02 Structure of Atom
Exercises
Question:
Give the number of electrons in the species H2+, H2 and O2+.
Answer:
H^{2}_{+}: 2 electrons H_{2}: 4 electrons O^{2}_{+}: 6 electrons
Question:
Among the following orbitals which orbital will experience the larger effective nuclear charge?
Answer:
 2p
 3s
Answer:
 Compare the effective nuclear charge experienced by the 2p orbital and the 3s orbital.
 The 3s orbital is lower in energy than the 2p orbital, so it will experience a larger effective nuclear charge.
Question:
2s, 3s, 4d, 4f, 3d, 3p
Answer:

2s: 2s means two steps forward. Therefore, move two steps forward.

3s: 3s means three steps forward. Therefore, move three steps forward.

4d: 4d means four steps backward. Therefore, move four steps backward.

4f: 4f means four steps forward. Therefore, move four steps forward.

3d: 3d means three steps backward. Therefore, move three steps backward.

3p: 3p means three steps to the right. Therefore, move three steps to the right.
Question:
Neon gas is generally used in the signboards. If it emits strongly at 616 nm
Answer:

Neon gas is a type of noble gas that is composed of two atoms of neon.

Neon gas is used in signboards because it emits a bright, visible light when an electric current is passed through it.

The light emitted from neon gas is generally in the visible spectrum and is most strongly emitted at 616 nm (nanometers).
Question:
Calculate:
(a) The frequency of emission,
(b) Distance travelled by this radiation in 30 s
(c) The energy of quantum and
(d) The number of quanta present, if it produces 2 J of energy.
Answer:
(a) The frequency of emission can be calculated using the equation: frequency = velocity of light/wavelength
(b) The distance travelled by the radiation in 30 seconds can be calculated using the equation: distance = velocity of light x time
(c) The energy of a quantum can be calculated using the equation: Energy = Planck’s Constant x frequency
(d) The number of quanta present can be calculated using the equation: Number of quanta = Energy/Energy of a quantum
Question:
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.
Answer:
Step 1: The element has a mass number of 81.
Step 2: This means there are 81 total particles in the element, including protons and neutrons.
Step 3: 31.7% of 81 is 25.5. This means there are 25.5 more neutrons than protons.
Step 4: Therefore, the element has 56 protons and 81 neutrons.
Step 5: The atomic symbol for the element is Bismuth, which is Bi.
Question:
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15×10−18J from the radiations of 600 nm, calculate the number of photons received by the detector.
Answer:
 Convert the energy (3.15×10−18J) to the equivalent wavelength of the radiation:
E = hc/λ
3.15×10^{18}J = (6.626 x 10^{34}J.s)(3.00 x 10^{8}m/s)/λ
λ = 4.33 x 10^{7}m
 Calculate the number of photons received by the detector:
Number of photons = Energy/Energy per photon
Number of photons = 3.15×10−18J / (6.626 x 10^{34}J.s)(3.00 x 10^{8}m/s)/(4.33 x 10^{7}m)
Number of photons = 1.98 x 10^{12} photons
Question:
Calculate the energy required for the process He^{+}(g)→He^{2+}(g)+e^{}. The ionization energy for the H atom in the ground state is 2.18×10^{18}J atom^{1}.
Answer:

Calculate the ionization energy for the He atom in the ground state. This is 4.52 × 10^{18} J atom^{1}.

Calculate the energy required for the process He+(g)→He2+(g)+e−. This is equal to the ionization energy of the He atom, which is 4.52 × 10^{18} J atom^{1}.
Question:
During Rutherford’s experiment, generally, the thin foil of heavy atoms like gold was used and bombarded by the αparticles. If the thin foil of light atoms like aluminium etc. is used, what difference would have been observed?
Answer:

If Rutherford had used a thin foil of light atoms such as aluminium instead of heavy atoms such as gold, the αparticles would have had a much easier time penetrating the foil.

This would have resulted in a higher number of αparticles passing through the foil, as well as a higher number of particles being deflected.

The experiment would have produced different results, as the amount of deflection would have been greater compared to using a foil of heavy atoms.

