Î½ ) =
= 5.17 Ã— 10^{14} S^{1}
(ii) Wave number() = = = 1.72 Ã— 10^{6} m^{1}.
Solution:
Î» = 580 mm
= 580 Ã— 10^{9} m
c = Î» v
where C = vel of light (3.0 Ã— 10^{8}ms^{1})
Frequency (Î½ ) =
= 5.17 Ã— 10^{14} S^{1}
(ii) Wave number() = = = 1.72 Ã— 10^{6} m^{1}.
Question5
Find energy of each of photons which
(i) Correspond to light of frequency 3 Ã— 10^{15} Hz
(ii) Have wavelength of 0.50 Ã…
Solution:
(i) E = hv = 6.6 Ã— 10^{34}Js Ã— 3 Ã— 10^{8}
= 1.98 Ã— 10^{18} J
(ii) E = h=
(i) Correspond to light of frequency 3 Ã— 10^{15} Hz
(ii) Have wavelength of 0.50 Ã…
Solution:
(i) E = hv = 6.6 Ã— 10^{34}Js Ã— 3 Ã— 10^{8}
= 1.98 Ã— 10^{18} J
(ii) E = h=
= 3.98 Ã— 10^{15} J.
Question6
Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 Ã— 10^{10} S.
Solution:
T = 2.0 Ã— 10^{10} s
Frequency, v = = = 5 Ã— 10^{9 }S^{1}
Velocity of light, C = vÎ» 3 Ã— 10^{8} = 5 Ã— 10^{9} Ã— Î»
Î» = = 6.0 Ã— 10^{2} m
wave number, = =
= 16.66 m^{1 }
Solution:
T = 2.0 Ã— 10^{10} s
Frequency, v = = = 5 Ã— 10^{9 }S^{1}
Velocity of light, C = vÎ» 3 Ã— 10^{8} = 5 Ã— 10^{9} Ã— Î»
Î» = = 6.0 Ã— 10^{2} m
wave number, = =
= 16.66 m^{1 }
Question7
What is the number of photons of light with a wavelength of 4000 pm that provide 1 Joule of energy?
Solution:
Wave length of proton = 4000 pm = 4000 Ã— 10^{12} m
E_{photon} = hv = where h = plankâ€™s constant
c = Velocity of light
V = frequency
= wave length
=
= 4.9687 Ã— 10^{17}J
Number of photons that provide 1 Joule of energy
=
= 2.012 Ã— 10^{16} photon.
Solution:
Wave length of proton = 4000 pm = 4000 Ã— 10^{12} m
E_{photon} = hv = where h = plankâ€™s constant
c = Velocity of light
V = frequency
= wave length
=
= 4.9687 Ã— 10^{17}J
Number of photons that provide 1 Joule of energy
=
= 2.012 Ã— 10^{16} photon.
Question8
A photon of wavelength 4 Ã— 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 (1 eV = 1.6020 Ã— 10^{19} J)
Solution:
(i) Energy of striking photon (E) =
=
= 4.97 Ã— 10^{19} j = 3.10eV.
(ii) work function (w_{o}) : 2.13 eV
K.E_{ }of the emitted electron = (Energy of striking photon(hv) = work function(w_{o})
= 3.10eV â€“ 2.13 eV
= 0.97 eV.
Solution:
(i) Energy of striking photon (E) =
=
= 4.97 Ã— 10^{19} j = 3.10eV.
(ii) work function (w_{o}) : 2.13 eV
K.E_{ }of the emitted electron = (Energy of striking photon(hv) = work function(w_{o})
= 3.10eV â€“ 2.13 eV
= 0.97 eV.
Question9
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionize the sodium atom.Calculate the ionisatin energy of sodium in KJ mol^{1}.
Solution:
Wavelength, Î» = 242 nm = 242 Ã— 10^{9} m
Energy of photon = hc/Î» =
= 0.0821 Ã— 10^{17} J atom^{1}
This energy is sufficient to cause ionization of one atom of Na
Therefore, ionization energy in KJ for 1 mole
=
= 494 KJ mol^{1}
Solution:
Wavelength, Î» = 242 nm = 242 Ã— 10^{9} m
Energy of photon = hc/Î» =
= 0.0821 Ã— 10^{17} J atom^{1}
This energy is sufficient to cause ionization of one atom of Na
Therefore, ionization energy in KJ for 1 mole
=
= 494 KJ mol^{1}
Question10
A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57Î¼ m. Calculate the rate of emission of quanta per second.
Solution:
100 watt = 100 J/sec
25 watt = 25 J/sec
Energy of one photon (E) = hv
= h
=
25 J of energy emitted in 1sec
Rate of emission of quanta per second
=
= 7.4 Ã— 10^{19} s^{1 }
Solution:
100 watt = 100 J/sec
25 watt = 25 J/sec
Energy of one photon (E) = hv
= h
=
25 J of energy emitted in 1sec
Rate of emission of quanta per second
=
= 7.4 Ã— 10^{19} s^{1 }
Question11
Electrons are emitted with zero velocity form a metal surface when it is exposed to radiation of wave length 6800 Ã…. Calculate threshold frequency(v_{0}) work function (W_{0}) of the metal.
Solution:
Î» = 6800Ã… = 6800 Ã— 10^{10} m
Energy of the incident light E = = joules
= = 1.83eV
E = hv_{o}
V_{0} = = = 4.41 Ã— 10^{14} s^{1}
W_{0} = hv_{0} = 6.63 Ã— 10^{34} Ã— 4.41 Ã— 10^{14}
= 2.91 Ã— 10^{19}J.
