JEE Chemistry Structure of Atom MCQs Set 20

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Multiple Choice Questions

Question 1: Which of the following statements about the electron is incorrect?
(a) It is a negatively charged particle
(b) The mass of the electron is equal to the mass of the neutron
(c) It is a basic constituent of all atoms
(d) It is a constituent of cathode rays
Show Answer & Explanation

Answer: (b) The mass of the electron is equal to the mass of the neutron

Explanation:
1. The mass of an electron (9.1 × 10-31 kg) is about 1/1840 of the mass of a neutron (1.675 × 10-27 kg).
2. The other three statements are correct.

Teacher's Note:
1. Compare masses: proton ≈ neutron ≈ 1840 × electron.
2. Electrons were discovered through cathode ray experiments.

Question 2: Which of the following will not show deflection from the path on passing through an electric field?
(a) proton
(b) cathode rays
(c) electron
(d) neutron
Show Answer & Explanation

Answer: (d) neutron

Explanation:
1. Deflection in an electric field requires an electric charge.
2. The neutron is neutral, so it is not deflected; protons, electrons and cathode rays (electrons) are.

Teacher's Note:
1. Cathode rays are streams of electrons.
2. Neutrons are also unaffected by magnetic fields.

Question 3: Which of the following conclusions could not be derived from Rutherford's α-particle scattering experiment?
(a) Most of the space in the atom is empty
(b) The radius of the atom is about 10-10 m while that of the nucleus is about 10-15 m
(c) Electrons move in circular paths of fixed energy called orbits
(d) Electrons and the nucleus are held together by electrostatic forces of attraction
Show Answer & Explanation

Answer: (c) Electrons move in circular paths of fixed energy called orbits

Explanation:
1. Rutherford's experiment showed a small dense nucleus and mostly empty space.
2. The concept of electrons in fixed circular orbits of definite energy came from Bohr, not from Rutherford.

Teacher's Note:
1. Bohr added quantised orbits to Rutherford's nuclear model.
2. Be careful to separate the contributions of different scientists.

Question 4: The total number of protons in 34 mg of NH3 at STP is:
(a) 12.044 × 1023
(b) 12.044 × 1022
(c) 12.044 × 1021
(d) 12.044 × 1020
Show Answer & Explanation

Answer: (c) 12.044 × 1021

Explanation:
1. Moles of NH3 = 34 × 10-3 g / 17 g mol-1 = 2 × 10-3 mol, which is 2 × 10-3 × 6.022 × 1023 = 1.2044 × 1021 molecules.
2. Each molecule has 7 + 3 = 10 protons, so total protons = 1.2044 × 1022 = 12.044 × 1021.

Teacher's Note:
1. Number of protons = number of molecules × protons per molecule.
2. Conditions such as STP do not matter for counting particles.

Question 5: The diameter of a zinc atom is 2.6 Å. The number of zinc atoms arranged side by side in a length of 1.6 cm is:
(a) 6.023 × 1023
(b) 6.15 × 107
(c) 1.62 × 1019
(d) 6.15 × 1012
Show Answer & Explanation

Answer: (b) 6.15 × 107

Explanation:
1. Length = 1.6 cm = 1.6 × 10-2 m and diameter = 2.6 × 10-10 m.
2. Number of atoms = 1.6 × 10-2 / 2.6 × 10-10 = 6.15 × 107.

Teacher's Note:
1. Convert both lengths to the same unit first.
2. 1 Å = 10-10 m.

Question 6: Two atoms are isobars if:
(a) they have the same atomic number but different mass number
(b) they have the same number of electrons but different number of neutrons
(c) they have the same number of neutrons but different number of electrons
(d) the sum of protons and neutrons is the same but the number of protons is different
Show Answer & Explanation

Answer: (d) the sum of protons and neutrons is the same but the number of protons is different

Explanation:
1. Isobars have the same mass number (protons + neutrons) but different atomic numbers.
2. Option (a) describes isotopes and option (c) describes isotones.

Teacher's Note:
1. Isotopes: same Z. Isobars: same A. Isotones: same number of neutrons.
2. Examples: 40Ar and 40Ca are isobars.

