JEE Chemistry Structure Of Atom MCQs Set 04

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Chapter-wise Objective Questions: Structure Of Atom

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Multiple Choice Questions

Question 1: The magnitude of the kinetic energy of an electron in a Bohr orbit is equal to
(a) Half of the potential energy
(b) Twice the potential energy
(c) One-fourth of the potential energy
(d) None of these
Show Answer & Explanation

Answer: (a) Half of the potential energy

Explanation:
1. In a Bohr orbit KE = +Ze²/2r (in CGS) and PE = −Ze²/r.
2. So
3. KE
4. = ½
5. PE
6. , and the total energy equals −KE.

Teacher's Note:
1. Remember the three relations: KE = −E, PE = 2E and PE = −2 KE.

Question 2: Rutherford's experiment on the scattering of α-particles showed for the first time that the atom has
(a) Electrons
(b) Protons
(c) A nucleus
(d) Neutrons
Show Answer & Explanation

Answer: (c) A nucleus

Explanation:
1. Most α-particles passed straight through but a few were deflected through large angles.
2. This showed that the positive charge and mass are concentrated in a tiny, dense nucleus.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 3: Rutherford's scattering experiment is related to the size of the
(a) Nucleus
(b) Atom
(c) Electron
(d) Neutron
Show Answer & Explanation

Answer: (a) Nucleus

Explanation:
1. The distance of closest approach of α-particles to the nucleus gave an estimate of the nuclear size (about 10−15 m).

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 4: Bohr's model can explain
(a) The spectrum of the hydrogen atom only
(b) The spectrum of an atom or ion containing one electron only
(c) The spectrum of the hydrogen molecule
(d) The solar spectrum
Show Answer & Explanation

Answer: (b) The spectrum of an atom or ion containing one electron only

Explanation:
1. Bohr's theory works for hydrogen-like species such as H, He+, Li2+, which have a single electron.
2. It fails for multi-electron atoms.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 5: When atoms are bombarded with α-particles, only a few in a million suffer deflection and the others pass out undeflected. This is because
(a) The force of repulsion on the moving α-particle is small
(b) The force of attraction on the α-particle to the oppositely charged electrons is very small
(c) There is only one nucleus and a large number of electrons
(d) The nucleus occupies a much smaller volume compared to the volume of the atom
Show Answer & Explanation

Answer: (d) The nucleus occupies a much smaller volume compared to the volume of the atom

Explanation:
1. Deflection occurs only when an α-particle passes very close to the nucleus.
2. The nucleus is about 10−5 times the size of the atom, so close approaches are rare.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 6: Positronium consists of an electron and a positron (same mass as an electron but opposite charge) orbiting about their common centre of mass. The Rydberg constant for this system is (R∞ is the value for hydrogen with an infinitely heavy nucleus)
(a) R∞/4
(b) R∞/2
(c) 2R∞
(d) R∞
Show Answer & Explanation

Answer: (b) R∞/2

Explanation:
1. The Rydberg constant is proportional to the reduced mass μ of the system.
2. For positronium μ = m·m/(m + m) = m/2.
3. So R = R∞/2.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 7: When an electron jumps from the L shell to the K shell
(a) Energy is absorbed
(b) Energy is released
(c) Energy is sometimes absorbed and sometimes released
(d) Energy is neither absorbed nor released
Show Answer & Explanation

Answer: (b) Energy is released

Explanation:
1. The K shell is lower in energy than the L shell.
2. A fall to a lower level releases energy as a photon.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 8: When beryllium is bombarded with α-particles, extremely penetrating radiations which cannot be deflected by electric or magnetic fields are given out. These are
(a) A beam of protons
(b) α-rays
(c) A beam of neutrons
(d) X-rays
Show Answer & Explanation

Answer: (c) A beam of neutrons

Explanation:
1. Be + He → C + n; the radiation is undeflected, so it carries no charge.
2. It is a beam of neutrons (Chadwick, 1932).

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 9: The energy of an orbit in a hydrogen-like atom
(a) Increases as we move away from the nucleus
(b) Decreases as we move away from the nucleus
(c) Remains the same as we move away from the nucleus
(d) None of these
Show Answer & Explanation

Answer: (a) Increases as we move away from the nucleus

Explanation:
1. E_n = −13.6/n² eV is negative and increases (becomes less negative) as n increases.
2. The energy is highest (zero) at infinite distance.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 10: Bohr's model of the atom could not account for the
(a) Emission spectrum
(b) Absorption spectrum
(c) Line spectrum of hydrogen
(d) Fine spectrum
Show Answer & Explanation

Answer: (d) Fine spectrum

Explanation:
1. Bohr's model explains the main lines of hydrogen. Fine structure, where each line is split into very close components, needs sub-shells (Sommerfeld) and relativistic treatment.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 11: The energy of the second Bohr orbit of the hydrogen atom is −328 kJ/mol. The energy of the fourth Bohr orbit would be
(a) −41 kJ/mol
(b) −1312 kJ/mol
(c) −164 kJ/mol
(d) −82 kJ/mol
Show Answer & Explanation

Answer: (d) −82 kJ/mol

Explanation:
1. E_n is proportional to 1/n².
2. E_4 = E_2 × (2²/4²) = −328 × 4/16 = −82 kJ/mol.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 12: The energy of an electron in the nth orbit of the hydrogen atom is
(a) −13.6/n4 eV
(b) −13.6/n3 eV
(c) −13.6/n² eV
(d) −13.6/n eV
Show Answer & Explanation