Solution:
Î» = 6800Ã… = 6800 Ã— 10^{10} m
Energy of the incident light E = = joules
= = 1.83eV
E = hv_{o}
V_{0} = = = 4.41 Ã— 10^{14} s^{1}
W_{0} = hv_{0} = 6.63 Ã— 10^{34} Ã— 4.41 Ã— 10^{14}
= 2.91 Ã— 10^{19}J.
Question12
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes a transition from an energy level with n = 4 to an energy level with n = 2? What is the colour corresponding to this wavelength ?
Solution:
=(109677x 3)/16
or
= 486nm ( visible light  blue colour).
Solution:
=(109677x 3)/16
or
= 486nm ( visible light  blue colour).
Question13
How much energy is required to ionize a H atom if electron occupies n = 5 orbit ? Compare your answer with the ionization energy of H atom(energy required to remove the electron from n = 1 orbit).
Solution:
E_{n} =
n=5
E_{n} = = J
=
The value of ionization energy of hydrogen atom is 2.18 10â€”18 J (n=1) which is greater than that in the 5^{th} orbit
Solution:
E_{n} =
n=5
E_{n} = = J
=
The value of ionization energy of hydrogen atom is 2.18 10â€”18 J (n=1) which is greater than that in the 5^{th} orbit
Question14
What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state ?
Solution:
Number of maximum number of emission lines
=
= n = 6 âˆ´ = 15 lines
Solution:
Number of maximum number of emission lines
=
= n = 6 âˆ´ = 15 lines
Question15
(i) The energy associated with the first orbit in the hydrogen atom is â€“2.17 Ã— 10^{18} J atom^{1}. What is the energy associated with 5^{th} orbit
(ii) Calculate the radius of 5^{th} orbit for Hydrogen atom
Solution:
(i) E_{n} = J atm^{1}
I.E = Eâˆž  E_{5} = 0 
= = 8.7 Ã— 10^{20}J
(ii) r_{n} = n^{2}r_{1}
radius of Bohrâ€™s 1^{st} orbit = 0.053nm
r_{5} = (5)^{2} Ã— 0.053 nm
= 25 Ã— 0.053 nm
= 1.325 nm.
(ii) Calculate the radius of 5^{th} orbit for Hydrogen atom
Solution:
(i) E_{n} = J atm^{1}
I.E = Eâˆž  E_{5} = 0 
= = 8.7 Ã— 10^{20}J
(ii) r_{n} = n^{2}r_{1}
radius of Bohrâ€™s 1^{st} orbit = 0.053nm
r_{5} = (5)^{2} Ã— 0.053 nm
= 25 Ã— 0.053 nm
= 1.325 nm.
Question16
Calculate the wave number for the shortest wavelength transition in the Balmer series of atomic hydrogen.
Solution:
According to Rydbergâ€™s formula
=
For the shortest wave length (Î» ) its wave number() is maximum. For this, n_{2} must be âˆž. = R_{H} where RH is Rydbergâ€™s coast which is equal to 109678.
= 109678 cm^{1}
= 109678 Ã— = 27419.5 cm^{1}
Solution:
According to Rydbergâ€™s formula
=
For the shortest wave length (Î» ) its wave number() is maximum. For this, n_{2} must be âˆž. = R_{H} where RH is Rydbergâ€™s coast which is equal to 109678.
= 109678 cm^{1}
= 109678 Ã— = 27419.5 cm^{1}
Question17
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 â€“2.18 Ã— 10^{11} ergs.
Solution:
The ground state electron energy = 2.18 Ã— 10^{11} erg
E_{n} =
n_{2} = 5
=
Energy, requiredâ€™s to shift for 1^{st} orbit to 5^{th} orbit ] = 8.72 Ã— 10^{20}J
Solution:
The ground state electron energy = 2.18 Ã— 10^{11} erg
E_{n} =
n_{2} = 5
=
Energy, requiredâ€™s to shift for 1^{st} orbit to 5^{th} orbit ] = 8.72 Ã— 10^{20}J
(iii) Î” E =
=
=
= 2.05 Ã— 10^{11} ergs
Frequency (v) = ; h = 6.62 Ã— 10^{27} erg
=
= 3.14 Ã— 10^{15} S^{1} Î» = = = 956 Ã…
Question18
The electron energy in hydrogen atom is given by E = (21.7 Ã— 10^{19})/n^{2} joules. Calculate the energy required to remove an electron from the n = 2 orbits. What is the longest wavelength in cm of light can be used to cause this transition ?
Solution:
The energy in the 2^{nd} Bohr orbit
E_{2} = J = 5.44 Ã— 10^{19} J
For the removal of electron is has to be moved to an infinite distance for which n = âˆž and
Eâˆž = = 0âˆ´ Î” E = Eâˆž  E_{2 }= 0(5.44) Ã— 10^{19}J)
_{ }= Energy needed to remove the electron from 2^{nd} Bohr orbit of H atom
_{ }= 5.44 Ã— 10^{19} J
Energy of photon, E = hÎ½ = hc/Î» = 5.44 Ã— 10^{_19} J
And_{ } Î» = =
_{ }= 3.639 Ã— 10^{7} m.