This would have allowed Rutherford to better understand the structure of the atom and the nature of the αparticles.
Question:
The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.
Answer:
 Convert the wavelength of each transition to meters:
Wavelength 1 = 589 nm = 5.89 x 10^{7} m
Wavelength 2 = 589.6 nm = 5.896 x 10^{7} m
 Calculate the frequency of each transition using the equation:
Frequency = c/wavelength
Frequency 1 = 3 x 10^{8} m/s / 5.89 x 10^{7} m = 5.09 x 10^{14} Hz
Frequency 2 = 3 x 10^{8} m/s / 5.896 x 10^{7} m = 5.08 x 10^{14} Hz
 Calculate the energy difference between two excited states using the equation:
Energy difference = h(frequency 2  frequency 1)
Energy difference = 6.626 x 10^{14} Js x (5.08 x 10^14 Hz  5.09 x 10^{14} Hz)
Energy difference = 1.31 x 10^{19} J
Question:
Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0×10^{10}s.
Answer:
Wavelength (λ) = Speed of Light (c) × Period (T)
λ = 3.0 × 10^{8} m/s × 2.0 × 10^{10} s
λ = 6.0 × 10^{2} m
Frequency (f) = 1/T
f = 1/2.0 × 10^{10} s
f = 5.0 × 10^{9} Hz
Wave number (k) = 2π/λ
k = 2π/6.0 × 10^{2} m
k = 3.3 × 10^{7} m^{1}
Question:
The wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen is 1.523×10^{6}m^{(1)} Enter 1 if the statement is True or 0 if False.
Answer:
Answer: 1
Question:
How much energy is required to ionise H atom if the electron occupies n = 5 orbits? Compare your answer with the ionization enthalpy of H atom
[ energy required to remove the electron from n =1 orbit].
Answer:
Step 1: Determine the energy required to ionise H atom if the electron occupies n = 5 orbits. This can be calculated using the formula E = 13.6 eV/n2, where E is the energy required in electron volts (eV) and n is the principle quantum number.
Step 2: Substitute n = 5 into the formula to calculate the energy required to ionise H atom if the electron occupies n = 5 orbits.
E = 13.6 eV/52 E = 13.6 eV/25 E = 0.544 eV
Step 3: Compare the calculated energy with the ionization enthalpy of H atom, which is 13.6 eV.
The energy required to ionise H atom if the electron occupies n = 5 orbits is 0.544 eV, which is much lower than the ionization enthalpy of H atom (13.6 eV). This is because the energy required to remove the electron from n = 1 orbit is much greater than the energy required to remove the electron from higher orbits.
Question:
A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second.
Answer:
Step 1: Convert 25 watts of power to joules per second. 25 watts = 25 joules/second
Step 2: Calculate the energy of one quantum of light. Energy of one quantum of light = hc/λ where h = Planck’s constant (6.626 x 10^{34} Js) c = speed of light (3 x 10^{8} m/s) λ = wavelength (0.57 x 10^{6} m)
Energy of one quantum of light = (6.626 x 10^{34} Js)(3 x 10^{8} m/s)/(0.57 x 10^{6} m) = 3.9 x 10^{19} J
Step 3: Calculate the rate of emission of quanta per second. Rate of emission of quanta per second = Power/Energy of one quantum of light = 25 J/s/3.9 x 10^{19} J = 6.4 x 10^{19} quanta/s
Question:
Explain giving reasons which of the following sets of quantum numbers are not possible.
(a )n=0 , l=0 , m1=0 , ms=+1/2
(b) n=1 , l=0 , m1=0 , ms=−1/2
(c) n=1 , l=1 , m1=0 , ms=+1/2
(d) n=2 , l=1 , m1=0 , ms=−1/2
(e) n=2 , l=3 , m1=−3 , ms=+1/2
(f) n=3 , l=1 , m1=0 , ms=+1/2
Answer:
(a) Not possible: The spin quantum number (ms) cannot be greater than the orbital quantum number (l).
(b) Not possible: The spin quantum number (ms) cannot be negative when the orbital quantum number (l) is 0.
(c) Possible
(d) Possible
(e) Not possible: The magnetic quantum number (m1) cannot be greater than the orbital quantum number (l).
(f) Possible
Question:
What is the lowest value of n that allows g orbitals to exist?
Answer:
Step 1: Identify the definition of an ‘g orbital’.
Step 2: Research the properties of ‘g orbitals’ to determine the lowest value of ’n’ that allows them to exist.
Question:
An ion with the 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:

The ion has one unit of negative charge, so it must have one electron.

Since the ion has 11.1% more neutrons than electrons, the number of neutrons must be 1.111 times the number of electrons.

Therefore, the number of neutrons must be 1.111 x 1 = 1.111.

The mass number is the sum of the number of protons and neutrons. So the number of protons must be 37  1.111 = 35.889.

The symbol of the ion is 35Cl^{} (Cl is the symbol for chlorine).
Question:
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:
Step 1: Determine the number of protons. Protons = 3 units of positive charge = 3
Step 2: Determine the number of neutrons. Neutrons = (30.4% more than electrons) x (number of electrons) = (30.4% more than electrons) x (number of protons) = (30.4%) x (3) = 0.912 = 1 (rounded to the nearest whole number)
Step 3: Assign the symbol to this ion. Symbol = 56 3+ (56 protons and 3 units of positive charge)
Question:
(i) Calculate the total number of electrons present in one mole of methane.
Answer:
Answer: (i) One mole of methane contains 6.022 x 1023 molecules of methane. Each molecule of methane contains one carbon atom and four hydrogen atoms. Carbon atom has 6 electrons and hydrogen atom has 1 electron. Therefore, the total number of electrons present in one mole of methane is 6.022 x 1023 x (6 + 4 x 1) = 6.022 x 1023 x 10 = 6.022 x 1024 electrons.
Question:
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C.
Answer:
(i) Given: Mass of 14C = 7 mg
(ii) Solution: (a) Total number of neutrons in 7 mg of 14C = 6
(b) Total mass of neutrons in 7 mg of 14C = 6 x 1.67 x 10^{27} kg = 1.002 x 10^{26} kg
Question:
(Assume that mass of a neutron =1.675×10^{27}kg).
Answer:
 Calculate the mass of 1 mole of neutrons:
Mass of 1 mole of neutrons = 1.675 x 10^{27} kg x 6.02 x 10^{23} = 1.019 x 10^{3} kg
 Calculate the number of neutrons in 1 mole of neutrons:
Number of neutrons in 1 mole of neutrons = 6.02 x 10^{23}
 Calculate the number of atoms in 1 mole of neutrons:
Number of atoms in 1 mole of neutrons = 6.02 x 10^{23} x 2 = 1.204 x 10^{24} atoms
Question:
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.
Answer:
(a) The total number of protons in 34 mg of NH3 at STP is 3.6 x 10^{22} protons.
(b) The total mass of protons in 34 mg of NH3 at STP is 2.3 x 10^{22} grams.
Question:
Will the answer change if the temperature and pressure are changed?
Answer:

Determine what type of question is being asked. Answer: This is a yes/no question.

Analyze the question to determine the answer. Answer: Yes, the answer may change if the temperature and pressure are changed.
Question:
The mass of an electron is 9.1×10^(−31)kg . If its K.E. is 3.0×10^{(25)}J . Calculate its wavelength.
Answer:
Step 1: Convert the K.E. from joules to electron volts (eV). 1 eV = 1.6 × 10^(19) J 3.0 × 10^(25) J = 3.0 × 10^{(25)} / (1.6 × 10^{(19)}) eV K.E. = 1.875 × 10^(6) eV
Step 2: Calculate the velocity of the electron using the equation v = (2K.E./m)^{(1/2)}. 2K.E. = 3.75 × 10^{(6)} eV m = 9.1 × 10^{(31)} kg v = (3.75 × 10^{(6)} eV / 9.1 × 10^{(31)} kg)^{(1/2)}. v = 3.2 × 10^{(6)}. m/s
Step 3: Calculate the wavelength of the electron using the equation λ = h/mv, where h is Planck’s constant (6.626 × 10^{(34)} J s). λ = 6.626 × 10^{(34)} J s / (9.1 × 10^{(31)} kg × 3.2 × 10^{6} m/s) λ = 2.1 × 10^{(11)} m
Question:
Calculate the wavelength of an electron moving with a velocity of 2.05×10^{(7)}ms^{(1)}.
Answer:
Step 1: Understand the question. The question is asking to calculate the wavelength of an electron moving with a velocity of 2.05×10^{(7)}ms^{(1)}.
Step 2: Identify the variables. The given variables are: Velocity (v) = 2.05×10^{(7)}ms^{(1)} Wavelength (λ) = Unknown
Step 3: Apply the formula for wavelength. The formula for wavelength is: λ = h/mv Where h is Planck’s constant (6.626×10^{(34)}Js), m is the mass of the electron (9.109×10^{(31)}kg) and v is the velocity of the electron.
Step 4: Calculate the wavelength. Substituting the values in the formula, λ = (6.626×10^−34Js)/(9.109×10^{(31)}kg × 2.05×10^{7}ms^{(1)}) λ = 3.18×10^{(12)}m
Therefore, the wavelength of an electron moving with a velocity of 2.05×10^{(7)}ms^{(1)} is 3.18×10^{(12)}m.
Question:
What is the maximum number of emission lines when the excited electron of H atom in n = 6 drops to the ground state?
Answer:
Answer:
Step 1: The maximum number of emission lines when an electron in an H atom drops from n = 6 to the ground state (n = 1) is 5.
Step 2: This is because the transition of an electron from n = 6 to n = 1 involves the emission of 5 quanta of energy.
Question:
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:
 The electrons in the 2p orbital experience the lowest effective nuclear charge.
Question:
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×^{(7)}ms^{(1)}, calculate the energy with which it is bound to the nucleus.
Answer:
Step 1: Calculate the Planck’s constant. Planck’s constant = 6.626 x 10^{(34)} Js
Step 2: Calculate the frequency of the photon. Frequency = Speed of light/wavelength Frequency = (3 x 10^{8} m/s)/(150 x 10^{(34)} m) Frequency = 2 x 10^14 Hz
Step 3: Calculate the energy of the photon. Energy = Planck’s constant x frequency Energy = (6.626 x 10^{(34)} Js) x (2 x 10^{(14)} Hz) Energy = 1.325 x 10^{(19)} J
Step 4: Calculate the energy with which the electron is bound to the nucleus. Energy = Kinetic energy + Photon energy Kinetic energy = (1/2)mv^{2} m = Mass of electron = 9.11 x 10^{(31)} kg v = Velocity of electron = 1.5 x 10^{7} ms^{1} Kinetic energy = (1/2)(9.11 x 10^{31} kg)(1.5 x 10^{7} ms^{1})^{2} Kinetic energy = 1.13 x 10^{17} J
Energy with which the electron is bound to the nucleus = Kinetic energy + Photon energy Energy with which the electron is bound to the nucleus = (1.13 x 10^{17} J) + (1.325 x 10^{19} J) Energy with which the electron is bound to the nucleus = 2.45 x 10^{17} J
Question:
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies.
 n=4,l=2,m1=−2,ms=−1/2
 n=3,l=2,m1=1,ms=+1/2
 n=4,l=1,m1=0,ms=+1/2
 n=3,l=2,m1=−2,ms=−1/2
 n=3,l=1,m1=−1,ms=+1/2
 n=4,l=1,m1=0,ms=+1/2
Answer:
 n=3,l=1,m1=−1,ms=+1/2
 n=3,l=2,m1=−2,ms=−1/2
 n=3,l=2,m1=1,ms=+1/2
 n=4,l=1,m1=0,ms=+1/2
 n=4,l=1,m1=0,ms=+1/2
 n=4,l=2,m1=−2,ms=−1/2
Question:
Lifetimes of the molecules in the excited state 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 2.5×10^15, calculate the energy of the source.
Answer:

Convert the pulse duration of 2 ns into seconds: 2 ns = 2 x 10^{9} s

Calculate the number of photons per second: 2.5 x 10^15 photons / 2 x 10^{9} s = 1.25 x 10^{24} photons/s

Calculate the energy of the source: Energy = (1.25 x 10^{24} photons/s) x (6.626 x 10^{34} J/photon) Energy = 8.28 x 10^{10} J
Question:
2×10^8 atoms of carbon are arranged side by side. Calculate the radius of the carbon atom if the length of this arrangement is 2.4 cm.
Answer:
Step 1: Convert 2.4 cm to meters. 2.4 cm = 0.024 m
Step 2: Calculate the number of atoms in a meter. Number of atoms in a meter = 2 × 10^{8} ÷ 0.024 = 8.33 × 10^{9} atoms/m
Step 3: Calculate the radius of the carbon atom. Radius of carbon atom = (1/8.33 × 10^{9})^{(1/3)} = 0.0053 m = 5.3 × 10^{3} m
Question:
Calculate the mass and charge of one mole of electrons. Given: Mass of one electron = 9.10×10^{31} kg Charge of one electron = 1.602×10^{19} coulomb
Answer:
Step 1: To calculate the mass of one mole of electrons, multiply the mass of one electron by Avogadro’s number (6.022x10^{23}).
Mass of one mole of electrons = 9.10×10^{(31)} kg x 6.022x10^{23}
= 5.48×10^{(8)} kg
Step 2: To calculate the charge of one mole of electrons, multiply the charge of one electron by Avogadro’s number (6.022x10^{23}).
Charge of one mole of electrons = 1.602×10^{19} coulomb x 6.022x10^{23}
= 9.65×10^{(5)} coulomb
Question:
What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Answer:
Step 1: Calculate the frequency of light with a wavelength of 4000 pm.
Frequency = c/λ
Frequency = (3 x 10^{8} m/s) / (4000 x 10^{(12)} m)
Frequency = 7.5 x 10^{14} Hz
Step 2: Calculate the energy of one photon of light with a wavelength of 4000 pm.
Energy = h x frequency
Energy = (6.626 x 10^{34} Js) x (7.5 x 10^{14} Hz)
Energy = 4.969 x 10^{19} J
Step 3: Calculate the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy.
Number of photons = Energy / Energy of one photon
Number of photons = (1 J) / (4.969 x 10^{19} J)
Number of photons = 2 x 10^{18} photons
Question:
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:
Step 1: Calculate the frequency of the neutron using the given wavelength. Frequency (f) = c/λ where c = speed of light = 3 x 10^{8} m/s λ = wavelength = 800 pm = 800 x 10^{12} m
Therefore, f = (3 x 10^8 m/s) / (800 x 10^{12} m) f = 3.75 x 10^{19} Hz
Step 2: Calculate the characteristic velocity associated with the neutron. Characteristic velocity (v) = f x λ where f = frequency = 3.75 x 10^{19} Hz λ = wavelength = 800 pm = 800 x 10^{12} m
Therefore, v = (3.75 x 10^19 Hz) x (800 x 10^{12} m) v = 3 x 10^7 m/s
Question:
What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition, n = 4 to n = 2 of He+ spectrum ?
Answer:
Step 1: Identify the Balmer transition of He+ spectrum, n = 4 to n = 2.
Step 2: Calculate the wavelength of the Balmer transition of He+ spectrum, n = 4 to n = 2.
Step 3: Find the transition in the hydrogen spectrum that has the same wavelength as the Balmer transition of He+ spectrum, n = 4 to n = 2.
Question:
(i) The energy associated with the first orbit in the hydrogen atom is 2.18×10−18J atom−1. What is the energy associated with the fifth orbit? (ii) Calculate the radius of Bohr’s fifth orbit for the hydrogen atom.
Answer:
(i) The energy associated with the fifth orbit in the hydrogen atom is 2.18 x 10^{18} J atom^{1} x 5^{2} = 5.90 x 10^{17} J atom^{1}.
(ii) The radius of Bohr’s fifth orbit for the hydrogen atom can be calculated using the formula E = 2.18 x 10^{18} J atom^{1} / (4πε₀r), where ε₀ is the permittivity of free space and r is the radius of the orbit. Solving for r, we get r = 2.18 x 10^{18} J atom^{1} / (4πε₀E), where E is the energy of the fifth orbit. Substituting the values, we get r = 2.18 x 10^{18} J atom^1 / (4πε₀ x 5.90 x 10^{17} J atom^{1}) = 0.529 x 10^{10} m.
Question:
The work function for the 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 of 500 nm, calculate the kinetic energy and the velocity of the photoelectron.
Answer:
a) The threshold wavelength can be calculated using the equation: Wavelength (λ) = hc/Work Function (W)
Where h is Planck’s constant (6.626 x 10^{34} Js), c is the speed of light (3 x 10^{8} ms^{1}) and W is the work function (1.9 eV).
Therefore, the threshold wavelength is: λ = (6.626 x 10^{34} Js) x (3 x 10^{8} ms^{1}) / (1.9 eV) λ = 8.76 x 10^7 m
b) The threshold frequency can be calculated using the equation: Frequency (f) = c/Wavelength (λ)