Question 7: Which of the following is isoelectronic with the carbon atom?
(a) N+
(b) O2-
(c) Na+
(d) Al3+
Show Answer & Explanation

Answer: (a) N+

Explanation:
1. Carbon has 6 electrons.
2. N+ has 7 - 1 = 6 electrons, while O2-, Na+ and Al3+ have 10 electrons each.

Teacher's Note:
1. Electrons in an ion = atomic number minus charge.
2. Isoelectronic species have the same number of electrons.

Question 8: Which of the following pairs has identical values of e/m?
(a) a proton and a neutron
(b) a proton and deuterium
(c) deuterium and an α-particle
(d) an electron and γ-rays
Show Answer & Explanation

Answer: (c) deuterium and an α-particle

Explanation:
1. Deuteron has charge +e and mass 2m (e/2m), and an α-particle has charge +2e and mass 4m (also e/2m).
2. A neutron and γ-rays are uncharged, and a proton (e/m) differs from a deuteron.

Teacher's Note:
1. The e/m ratio is the specific charge.
2. Neutral particles have e/m = 0.

Question 9: The arrangement showing the increasing value of e/m for e, p, n and α-particles is:
(a) n < α < p < e
(b) e < p < α < n
(c) n < p < e < α
(d) p < n < α < e
Show Answer & Explanation

Answer: (a) n < α < p < e

Explanation:
1. The neutron has zero charge, so e/m = 0.
2. Then α has e/m = 2e/4m = e/2m, a proton has e/m, and the electron has the largest e/m because its mass is very small.

Teacher's Note:
1. Order of specific charge: n < α < p < e.
2. Specific charge was used by Thomson to discover the electron.

Question 10: Which of the following transitions is not allowed in the normal electronic emission spectrum of an atom?
(a) 2s → 1s
(b) 2p → 1s
(c) 3d → 4p
(d) 5p → 3s
Show Answer & Explanation

Answer: (a) 2s → 1s

Explanation:
1. The selection rule for an allowed transition is Δl = ±1.
2. The 2s → 1s transition has Δl = 0, so it is forbidden; the other three have Δl = ±1.

Teacher's Note:
1. Selection rule: Δl = ±1.
2. Remember that s → s and p → p are forbidden.

Question 11: Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n, and can be represented as ν = 3.29 × 1015 Hz [1/32 - 1/n2]. If the transition is observed at 1285 nm, the value of n and the region of the spectrum are:
(a) 4, infrared
(b) 5, infrared
(c) 6, visible
(d) 3, infrared
Show Answer & Explanation

Answer: (b) 5, infrared

Explanation:
1. ν = c/λ = 3 × 108 / 1285 × 10-9 = 2.33 × 1014 Hz.
2. Then 1/9 - 1/n2 = 2.33 × 1014/3.29 × 1015 = 0.0709, so 1/n2 = 0.0402 and n = 5; 1285 nm lies in the infrared.

Teacher's Note:
1. The Paschen series lies in the infrared region.
2. Lyman is UV, Balmer is visible, and Brackett and Pfund are IR.

Question 12: The yellow light emitted from a sodium lamp has a wavelength of 580 nm. Its frequency and wave number are:
(a) 5.17 × 1014 s-1, 1.724 × 104 cm-1
(b) 4.27 × 1014 s-1, 3.245 × 104 cm-1
(c) 1.26 × 1015 s-1, 8.37 × 103 cm-1
(d) 51.7 × 1014 s-1, 1.724 × 105 cm-1
Show Answer & Explanation

Answer: (a) 5.17 × 1014 s-1, 1.724 × 104 cm-1

Explanation:
1. ν = c/λ = 3 × 108 / 580 × 10-9 = 5.17 × 1014 s-1.
2. Wave number = 1/λ = 1/(580 × 10-7 cm) = 1.724 × 104 cm-1.

Teacher's Note:
1. Wave number is the reciprocal of wavelength.
2. Be careful with units: nm to cm requires multiplying by 10-7.

Question 13: The number of photons of light of wavelength 4000 pm that provide 1 J of energy is:
(a) 4.9695 × 1017
(b) 4.9695 × 108
(c) 2.0122 × 1015
(d) 2.0122 × 1016
Show Answer & Explanation

Answer: (d) 2.0122 × 1016

Explanation:
1. Energy of one photon = hc/λ = (6.626 × 10-34 × 3 × 108) / (4 × 10-9) = 4.9695 × 10-17 J.
2. Number of photons = 1 / 4.9695 × 10-17 = 2.0122 × 1016.