1 Ã— 10^{10} m = 1 Ã…
3.64 Ã— 10^{7} = = 3639 Ã…
Solution:
The energy in the 2^{nd} Bohr orbit
E_{2} = J = 5.44 Ã— 10^{19} J
For the removal of electron is has to be moved to an infinite distance for which n = âˆž and
Eâˆž = = 0âˆ´ Î” E = Eâˆž  E_{2 }= 0(5.44) Ã— 10^{19}J)
_{ }= Energy needed to remove the electron from 2^{nd} Bohr orbit of H atom
_{ }= 5.44 Ã— 10^{19} J
Energy of photon, E = hÎ½ = hc/Î» = 5.44 Ã— 10^{_19} J
And_{ } Î» = =
_{ }= 3.639 Ã— 10^{7} m.
1 Ã— 10^{10} m = 1 Ã…
3.64 Ã— 10^{7} = = 3639 Ã…
Question19
Calculate the wavelength of an electron moving with a velocity of 2.05 Ã— 10^{7} ms^{1}.
Solution:
According to de Broglieâ€™s equation, l =
mass (m) of electron = 9.1 Ã— 10^{31} kg
=
= 2.91 Ã— 10^{11} m.
Solution:
According to de Broglieâ€™s equation, l =
mass (m) of electron = 9.1 Ã— 10^{31} kg
=
= 2.91 Ã— 10^{11} m.
Question20
The mass of an electron is 9.1 Ã— 10^{31} kg. If its K.E is 3.0 Ã— 10^{25} J, calculate its wavelength.
Solution:
Kinetic energy = = 3.0 Ã— 10^{25} J
V^{2} = 3.0 Ã— 10^{25} J Ã—
V =
V = 812ms^{1} Î» = Î» =
= 8967 Ã— 10^{10} m
= 8967 Ã…
Solution:
Kinetic energy = = 3.0 Ã— 10^{25} J
V^{2} = 3.0 Ã— 10^{25} J Ã—
V =
V = 812ms^{1} Î» = Î» =
= 8967 Ã— 10^{10} m
= 8967 Ã…
Question21
Which of the following are isoelectronic species i.e. those having the same number of electrons? Na^{+}, K^{+}, Mg^{2+},Ca^{2+},S^{2},Ar
Solution:
(Na^{+}, Mg^{2+}) Na^{+} = 2, 8
Mg^{2+ }= 2, 8
Ca^{2+}, Ar Ca^{2+} = 2, 8,8
Ar = 2,8,8
S^{2}, K^{+} S^{2} = 2, 8,8
K^{+} = 2, 8,8
Solution:
(Na^{+}, Mg^{2+}) Na^{+} = 2, 8
Mg^{2+ }= 2, 8
Ca^{2+}, Ar Ca^{2+} = 2, 8,8
Ar = 2,8,8
S^{2}, K^{+} S^{2} = 2, 8,8
K^{+} = 2, 8,8
Question22
(i) Write the electronic configurations of the following ions :
a) H^{} b) Na^{+} c) O^{2} d) F^{}
ii) What are the atomic numbers of element whose outermost electrons are represented by a)2p^{3} b) 3p^{5? }
(iii) What atoms are indicated by the following configurations:
a) [He] 2s^{1} b) [Ne] 3s^{2} 3p^{3} c) [Ar] 4s^{2} 3d^{1}
Solution:
i) g a) H = 1s^{2}
b) Na^{+} = 1s^{2}2s^{2}2p^{6}
c) O^{2} = 1s^{2}2s^{2}2p^{6}
d) F^{} = 1s^{2}2s^{2}2p^{6}
ii)The atomic numbers are a) 2p^{3} is 7 and b) is 17.
iii) a) [He] 2s^{1} = Li
b) [Ne] 3s^{2} 3p^{3} = P
c) [Ar] 4s^{2} 3d^{1} = Sc
a) H^{} b) Na^{+} c) O^{2} d) F^{}
ii) What are the atomic numbers of element whose outermost electrons are represented by a)2p^{3} b) 3p^{5? }
(iii) What atoms are indicated by the following configurations:
a) [He] 2s^{1} b) [Ne] 3s^{2} 3p^{3} c) [Ar] 4s^{2} 3d^{1}
Solution:
i) g a) H = 1s^{2}
b) Na^{+} = 1s^{2}2s^{2}2p^{6}
c) O^{2} = 1s^{2}2s^{2}2p^{6}
d) F^{} = 1s^{2}2s^{2}2p^{6}
ii)The atomic numbers are a) 2p^{3} is 7 and b) is 17.
iii) a) [He] 2s^{1} = Li
b) [Ne] 3s^{2} 3p^{3} = P
c) [Ar] 4s^{2} 3d^{1} = Sc
Question23
What is the lowest value of n that allows g orbitals to exist?
Solution:
n = 5 l = (n1) = 4
If l = 4 it is g orbitals
Solution:
n = 5 l = (n1) = 4
If l = 4 it is g orbitals
Question24
An electron is in one of the 3d orbitals. Give the possible values of n, l and m for this electron.
Solution:
For 3d orbitals,
n = 3
l = (n1) = (31) =2
m = (2l+1) values
m = 2,1,0,+1,+2(any one value)
Solution:
For 3d orbitals,
n = 3
l = (n1) = (31) =2
m = (2l+1) values
m = 2,1,0,+1,+2(any one value)
Question25
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and ii) electronic configuration of the element.
Solution:
i) Number of protons = 29
ii) Electronic configuration
1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10 }
Solution:
i) Number of protons = 29
ii) Electronic configuration
1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}4s^{1}3d^{10 }
Question26
Give the number of electrons in H_{2}^{+}, H_{2}
Solution:
H_{2}^{+} = 1
H_{2} = 2
Solution:
H_{2}^{+} = 1
H_{2} = 2
Question27
(i) An atomic orbital has n = 3 what are possible values of l and m?