Answer: (c) −13.6/n² eV

Explanation:
1. For hydrogen, E_n = −13.6/n² eV.
2. For example, n = 1 gives −13.6 eV and n = 2 gives −3.4 eV.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 13: If the wavelength of a photon is 2.2 × 10−11 m and h = 6.6 × 10−34 J s, the momentum of the photon is
(a) 3 × 10−23 kg m s−1
(b) 3.33 × 1022 kg m s−1
(c) 1.452 × 10−44 kg m s−1
(d) 6.89 × 1043 kg m s−1
Show Answer & Explanation

Answer: (a) 3 × 10−23 kg m s−1

Explanation:
1. Momentum p = h/λ.
2. p = 6.6 × 10−34/2.2 × 10−11 = 3 × 10−23 kg m s−1.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 14: The different lines of the Lyman series of the hydrogen spectrum are present in the
(a) UV field
(b) IR field
(c) Visible field
(d) Far IR field
Show Answer & Explanation

Answer: (a) UV field

Explanation:
1. Lyman lines arise from transitions to n = 1 and have the shortest wavelengths.
2. They lie in the ultraviolet region.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 15: In the Balmer series of the hydrogen atom spectrum, the electronic transition that causes the third line is
(a) Fifth Bohr orbit to second one
(b) Fifth Bohr orbit to first one
(c) Fourth Bohr orbit to second one
(d) Fourth Bohr orbit to first one
Show Answer & Explanation

Answer: (a) Fifth Bohr orbit to second one

Explanation:
1. Balmer lines end at n = 2.
2. First line 3 → 2, second 4 → 2, third 5 → 2.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 16: The energy of an electron in the second Bohr orbit of the hydrogen atom is
(a) −5.44 × 10−19 J
(b) −5.44 × 10−19 kJ
(c) −5.44 × 10−19 cal
(d) −5.44 × 10−19 eV
Show Answer & Explanation

Answer: (a) −5.44 × 10−19 J

Explanation:
1. E_2 = −13.6/4 = −3.4 eV.
2. In joules: −3.4 × 1.6 × 10−19 = −5.44 × 10−19 J.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 17: The radius of the first Bohr orbit of hydrogen is 0.53 Å. The radius of the third Bohr orbit would be
(a) 0.79 Å
(b) 1.59 Å
(c) 3.18 Å
(d) 4.77 Å
Show Answer & Explanation

Answer: (d) 4.77 Å

Explanation:
1. r_n = 0.53 n² Å.
2. r_3 = 0.53 × 9 = 4.77 Å.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 18: The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom is (ionisation energy of H = 2.18 × 10−18 J per atom and h = 6.625 × 10−34 J s)
(a) 3.08 × 1015 s−1
(b) 2.00 × 1015 s−1
(c) 1.54 × 1015 s−1
(d) 1.03 × 1015 s−1
Show Answer & Explanation

Answer: (a) 3.08 × 1015 s−1

Explanation:
1. ΔE = 2.18 × 10−18 (1/1² − 1/4²) = 2.18 × 10−18 × 15/16 = 2.044 × 10−18 J.
2. ν = ΔE/h = 2.044 × 10−18/6.625 × 10−34 = 3.08 × 1015 s−1.

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 19: The wavelength of the radiation emitted when a hydrogen atom electron falls from infinity to stationary state 1 would be (Rydberg constant = 1.097 × 107 m−1)
(a) 406 nm
(b) 192 nm
(c) 91 nm
(d) 9.1 × 10−8 nm
Show Answer & Explanation

Answer: (c) 91 nm

Explanation:
1. 1/λ = R(1/1² − 1/∞²) = R.
2. λ = 1/1.097 × 107 = 9.12 × 10−8 m = 91 nm.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

Question 20: In the hydrogen atom spectrum, a transition that belongs to the Paschen series has
(a) n1 = 1 and n2 = 2, 3, 4
(b) n1 = 3 and n2 = 4, 5, 6
(c) n1 = 1 and n2 = 3, 4, 5
(d) n1 = 2 and n2 = 3, 5
Show Answer & Explanation

Answer: (b) n1 = 3 and n2 = 4, 5, 6

Explanation:
1. The Paschen series ends on the third orbit, so n1 = 3.
2. The electron falls from n2 = 4, 5, 6, ... to n1 = 3.

Teacher's Note:
1. Convert every quantity to moles or equivalents first.
2. Then apply the stoichiometric ratio or the definition asked for.

Question 21: The radius of which of the following orbits is the same as that of the first Bohr orbit of the hydrogen atom?
(a) He+ (n = 2)
(b) Li2+ (n = 2)
(c) Li2+ (n = 3)
(d) Be3+ (n = 2)
Show Answer & Explanation

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

Explanation:
1. r ∝ n²/Z.
2. He+: 4/2 = 2; Li2+ (n = 2): 4/3; Li2+ (n = 3): 9/3 = 3; Be3+: 4/4 = 1.
3. Only Be3+ (n = 2) matches hydrogen n = 1 (ratio 1).

Teacher's Note:
1. Eliminate options that disagree with the basic definition.
2. Confirm the remaining option by a quick calculation.

Question 22: The time taken for an electron to complete one revolution in the Bohr orbit of hydrogen is
(a) 4π²mr²/nh
(b) nh/4π²mr
(c) nh/4π²mr²
(d) h/2πmr
Show Answer & Explanation

Answer: (a) 4π²mr²/nh

Explanation:
1. Angular momentum mvr = nh/2π, so v = nh/2πmr.
2. Time period T = 2πr/v = 2πr × 2πmr/nh = 4π²mr²/nh.

Teacher's Note:
1. Read the data carefully and write the relevant formula before substituting.
2. Check units and the order of magnitude of the answer.

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