Teacher's Note:
1. 4000 pm = 4 × 10-9 m.
2. Number of photons = total energy / energy of one photon.

Question 14: A 25 watt bulb emits monochromatic yellow light of wavelength 0.57 μm. The rate of emission of quanta per second is:
(a) 7.169 × 1020
(b) 8.29 × 1019
(c) 7.289 × 1021
(d) 7.169 × 1019
Show Answer & Explanation

Answer: (d) 7.169 × 1019

Explanation:
1. Energy of one photon = hc/λ = 6.626 × 10-34 × 3 × 108 / 0.57 × 10-6 = 3.487 × 10-19 J.
2. Rate = 25 / 3.487 × 10-19 = 7.169 × 1019 photons per second.

Teacher's Note:
1. Power is energy per second.
2. 1 W = 1 J/s.

Question 15: The maximum kinetic energy of photoelectrons is 6.63 × 10-19 J when a metal is irradiated with radiation of frequency 2 × 1015 Hz. The threshold frequency of the metal is about:
(a) 2 × 1015 s-1
(b) 1 × 1015 s-1
(c) 2.5 × 1015 s-1
(d) 4 × 1015 s-1
Show Answer & Explanation

Answer: (b) 1 × 1015 s-1

Explanation:
1. h(ν - ν0) = KE, so ν - ν0 = 6.63 × 10-19 / 6.63 × 10-34 = 1 × 1015 Hz.
2. Therefore ν0 = 2 × 1015 - 1 × 1015 = 1 × 1015 s-1.

Teacher's Note:
1. Einstein's equation: hν = hν0 + KE.
2. Threshold frequency is the minimum frequency to eject electrons.

Question 16: The minimum energy that photons must possess to produce the photoelectric effect with platinum, whose threshold frequency is 1.3 × 1015 s-1, is:
(a) 3.6 × 10-13 erg
(b) 8.2 × 10-13 erg
(c) 8.2 × 10-14 erg
(d) 8.6 × 10-12 erg
Show Answer & Explanation

Answer: (d) 8.6 × 10-12 erg

Explanation:
1. E = hν0 = 6.626 × 10-34 J s × 1.3 × 1015 s-1 = 8.61 × 10-19 J.
2. Since 1 J = 107 erg, E = 8.6 × 10-12 erg.

Teacher's Note:
1. Convert joules to ergs by multiplying by 107.
2. Minimum energy equals the work function.

Question 17: An emission transition starts from the orbit of radius 1.3225 nm and ends at 211.6 pm. The series and the region of the spectrum are:
(a) Lyman, UV
(b) Balmer, visible
(c) Paschen, IR
(d) Brackett, IR
Show Answer & Explanation

Answer: (b) Balmer, visible

Explanation:
1. r = 52.9 pm × n2: for 1322.5 pm, n2 = 25, so n = 5; for 211.6 pm, n2 = 4, so n = 2.
2. A transition ending at n = 2 belongs to the Balmer series, which lies in the visible region.

Teacher's Note:
1. The radius of the nth Bohr orbit of hydrogen = 52.9 n2 pm.
2. Balmer series ends at n = 2.

Question 18: The radius of which of the following orbits is the same as that of the first Bohr orbit of hydrogen?
(a) Li2+ (n = 2)
(b) Li2+ (n = 3)
(c) Be3+ (n = 2)
(d) He+ (n = 2)
Show Answer & Explanation

Answer: (c) Be3+ (n = 2)

Explanation:
1. Radius is r = 0.529 n2/Z Å, so we need n2/Z = 1.
2. For Be3+ (Z = 4) with n = 2, n2/Z = 4/4 = 1.

Teacher's Note:
1. For Li2+ n = 2: 4/3; n = 3: 9/3 = 3; He+ n = 2: 4/2 = 2.
2. Only Be3+ gives 1.

Question 19: Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise a sodium atom. The ionisation energy of sodium in kJ/mol is:
(a) 4.945 × 105
(b) 4.945 × 104
(c) 4.945 × 102
(d) 4.945 × 103
Show Answer & Explanation

Answer: (c) 4.945 × 102

Explanation:
1. Energy per atom = hc/λ = 6.626 × 10-34 × 3 × 108 / 242 × 10-9 = 8.21 × 10-19 J.
2. Per mole = 8.21 × 10-19 × 6.022 × 1023 = 4.945 × 105 J = 494.5 kJ/mol.