(ii) List the quantum numbers of electrons for 3d orbital.
(iii) Which of the following orbitals are possible 
lp, 2s, 2p and 3f?
Solution:
(ii) List the quantum numbers of electrons for 3d orbital.
(iii) Which of the following orbitals are possible 
lp, 2s, 2p and 3f?
Solution:
(i) n 
L[0 to (n1)] 
m(l to +l) through Zero 



3 
33 = 0 (3s) 
m = 0 
(ii) For 3d 
l = 2(3d) 
m = 2; 1;0; +1; +2 
Question28
Using the s, p, d notation, 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
Solution:
(a) 1s
(b) 3p
(c) 4d
(d) 4f.
(a) n = 1, l = 0
(b) n = 3, l = 1
(c) n = 4, l = 2
(d) n = 4, l = 3
Solution:
(a) 1s
(b) 3p
(c) 4d
(d) 4f.
Question29
Explain giving reasons, which following sets of quantum number are not possible.
a) n = 0, l = 0, m = 0, s = +1/2
b) n = 1, l = 0, m = 0, s = 1/2
c) n = 1, l = 1, m = 0, s = +1/2
d) n = 2, l = 1, m = 0, s = 1/2
e) n = 3, l = 3, m = 3, s = +1/2
f) n = 3, l = 1, m = 0, s = +1/2
Solution:
a) It is not possible, because principal quantum number â€˜nâ€™ cannot be zero. It should be positive integer
c) It is not possible because for given value of n, l can be (n1) i.e. if n = 1, l = 0
e) If n =3, l can be(3,3) = 0; (32) = 1 and (31) = 2. l cannot be 3.
a) n = 0, l = 0, m = 0, s = +1/2
b) n = 1, l = 0, m = 0, s = 1/2
c) n = 1, l = 1, m = 0, s = +1/2
d) n = 2, l = 1, m = 0, s = 1/2
e) n = 3, l = 3, m = 3, s = +1/2
f) n = 3, l = 1, m = 0, s = +1/2
Solution:
a) It is not possible, because principal quantum number â€˜nâ€™ cannot be zero. It should be positive integer
c) It is not possible because for given value of n, l can be (n1) i.e. if n = 1, l = 0
e) If n =3, l can be(3,3) = 0; (32) = 1 and (31) = 2. l cannot be 3.
Question30
How many electrons in an atom may have the following quantum numbers
a) n = 4; m_{l} = 2 m_{S} = +
b) n = 3; l = 0
Solution:
a) 4d electrons have + spins. So 5 electrons possess this quantum number
b) n = 3 ; l = 0 it is 3s which has 2 electrons
a) n = 4; m_{l} = 2 m_{S} = +
b) n = 3; l = 0
Solution:
a) 4d electrons have + spins. So 5 electrons possess this quantum number
b) n = 3 ; l = 0 it is 3s which has 2 electrons
Question31
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.
Solution:
According to Bohr an electron is allowed to move in atomic orbits whose angular momentum (mvn) is an integral multiple of h/2Ï€ mvr = h/2Ï€ or 2Ï€ r = n Ã— h/mv â€¦â€¦â€¦.. (i)
According to de Broglie Î» = h/mv â€¦â€¦â€¦â€¦ (ii)
Comparing (i) and (ii)
2Ï€ r = nÎ» â€¦â€¦â€¦â€¦. (iii)
Equation (iii) tells that integral number of n of wave lengths can be accommodated is Bohrâ€™s orbit, having circumference, 2Ï€ r. It means that there will be whole number of waves in an orbit, i.e; an electron wave is in phase.
Solution:
According to Bohr an electron is allowed to move in atomic orbits whose angular momentum (mvn) is an integral multiple of h/2Ï€ mvr = h/2Ï€ or 2Ï€ r = n Ã— h/mv â€¦â€¦â€¦.. (i)
According to de Broglie Î» = h/mv â€¦â€¦â€¦â€¦ (ii)
Comparing (i) and (ii)
2Ï€ r = nÎ» â€¦â€¦â€¦â€¦. (iii)
Equation (iii) tells that integral number of n of wave lengths can be accommodated is Bohrâ€™s orbit, having circumference, 2Ï€ r. It means that there will be whole number of waves in an orbit, i.e; an electron wave is in phase.
Question32
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He^{+} spectrum ?
Solution:
Atomic number, Z for H_{2} = 2
n_{2} = 4 ; n_{1} = 2
Whole number =
Where RH is Rydbergâ€™s Constant
= =
For hydrogen spectrum n_{1} and n_{2} to be calculated
= and Z = 1
or = = ( for He and Hydrogen are the same)
or =
This means that n_{1} = 1 and n_{2} = 2
Thus, the transition is form 2^{nd} level to 1^{st} level, in case of hydrogen spectrum.
Solution:
Atomic number, Z for H_{2} = 2
n_{2} = 4 ; n_{1} = 2
Whole number =
Where RH is Rydbergâ€™s Constant
= =
For hydrogen spectrum n_{1} and n_{2} to be calculated
= and Z = 1
or = = ( for He and Hydrogen are the same)
or =
This means that n_{1} = 1 and n_{2} = 2
Thus, the transition is form 2^{nd} level to 1^{st} level, in case of hydrogen spectrum.