Where c is the speed of light (3 x 10^{8} ms^{1}) and λ is the threshold wavelength (8.76 x 10^{7} m).
Therefore, the threshold frequency is: f = (3 x 10^{8} ms^{1}) / (8.76 x 10^{7} m) f = 3.43 x 10^{14} Hz
If the caesium element is irradiated with a wavelength of 500 nm, the kinetic energy of the photoelectron can be calculated using the equation: Kinetic Energy (KE) = hf  Work Function (W)
Where h is Planck’s constant (6.626 x 10^{34} Js), f is the frequency of the radiation (6.00 x 10^{14} Hz) and W is the work function (1.9 eV).
Therefore, the kinetic energy of the photoelectron is: KE = (6.626 x 10^{34} Js) x (6.00 x 10^{14} Hz)  (1.9 eV) KE = 1.18 eV
The velocity of the photoelectron can then be calculated using the equation: Velocity (v) = √(2KE/m)
Where KE is the kinetic energy of the photoelectron (1.18 eV) and m is the mass of the electron (9.11 x 10^{31} kg).
Therefore, the velocity of the photoelectron is: v = √((2 x 1.18 eV) / (9.11 x 10^{31} kg)) v = 2.64 x 10^{6} ms^{1}
Question:
The diameter of the zinc atom is 2.6Ao. Calculate (a) radius of zinc atom in pm and (b) a 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 in pm = 2.6Ao/2 = 1.3Ao = 1.3 x 10^{10}m = 1.3 x 10^{10} x 10^{12}m/pm = 1.3 x 10^{22} pm
(b) Number of atoms present in a length of 1.6 cm = 1.6 cm/diameter of zinc atom = 1.6 x 10^{2}m/2.6 x 10^{10}m = 6.15 x 10^{17} atoms
Question:
A photon of wavelength 4×10^{7} m strikes on a metal surface, the work function of the metal is 2.13 eV. Calculate: (i) The energy of the photon (eV) (ii) The kinetic energy of the emission (iii) The velocity of the photoelectron. [1eV=1.6020×10^{(19)}J].
Answer:
(i) Energy of the photon (eV) = hc/λ = (6.62610^{34} J.s)(310^{8} m/s)/(410^{7} m) = 4.90510^{19} J = 4.90510^{19} J / (1.602010^{19} J/eV) = 3.06 eV
(ii) Kinetic energy of the emission = Energy of the photon  Work function of the metal = 3.06 eV  2.13 eV = 0.93 eV
(iii) Velocity of the photoelectron = √(2Kinetic energy of the emission/mass of electron) = √(20.93 eV/9.1110^{31} kg) = 1.8610^{6} m/s
Question:
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) Cl17 (ii) U233 (iii) Be9
Question:
Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate the frequency (v) and wavenumber (vˉ) of the yellow light.
Answer:
Step 1: Calculate the frequency (v) of the yellow light.
Frequency (v) = Speed of Light (c) / Wavelength (λ)
Frequency (v) = 3 x 10^{8} m/s / 580 nm
Frequency (v) = 5.17 x 10^{14} Hz
Step 2: Calculate the wavenumber (vˉ) of the yellow light.
Wavenumber (vˉ) = 1 / Wavelength (λ)
Wavenumber (vˉ) = 1 / 580 nm
Wavenumber (vˉ) = 1.72 x 10^{7} mˉ1
Question:
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:
Step 1: Convert 20 cm to nanometer. 20 cm = 20 x 10^{2} m = 2 x 10^{6} nm
Step 2: Calculate the number of carbon atoms that can fit in 2 x 10^{6} nm. Number of carbon atoms = 2 x 10^{6} nm / 0.15 nm = 1.33 x 10^{7} carbon atoms
Question:
(i) Calculate the number of electrons which will together weighs one gram. (ii) Calculate the mass and charge of one mole of electrons.
Answer:
(i) Since the mass of one electron is 9.109 x 10^31 grams, the number of electrons that together weigh one gram is 1.1 x 10^{30}.
(ii) The mass of one mole of electrons is 9.109 x 10^31 grams and the charge of one mole of electrons is 6.022 x 10^{23} coulombs.
Question:
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800A∘. Calculate threshold frequency v0 and work function w0 of the metal.
Answer:
Step 1: Calculate the threshold wavelength λ0.
λ0 = 6800A∘
Step 2: Calculate the threshold frequency v0.
v0 = c/λ0
v0 = (3 x 10^{8} m/s)/(6800A∘)
v0 = 4.41 x 10^{14} Hz
Step 3: Calculate the work function w0.
w0 = hv0
w0 = (6.626 x 10^{34} J.s) x (4.41 x 10^{14} Hz)
w0 = 2.91 x 10^{19} J
Question:
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
Question:
(i) An atomic orbital has n = 3. What are the possible values of l and m1 ? (ii) List the quantum numbers (m1 and l ) of electrons for 3d orbital. (iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f
Answer:
(i) For an atomic orbital with n = 3, the possible values of l and m1 are l = 0, 1, 2 and m1 = 1, 0, 1.
(ii) The quantum numbers (m1 and l) of electrons for 3d orbital are m1 = 2, 1, 0, 1, 2 and l = 2.
(iii) The possible orbitals are 2s, 2p and 3f. 1p is not possible.
Question:
The unpaired electrons in Al and Si are present in 3p orbital. In which element (Al or Si), electrons will experience more effective nuclear charge from the nucleus?
Answer:

The unpaired electrons in Al and Si are present in 3p orbital.

The effective nuclear charge experienced by electrons in an atom can be calculated using the equation Zeff = Z  S, where Z is the atomic number and S is the number of core electrons.

Since Al has an atomic number of 13 and Si has an atomic number of 14, the effective nuclear charge experienced by electrons in Al will be higher than that experienced by electrons in Si.

Therefore, electrons in Al will experience a more effective nuclear charge from the nucleus than electrons in Si.
Question:
Indicate the number of unpaired electrons in (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Answer:
(a) P: 5 unpaired electrons (b) Si: 4 unpaired electrons (c) Cr: 4 unpaired electrons (d) Fe: 4 unpaired electrons (e) Kr: 8 unpaired electrons
Question:
A certain particle carries 2.5×10^{(16)}C of static electric charge. Calculate the number of electrons present in it.
Answer:
Answer: Step 1: Convert 2.5×10^{(16)}C of electric charge into Coulombs.
2.5×10^(16)C = 2.5×10^{(16)} x 1C/1C = 2.5×10^{(16)}C
Step 2: Calculate the number of electrons present in 2.5×10^{(16)}C of electric charge.
Number of electrons = 2.5×10^{(16)}C / (1.6×10^{(19)}C)
Number of electrons = 1.5625×10^{(7)} electrons
Question:
In Millikan’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 −1.282×10^{(18)}C. Calculate the number of electrons present on it.
Answer:
Answer:
Step 1: Calculate the charge of one electron. Charge of one electron = 1.602 × 10^{(19)}C
Step 2: Calculate the number of electrons on the oil drop. Number of electrons = (1.282 × 10^{(18)}) / (1.602 × 10^{(19)})
Step 3: Calculate the result. Number of electrons = 7.99 × 10^{(1)} Therefore, the number of electrons present on the oil drop is 7.99 × 10^{(1)}.
Question:
Symbols 35 79Br and 79Br can be written, whereas symbols 79 35Br and 35Br are not acceptable. Answer briefly:
Answer:
The symbols 79Br and 35Br are acceptable because they represent the atomic numbers of the elements, which are 79 (Gold) and 35 (Bromine). However, symbols 79 35Br and 35Br are not acceptable because they are not in the correct order. The atomic number must come first, followed by the element’s symbol.
Question:
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.
Answer:
(b) Amber light from traffic signal (c) Radiation from FM radio (a) Radiation from microwave oven (e) Xrays (d) Cosmic rays from outer space
Question:
Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6×10^24, calculate the power of this laser.
Answer:
Step 1: Calculate the energy of the photons emitted.
Energy = (Planck’s constant × speed of light) / wavelength
Energy = (6.626 × 10^{34} Js × 3.00 × 10^{8} m/s) / 337.1 × 10^{9} m
Energy = 2.01 × 10^{19} J
Step 2: Calculate the total energy emitted.
Total energy = Energy of each photon × number of photons
Total energy = 2.01 × 10^{19} J × 5.6 × 10^{24}
Total energy = 1.12 × 10^{6} J
Step 3: Calculate the power of the laser.
Power = Total energy / Time
Power = 1.12 × 10^{6} J / 1 s
Power = 1.12 × 10^{6} W
Question:
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800A∘. Calculate threshold frequency v0 and work function w0 of the metal.
Answer:
Step 1: Calculate the threshold wavelength λ0 using the equation λ0 = h/mv0, where h is Planck’s constant, m is the mass of the electron, and v0 is the threshold frequency.
Step 2: Calculate the threshold frequency v0 using the equation v0 = c/λ0, where c is the speed of light.
Step 3: Calculate the work function w0 using the equation w0 = hv0  KE, where h is Planck’s constant, v0 is the threshold frequency, and KE is the kinetic energy of the electron.
Question:
The electron energy in hydrogen atom is given by En=(−2.18×10^{(18)})n^{2} joules.Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength (in A ) of light that can be used to cause this transition.
Answer:
Answer:
 The energy required to remove an electron completely from the n = 2 orbit is given by En=(−2.18×10^{(18)})n^{2} joules.
Therefore, the energy required to remove an electron completely from the n = 2 orbit is En=(−2.18×10^{(18)})n^{2} joules = 8.72×10^{(18)} joules.
 The longest wavelength (in A ) of light that can be used to cause this transition can be calculated using the equation E=hc/λ, where E is the energy required to remove an electron completely from the n = 2 orbit, h is Planck’s constant, c is the speed of light and λ is the wavelength of light.
Therefore, the longest wavelength (in A ) of light that can be used to cause this transition is λ = hc/E = (6.626×10^{(34)} Js) (3.0×10^{8} m/s)/(8.72×10^{(18)} J) = 7.45×10^{(7)} m = 745 A.
Question:
(i) An atomic orbital has n = 3. What are the possible values of l and m1 ?
(ii) List the quantum numbers (m1 and l ) of electrons for 3d orbital.
(iii) Which of the following orbitals are possible?
1p, 2s, 2p and 3f
Answer:
(i) For an atomic orbital with n = 3, the possible values of l and m1 are l = 0, 1, 2 and m1 = 2, 1, 0, 1, 2.
(ii) The quantum numbers (m1 and l) of electrons for 3d orbital are m1 = 2, 1, 0, 1, 2 and l = 2.
(iii) The possible orbitals are 1p, 2s, 2p and 3f.
Question:
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
Question:
(i) Calculate the number of electrons which will together weighs one gram. (ii) Calculate the mass and charge of one mole of electrons.
Answer:
(i) One gram of electrons is equal to 6.02 x 10^{23} electrons.
(ii) The mass of one mole of electrons is equal to 9.11 x 10^31 kg and the charge of one mole of electrons is equal to 1.602 x 10^{19} C.
Question:
Find energy of each of the photons which (i) correspond to light of frequency 3×10^{15}Hz. (ii) have wavelength of 0.50A∘.
Answer:
(i) Energy of each of the photons which correspond to light of frequency 3×10^15Hz
E = hf
E = (6.626 x 10^{34}) x (3 x 10^{15})
E = 1.988 x 10^{18} Joules
(ii) Energy of each of the photons which have wavelength of 0.50A∘
E = hc/λ
E = (6.626 x 10^{34}) x (3 x 10^{8}) / 0.50
E = 3.976 x 10^{19} Joules
Question:
Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJmol^{(1)}.
Answer:
Step 1: Calculate the energy of the electromagnetic radiation in Joules.
Energy = (Planck’s constant x speed of light)/wavelength
Energy = (6.626 x 10^{(34)} x 3 x 10^{8})/242 x 10^{9}
Energy = 3.86 x 10^{19} Joules
Step 2: Calculate the ionisation energy of sodium in Joules.
Ionisation energy = Energy of electromagnetic radiation/Avogadro’s number
Ionisation energy = 3.86 x 10^{19}/6.02 x 10^{23}
Ionisation energy = 6.42 x 10^{43} Joules
Step 3: Convert the ionisation energy of sodium from Joules to kJmol^{(1)}.
Ionisation energy (kJmol^{(1)}) = Ionisation energy (Joules) x (1000/Avogadro’s number)
Ionisation energy (kJmol^{(1)}) = 6.42 x 10^{43} x (1000/6.02 x 10^{23})
Ionisation energy (kJmol^{(1)}) = 4.07 x 10^{19} kJmol^{(1)}
Question:
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:
Step 1: Calculate the energy difference between the two energy levels.
E2  E4 = (13.6 eV) (1/2)2  (13.6 eV) (1/4)2
Step 2: Calculate the frequency of the light emitted.
f = E2  E4 / h, where h is Planck’s constant (6.626 x 10^{34} J s)
f = (13.6 eV) (1/2)2  (13.6 eV) (1/4)2 / (6.626 x 10^{34} J s)
f = 3.4 x 10^{15} s^{1}
Step 3: Calculate the wavelength of the light emitted.
λ = c/f, where c is the speed of light (3 x 10^{8} ms^{1})
λ = (3 x 108 ms^{1}) / (3.4 x 1015 s^{1})
λ = 8.82 x 10^{7} m
Question:
What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit tothe 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.
Answer:

Calculate the energy required to shift the electron from the first Bohr orbit to the fifth Bohr orbit: Energy = 2.18 x 10^{(11)} ergs x (1 Joule/10^{7} ergs) x (1/4) = 5.45 x 10^{(19)} Joules

Calculate the wavelength of the light emitted when the electron returns to the ground state: Wavelength = (Planck’s Constant x Speed of Light)/Energy = (6.626 x 10^{(34)} Joulessec x 3 x 10^{8} m/sec)/(5.45 x 10^{(19)} Joules) = 6.5 x 10^{(7)} m = 650 nm
Question:
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
Answer:

Count the number of electrons in each species: \n Na+ = 10 electrons, K^{+} = 10 electrons, Mg^{2+} = 12 electrons, Ca^{2+} = 20 electrons, S^{2−} = 16 electrons, Ar = 18 electrons

Compare the number of electrons in each species: \n Na+ and K+ have the same number of electrons, so they are isoelectronic species. Mg^{2+}, Ca^{2+}, S^{2−} and Ar all have different numbers of electrons, so they are not isoelectronic.
Question:
(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 elements whose outermost electrons are represented by (a)3s^{1}(b)2p^{3} and (c)3d^{1} ?
(iii) Which atoms are indicated by the following configurations ?
(a) [He]2s^{1} (b) [Ne]3s^{2}3p^{3} (c) [Ar]4s^{2}3d^{1}
Answer:
(i) (a) H−: 1s^{2} (b) Na+: 1s^{22}s^{22}p^{63}s^{1} (c) O2−: 1s^{22}s^{22}p^{6} (d) F−: 1s^{22}s^{22}p^{5}
(ii) (a) Atomic number = 2 (b) Atomic number = 6 (c) Atomic number = 10
(iii) (a) Atom = He (b) Atom = Ne (c) Atom = Ar
Question:
An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for this electron.
Answer:

n: The principal quantum number (n) can have any value from 1 to ∞ (infinity).

l: The angular momentum quantum number (l) can have any value from 0 to n1.

ml: The magnetic quantum number (ml) can have any value from l to +l.
Question:
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) The number of protons = 29 + 35 = 64
(ii) The electronic configuration of the element = 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p4
Question:
Show that the circumference of the Bohr’s orbit for the hydrogen atom is an integral multiple of the de Broglie’s wavelength associated with the electron revolving around the orbit.
Answer:

Calculate the circumference of Bohr’s orbit for the hydrogen atom using the equation C = 2πr, where r is the radius of the orbit.

Calculate the de Broglie’s wavelength associated with the electron revolving around the orbit using the equation λ = h/p, where h is Planck’s constant and p is the momentum of the electron.

Determine if the circumference of Bohr’s orbit is an integral multiple of the de Broglie’s wavelength by dividing the circumference by the wavelength and checking if the result is an integer.
Question:
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v=3.29×10^15(Hz)[1/3^2−1/n^2] Calculate the value of n if the transition is observed at 1.285 nm. Find the region of the spectrum.
Answer:

Convert 1.285 nm to frequency: v = c/λ = (3.00 x 10^{8} m/s)/(1.285 x 10^{9} m) = 2.33 x 10^{17} Hz

Solve for n: 2.33 x 10^{17} Hz = 3.29 x 10^{15}(Hz)[1/3^2−1/n^2]

Calculate n: n = 3.24

Determine the region of the spectrum: The region of the spectrum is the infrared region.
Question:
Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at the one having 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer:
Answer:
 Calculate the energy difference between the two orbits: Energy difference = hc/λ Where, h = Planck’s constant c = Speed of light λ = Wavelength
Energy difference = (6.626 x 10^{34} x 3 x 10^{8}) / (1.3225 x 10^{9}  2.116 x 10^{12})
Energy difference = 2.566 x 10^{19} J
 Calculate the wavelength: λ = hc/Energy difference
λ = (6.626 x 10^34 x 3 x 10^{8}) / 2.566 x 10^{19}
λ = 486.1 nm

Name the series to which this transition belongs: This transition belongs to the Balmer series.

Name the region of the spectrum: The region of the spectrum for this transition is the visible light spectrum.
Question:
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 1.6×10^6 ms^{(1)}, calculate de Broglie wavelength associated with this electron.
Answer:
Step 1: Calculate the momentum of the electron. Momentum = mass x velocity Momentum = 9.1 x 10^{31} kg x 1.6 x 10^{6} m s^{(1)} Momentum = 1.4576 x 10^{24} kg m s^{(1)}
Step 2: Calculate the de Broglie wavelength. De Broglie wavelength = h/p De Broglie wavelength = 6.626 x 10^{34} J s/1.4576 x 10^{24} kg m s^{(1)} De Broglie wavelength = 4.54 x 10^{11} m
Question:
The velocity associated with a proton moving in a potential difference of 1000 V is 4.37×10^5 ms^{(1)}. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Answer:

Find the energy of the proton: Energy = Potential Difference x Charge of Proton = 1000 V x (1.6 x 10^{19} C) = 1.6 x 10^16 J

Find the momentum of the proton: Momentum = Energy/c = 1.6 x 10^{16} J / (3 x 10^{8} m/s) = 5.3 x 10^{25} kg m/s

Find the momentum of the hockey ball: Momentum = Mass x Velocity = 0.1 kg x (4.37 x 10^{5} m/s) = 4.37 x 10^{4} kg m/s

Find the wavelength associated with this momentum: Wavelength = h/Momentum = (6.626 x 10^{34} Js) / (4.37 x 10^{4} kg m/s) = 1.51 x 10^{38} m
Question:
(a) How many subshells are associated with n = 4 ? (b) How many electrons will be present in the subshells having ms value of 1/2 for n = 4 ?
Answer:
a) For n = 4, there are four subshells: s, p, d, and f.
b) For n = 4 and ms = 1/2, there will be two electrons present in the subshells. The two electrons will be in the s subshell and the p subshell. In the s subshell, there is one electron with ms = 1/2, and in the p subshell, there is one electron with ms = 1/2.
01 Some Basic Concepts of Chemistry
02 Structure of Atom
03 Classification of Elements and Periodicity in Properties
04 Chemical Bonding and Molecular Structure
05 States of Matter Gases and Liquids
06 Thermodynamics
07 Equilibrium
08 Redox Reactions
09 Hydrogen
10 The s block elements
11 Some p block elements
12 Organic Chemistry some basic principles and techniques
13 Hydrocarbons
14 Environmental