Teacher's Note:
1. Convert J to kJ by dividing by 1000.
2. Ionisation energy of sodium is about 495 kJ/mol.

Question 20: The maximum number of emission lines when the excited electron of a hydrogen atom in n = 6 drops to the ground state is:
(a) 2
(b) 10
(c) 15
(d) 20
Show Answer & Explanation

Answer: (c) 15

Explanation:
1. The number of lines from level n is n(n - 1)/2 for all possible transitions.
2. For n = 6, 6 × 5/2 = 15.

Teacher's Note:
1. This counts all possible transitions between levels, not only those ending in n = 1.
2. The formula applies to a collection of atoms.

Question 21: The velocity of an electron in the 3rd orbit of a hydrogen atom is:
(a) 2.79 × 107 cm/s
(b) 9.27 × 1027 cm/s
(c) 7.29 × 107 cm/s
(d) 92.7 × 107 cm/s
Show Answer & Explanation

Answer: (c) 7.29 × 107 cm/s

Explanation:
1. The velocity is v = 2.18 × 106 m/s × Z/n.
2. For n = 3, v = 2.18 × 106/3 = 7.27 × 105 m/s = 7.29 × 107 cm/s.

Teacher's Note:
1. Velocity in Bohr orbits is inversely proportional to n.
2. In the first orbit it is 2.18 × 106 m/s.

Question 22: A table-tennis ball has a mass of 10 g and a speed of 90 m/s. If the speed can be measured to within 4%, the uncertainty in position is:
(a) 3.6 × 10-23 m
(b) 1.46 × 10-33 m
(c) 2.8 × 10-30 m
(d) 1.00 × 10-34 m
Show Answer & Explanation

Answer: (b) 1.46 × 10-33 m

Explanation:
1. Δv = 4% of 90 = 3.6 m/s and Δp = mΔv = 0.01 × 3.6 = 0.036 kg m/s.
2. Δx = h/(4πΔp) = 6.626 × 10-34 / (4π × 0.036) = 1.46 × 10-33 m.

Teacher's Note:
1. Heisenberg's principle: Δx · Δp ≥ h/4π.
2. Use SI units throughout.

Question 23: The wavelength of the neutron used in a neutron diffraction microscope is 800 pm. The characteristic velocity of the neutron (mass 1.675 × 10-27 kg) is about:
(a) 4.92 × 102 m/s
(b) 2.46 × 103 m/s
(c) 9.84 × 102 m/s
(d) 0.246 × 103 m/s
Show Answer & Explanation

Answer: (a) 4.92 × 102 m/s

Explanation:
1. From de Broglie's relation v = h/(mλ) = 6.626 × 10-34 / (1.675 × 10-27 × 800 × 10-12).
2. v = 4.9 × 102 m/s.

Teacher's Note:
1. λ = h/mv.
2. 800 pm = 8 × 10-10 m.

Question 24: Out of electron, proton, neutron and α-particle, which one will need the highest velocity to produce matter waves of the same wavelength?
(a) electron
(b) proton
(c) neutron
(d) α-particle
Show Answer & Explanation

Answer: (a) electron

Explanation:
1. For the same wavelength, mv is constant, so v ∝ 1/m.
2. The electron has the least mass and therefore needs the highest velocity.

Teacher's Note:
1. λ = h/mv.
2. A lighter particle moves faster for the same wavelength.

Question 25: What is the wavelength of an α-particle of mass 6.6 × 10-27 kg moving with a speed of 105 cm/s? (h = 6.6 × 10-34 kg m2/s)
(a) 2 × 10-12 m
(b) 3 × 10-10 m
(c) 1 × 10-10 m
(d) 2 × 10-10 m
Show Answer & Explanation

Answer: (c) 1 × 10-10 m

Explanation:
1. Speed = 105 cm/s = 103 m/s.
2. λ = h/mv = 6.6 × 10-34 / (6.6 × 10-27 × 103) = 1 × 10-10 m.

Teacher's Note:
1. Always convert cm/s to m/s.
2. De Broglie wavelength decreases with mass and speed.

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