Question33
alculate the energy required for the process He^{+}_{(g) }He^{2+}_{(g)} + e The ionization energy for the Hatom in the ground state is 2.18 Ã— 10^{18}J atom^{1}.
Solution:
For He^{+}; Li^{2+} etc which resemble H^{+}, contain are electron and therefore known as Hydrogen like system.
E_{n} = atom^{1}
Hence n = 1 ; Z = 2
E_{n} for He^{+}_{(g)} =
= 8.72 Ã— 10^{18}J
Solution:
For He^{+}; Li^{2+} etc which resemble H^{+}, contain are electron and therefore known as Hydrogen like system.
E_{n} = atom^{1}
Hence n = 1 ; Z = 2
E_{n} for He^{+}_{(g)} =
= 8.72 Ã— 10^{18}J
Question34
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 length20 cm long.
Solution:
Diameter of a carbon atom =0.15 nm
=0.15 10 ^{â€“9} m
=0.15 10 ^{â€“9}10 ^{3}
=0.15 10 ^{â€“6}cm
0.15 10 ^{â€“6}cm is covered by one carbon atom
20 cm length will be covered by
=1.33333 10 ^{8} atoms of carbon
Solution:
Diameter of a carbon atom =0.15 nm
=0.15 10 ^{â€“9} m
=0.15 10 ^{â€“9}10 ^{3}
=0.15 10 ^{â€“6}cm
0.15 10 ^{â€“6}cm is covered by one carbon atom
20 cm length will be covered by
=1.33333 10 ^{8} atoms of carbon
Question35
2 Ã—10^{8} atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
Solution:
2 Ã—10^{8} atoms of carbon occupy 2.4 cm
1 atom will occupy
= 1.2 10^{8} cm
Diameter of a carbon atom =1.2 10^{8} cm
Radius of a carbon atom = 1.2 10^{8} cm /2
=0.6 10^{8}cm
=0.6 10^{8}^{ }10^{3 }cm
=0.006 10^{9 }m
Radius of a carbon atom =0.006 nm
Solution:
2 Ã—10^{8} atoms of carbon occupy 2.4 cm
1 atom will occupy
= 1.2 10^{8} cm
Diameter of a carbon atom =1.2 10^{8} cm
Radius of a carbon atom = 1.2 10^{8} cm /2
=0.6 10^{8}cm
=0.6 10^{8}^{ }10^{3 }cm
=0.006 10^{9 }m
Radius of a carbon atom =0.006 nm
Question36
The diameter of zinc atom is 2.6 Ã…. 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.
Solution:
a)Radius of zinc atom in pm = A^{o}
^{=}1.3 10^{10} m
= 1.3 10^{10 }10^{2}10^{2} m
= 130 10^{12} m
=130 pm
b) Diameter of Zn atom = 2.6A^{0}
= 2.6 10^{10} m
=2.6 10^{10}10^{3} cm
= 2.6 10^{7 cm}
2.6 10^{7 cm} length is occupied by one Zn atom
1.6 cm length will be occupied by
=6.154 10^{6} atoms
Solution:
a)Radius of zinc atom in pm = A^{o}
^{=}1.3 10^{10} m
= 1.3 10^{10 }10^{2}10^{2} m
= 130 10^{12} m
=130 pm
b) Diameter of Zn atom = 2.6A^{0}
= 2.6 10^{10} m
=2.6 10^{10}10^{3} cm
= 2.6 10^{7 cm}
2.6 10^{7 cm} length is occupied by one Zn atom
1.6 cm length will be occupied by
=6.154 10^{6} atoms
Question37
A certain particle carries 2.5 Ã— 10^{â€“16}C of static electric charge. Calculate the number of electrons present in it.
Solution:
1.602 10^{19} charge is carried by one electron.
2.5 10^{19} charge will be carried by =
=1.602
= 1.605 electrons.
Solution:
1.602 10^{19} charge is carried by one electron.
2.5 10^{19} charge will be carried by =
=1.602
= 1.605 electrons.
Question38
In Milikanâ€™s experiment, static electric charge on the oil drops has been obtained by shining Xrays. If the static electric charge on the oil drop is â€“1.282 Ã— 10^{â€“18}C,calculate the number of electrons present on it.
Solution:
The static electric charge on the coil droplet =  1.282 Ã— 10^{18} C
Charge on the electron = 1.602 Ã— 10^{19} C
Number of electrons on the oil droplet =
Solution:
The static electric charge on the coil droplet =  1.282 Ã— 10^{18} C
Charge on the electron = 1.602 Ã— 10^{19} C
Number of electrons on the oil droplet =
Question39
In Rutherfordâ€™s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the aparticles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?
Solution:
Aluminium is less malleable and ductile than gold. The minimum thickness of sheets of Aluminium is 0.01mm.Hence aluminium sheets will block the path of particles as they low penetrating power. particles also have high ionising power and hence they many ionise aluminium atoms whereas gold have greater ionisation energy it shows contain extent of inertness to the particles.
Solution:
Aluminium is less malleable and ductile than gold. The minimum thickness of sheets of Aluminium is 0.01mm.Hence aluminium sheets will block the path of particles as they low penetrating power. particles also have high ionising power and hence they many ionise aluminium atoms whereas gold have greater ionisation energy it shows contain extent of inertness to the particles.
Question40
Symbols and ^{79}Br can be written, whereas symbols and ^{35}Br are not acceptable. Answer briefly.
Solution:
and ^{79}Br are acceptable way of representation whereas and ^{35}Br are not acceptable because there is a rule in IUPAC nomenclature to write the higher number(mass number) on top and the lower number (atomic number ) at the bottom. This type of expressing the elements is very useful in writing radioactive disintegration reactions where neutrons are emitted.
Question41
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Solution:
Number of protons + Number of neutrons = mass number = 81
Let number of protons in the atom be = x
âˆ´ Number of neutrons in the atom = = 1.317 x
x + 1.317x = 81
2.317x = 81
x = 34.96 = 35 (nearest whole number)
âˆ´ Number of protons = 35
Atomic number of the element = 35
The element with atomic number 35 is Br.
âˆ´ Atomic symbol is _{35}^{81}Br.
Solution:
Number of protons + Number of neutrons = mass number = 81
Let number of protons in the atom be = x
âˆ´ Number of neutrons in the atom = = 1.317 x
x + 1.317x = 81
2.317x = 81
x = 34.96 = 35 (nearest whole number)
âˆ´ Number of protons = 35
Atomic number of the element = 35
The element with atomic number 35 is Br.
âˆ´ Atomic symbol is _{35}^{81}Br.
Question42
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) Xrays.
Solution:
Radiation from FM radio < radiation from microwave oven < amber light from traffic signal < xrays < cosmic rays from outer space.
Solution:
Radiation from FM radio < radiation from microwave oven < amber light from traffic signal < xrays < cosmic rays from outer space.
Question43
Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 Ã— 1024, calculate the power of this laser.
Solution:
E = NhÏ…
= Nh
=
=
= 0.3302 Ã— 10^{7}
E = 3.302 Ã— 10^{6} kg m^{2} s^{2}.
Solution:
E = NhÏ…
= Nh
=
=
= 0.3302 Ã— 10^{7}
E = 3.302 Ã— 10^{6} kg m^{2} s^{2}.
Question44
Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance travelled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.
Solution:
(a) Î» = 616 nm = 616 Ã— 10^{9} m
c = 3 Ã— 10^{8} ms^{1}
v = = 4.87 Ã— 10^{14} s^{1}.
(b) Distance travelled by the radiation in 30 s = c Ã— t
= 3 Ã— 10^{8} ms^{1} Ã— 30 s
= 9 Ã— 10^{9} m.
(c) Energy of quantum, E = hv
= 6.626 Ã— 10^{34} J s Ã— 4.87 Ã— 10^{14} s^{1}.
= 3.23 Ã— 10^{19} J
(d) Number of quanta =
= = 6.19 Ã— 10^{8}.
Solution:
(a) Î» = 616 nm = 616 Ã— 10^{9} m
c = 3 Ã— 10^{8} ms^{1}
v = = 4.87 Ã— 10^{14} s^{1}.
(b) Distance travelled by the radiation in 30 s = c Ã— t
= 3 Ã— 10^{8} ms^{1} Ã— 30 s
= 9 Ã— 10^{9} m.
(c) Energy of quantum, E = hv
= 6.626 Ã— 10^{34} J s Ã— 4.87 Ã— 10^{14} s^{1}.
= 3.23 Ã— 10^{19} J
(d) Number of quanta =
= = 6.19 Ã— 10^{8}.
Question45
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 Ã— 10^{â€“18} J from the radiations of 600 nm, calculate the number of photons received by the detector.
Solution:
Wavelength Î» = 600 nm = 600 Ã— 10^{9} m
Energy per photon = = 3.31 Ã— 10^{19} J.
Number of photons =
Solution:
Wavelength Î» = 600 nm = 600 Ã— 10^{9} m
Energy per photon = = 3.31 Ã— 10^{19} J.
Number of photons =
= = 9.52 â‰ˆ 10 photons.
Question46
The longest wavelength doublet absorption transition is observed at 589 and 589.6nm. Calculate the frequency of each transition and energy difference between two excited states.
Solution:
Î» _{1} = 589 nm Ï… _{1} = ? E_{1} = ?Î» _{2} = 589.6 nm Ï… _{2} = ? E_{2} = ?
E_{1} = hÏ… _{1}
Ï… _{1} = = 0.005093 Ã— 10^{8} Ã— 10^{9} = 5.093 Ã— 10^{14} s^{1}.
E_{1} = 6.626 Ã— 10^{34} Ã— 5 Ã— 10^{14} = 33.13 Ã— 10^{20} = 3.313 Ã— 10^{19} kg m^{2} s^{2}.
Ï… _{2} = = 0.00509 Ã— 10^{8} Ã— 10^{9}
E_{2} = 6.626 Ã— 10^{34}
Solution:
Î» _{1} = 589 nm Ï… _{1} = ? E_{1} = ?Î» _{2} = 589.6 nm Ï… _{2} = ? E_{2} = ?
E_{1} = hÏ… _{1}
Ï… _{1} = = 0.005093 Ã— 10^{8} Ã— 10^{9} = 5.093 Ã— 10^{14} s^{1}.
E_{1} = 6.626 Ã— 10^{34} Ã— 5 Ã— 10^{14} = 33.13 Ã— 10^{20} = 3.313 Ã— 10^{19} kg m^{2} s^{2}.
Ï… _{2} = = 0.00509 Ã— 10^{8} Ã— 10^{9}
E_{2} = 6.626 Ã— 10^{34}
Question47
The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Solution:
Work function, W_{0} = 1.9 eV
= 1.9 Ã— 1.6 Ã— 10^{19} J = 3.04 Ã— 10^{19} J
(a) W_{0} =
Î» _{0} =
Î» _{0} = 6.54 Ã— 10^{7} m = 654 nnm.
Thus, threshold wavelength of caesium is 654 nm.
(b) Threshold frequency, v_{0} = = 4.59 Ã— 10^{14} s^{1}.
(c) Energy of photon of radiation of wavelength 500 nm.
= = 3.98 Ã— 10^{19} J.
K. E. = hv â€“ hv_{0} = 3.98 Ã— 10^{19} J â€“ 3.04 Ã— 10^{19} J = 9.4 Ã— 10^{20} J.
mv^{2} = 9.4 Ã— 10^{20} J.
Solution:
Work function, W_{0} = 1.9 eV
= 1.9 Ã— 1.6 Ã— 10^{19} J = 3.04 Ã— 10^{19} J
(a) W_{0} =
Î» _{0} =
Î» _{0} = 6.54 Ã— 10^{7} m = 654 nnm.
Thus, threshold wavelength of caesium is 654 nm.
(b) Threshold frequency, v_{0} = = 4.59 Ã— 10^{14} s^{1}.
(c) Energy of photon of radiation of wavelength 500 nm.
= = 3.98 Ã— 10^{19} J.
K. E. = hv â€“ hv_{0} = 3.98 Ã— 10^{19} J â€“ 3.04 Ã— 10^{19} J = 9.4 Ã— 10^{20} J.
mv^{2} = 9.4 Ã— 10^{20} J.
v^{2} = = 2.06 Ã— 10^{11}
v = 4.54 Ã— 10^{5} ms^{1}.
Question48
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.
Solution:
Since the photoelectron can be stopped by applying a voltage of 0.35 eV therefore, K. E. of the photoelectron is equal to 0.35 eV.
K. E. = 0.35 eV = 0.35 Ã— 1.6 Ã— 10^{19} J
= 0.56 Ã— 10^{19} J.
hv = = 7.74 Ã— 10^{19} J
hv = W_{0} + K. E.
W_{0} = hv â€“ K. E.
= 7.74 Ã— 10^{19} J â€“ 0.56 Ã— 10^{19} J = 7.18 Ã— 10^{18} J.
Thus, work function of silver is 7.18 Ã— 10^{18} J.
Solution:
Since the photoelectron can be stopped by applying a voltage of 0.35 eV therefore, K. E. of the photoelectron is equal to 0.35 eV.
K. E. = 0.35 eV = 0.35 Ã— 1.6 Ã— 10^{19} J
= 0.56 Ã— 10^{19} J.
hv = = 7.74 Ã— 10^{19} J
hv = W_{0} + K. E.
W_{0} = hv â€“ K. E.
= 7.74 Ã— 10^{19} J â€“ 0.56 Ã— 10^{19} J = 7.18 Ã— 10^{18} J.
Thus, work function of silver is 7.18 Ã— 10^{18} J.
Question49
If the photon of the wavelength 150 pm strikes an atom and one of this inner bound electrons is ejected out with a velocity of 1.5 Ã— 10^{7}ms^{â€“1}, calculate the energy with which it is bound to the nucleus.
Solution:
Î» = 150 pm = 150 Ã— 10^{12} m
Energy of photon = = 1.33 Ã— 10^{15} J.
Kinetic energy of the ejected electron = mv^{2}
= Ã— 9.11 Ã— 10^{31} Ã— (1.5 Ã— 10^{7})^{2}
= 1.025 Ã— 10^{16} J
Binding energy = Energy of incident photon â€“ Kinetic energy
= 1.33 Ã— 10^{15} J â€“ 1.025 Ã— 10^{16} J = 1.23 Ã— 10^{15} J.
Solution:
Î» = 150 pm = 150 Ã— 10^{12} m
Energy of photon = = 1.33 Ã— 10^{15} J.
Kinetic energy of the ejected electron = mv^{2}
= Ã— 9.11 Ã— 10^{31} Ã— (1.5 Ã— 10^{7})^{2}
= 1.025 Ã— 10^{16} J
Binding energy = Energy of incident photon â€“ Kinetic energy
= 1.33 Ã— 10^{15} J â€“ 1.025 Ã— 10^{16} J = 1.23 Ã— 10^{15} J.
Question50
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 Ã— 10^{6 }ms^{â€“1}, calculate de Broglie wavelength associated with this electron.
Solution:
v = 1.6 Ã— 10^{6} ms^{1}Î» = ?
Î» = where m is the mass of the electron
=
= 0.4546 Ã— 10^{9} m
= 4.546 Ã— 10^{10} m = 4.546 AÂ°
âˆ´ Î» = 4.546 AÂ° .
Solution:
v = 1.6 Ã— 10^{6} ms^{1}Î» = ?
Î» = where m is the mass of the electron
=
= 0.4546 Ã— 10^{9} m
= 4.546 Ã— 10^{10} m = 4.546 AÂ°
âˆ´ Î» = 4.546 AÂ° .
Question51
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.
Solution:
Î» = 800 pm = 800 Ã— 10^{10} m
Mass of neutron, m = 1.675 Ã— 10^{27} kg
Î» =
Ï… = = 4.95 Ã— 10^{2} m s^{1}.
Solution:
Î» = 800 pm = 800 Ã— 10^{10} m
Mass of neutron, m = 1.675 Ã— 10^{27} kg
Î» =
Ï… = = 4.95 Ã— 10^{2} m s^{1}.
Question52
If the velocity of the electron in Bohrâ€™s first orbit is 2.19 Ã— 10^{6} ms^{â€“1}, calculate the de Broglie wavelength associated with it.
Solution:
Bohrâ€™s first orbit electron velocity = 2.19 Ã— 10^{6} ms^{1}
Î» = ?
Î» =
=
= Ã— 10^{34} Ã— 10^{25} m
= 0.7273 Ã— 10^{9} m
= 7.273 AÂ° .
Solution:
Bohrâ€™s first orbit electron velocity = 2.19 Ã— 10^{6} ms^{1}
Î» = ?
Î» =
=
= Ã— 10^{34} Ã— 10^{25} m
= 0.7273 Ã— 10^{9} m
= 7.273 AÂ° .
Question53
The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 Ã— 105 ms^{â€“1}. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Solution:
Potential difference = 1000 v
velocity = 4.37 Ã— 10^{5} ms ^{â€“1}
Hockey ball mass = 0.1 kg
v = 4.37 Ã— 10^{5} ms^{1}
Î» =
=
= Ã— 10^{34} Ã— 10^{4} m
= 1.516 Ã— 10^{38} m.
Solution:
Potential difference = 1000 v
velocity = 4.37 Ã— 10^{5} ms ^{â€“1}
Hockey ball mass = 0.1 kg
v = 4.37 Ã— 10^{5} ms^{1}
Î» =
=
= Ã— 10^{34} Ã— 10^{4} m
= 1.516 Ã— 10^{38} m.
Question54
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:
1. n = 4, l = 2, ml = 2 , ms = 1/2
2. n = 3, l = 2, ml = 1 , ms = +1/2
3. n = 4, l = 1, ml = 0 , ms = +1/2
4. n = 3, l = 2, ml = 2 , ms = 1/2
5. n = 3, l = 1, ml = 1 , ms = +1/2
6. n = 4, l = 1, ml = 0 , ms = +1/2
Solution:
5 < 2 = 4 < 3 = 6 < 1.
1. n = 4, l = 2, ml = 2 , ms = 1/2
2. n = 3, l = 2, ml = 1 , ms = +1/2
3. n = 4, l = 1, ml = 0 , ms = +1/2
4. n = 3, l = 2, ml = 2 , ms = 1/2
5. n = 3, l = 1, ml = 1 , ms = +1/2
6. n = 4, l = 1, ml = 0 , ms = +1/2
Solution:
5 < 2 = 4 < 3 = 6 < 1.
Question55
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 ?
Solution:
For the orbitals of the same type (same value of l), greater the value of principal quantum number (n), smaller is effective nuclear charge experienced. In the present case the electrons in the 4p orbital would experience lowest effective nuclear charge.
Solution:
For the orbitals of the same type (same value of l), greater the value of principal quantum number (n), smaller is effective nuclear charge experienced. In the present case the electrons in the 4p orbital would experience lowest effective nuclear charge.
Question56
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.
Solution:
i) 2s and 3s â€“ 2s
(ii) 4d and 4f â€“ 4d
(iii) 3d and 3p â€“ 3p
Solution:
i) 2s and 3s â€“ 2s
(ii) 4d and 4f â€“ 4d
(iii) 3d and 3p â€“ 3p
Question57
The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus ?
Solution:
The 3p orbital in Si will experience more effective nuclear charge due to greater positive charge on nucleus on it.
Solution:
The 3p orbital in Si will experience more effective nuclear charge due to greater positive charge on nucleus on it.
Question58
Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Solution:
(i) _{15}P: 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{1}_{x}, 3p^{1}_{y}, 3p^{1}_{z}.
There are 3 unpaired electrons in P atom.
(ii) _{14}Si: 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{1}_{x}, 3p^{1}_{y}.
There are 2 unpaired electrons in Si atom.
(iii) _{24}Cr: 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{1}, 3d^{5}.
There are 6 unpaired electrons in Cr atom.
Solution:
(i) _{15}P: 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{1}_{x}, 3p^{1}_{y}, 3p^{1}_{z}.
There are 3 unpaired electrons in P atom.
(ii) _{14}Si: 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{1}_{x}, 3p^{1}_{y}.
There are 2 unpaired electrons in Si atom.
(iii) _{24}Cr: 1s^{2}, 2s^{2}, 2p^{6}, 3s^{2}, 3p^{6}, 4s^{1}, 3d^{5}.
There are 6 unpaired electrons in Cr atom.
Question59
(a) How many subshells are associated with n = 4? (b) How many electrons will be present in the subshells having ms value of â€“1/2 for n = 4?
Solution:
a) If n = 4, l = 0, 1, 2, 3.
2n^{2} = 2 Ã— 4^{2} = 32 electrons.
b) 16 electrons will have ms value of â€“1/2 for n = 4.
Solution:
a) If n = 4, l = 0, 1, 2, 3.
2n^{2} = 2 Ã— 4^{2} = 32 electrons.
b) 16 electrons will have ms value of â€“1/2 for n = 4.
Question60
Write the complete symbol for the atom with the given atomic(Z) and atomic mass(A)
(i) Z = 17; A = 35
(ii) Z = 92 A = 233
(iii) Z = 4 A = 9
Solution:
i)_{ 17}Cl^{35 }ii)_{ 92}U^{235} iii)_{ 4}Be^{9}
(i) Z = 17; A = 35
(ii) Z = 92 A = 233
(iii) Z = 4 A = 9
Solution:
i)_{ 17}Cl^{35 }ii)_{ 92}U^{235} iii)_{ 4}Be^{9}