JEE Chemistry Structure of Atom MCQs Set 17

Chemistry Objective Questions and Answers: Structure Of Atom

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Multiple Choice Questions

Question 1: What is the angular velocity (\( \omega \)) of an electron occupying the second orbit of a \( \ce{Li^2+} \) ion? (K = Coulomb's constant)
(a) \( \dfrac{8\pi^3 m e^4}{h^3}K^2 \)
(b) \( \dfrac{8\pi^3 m e^4}{9h^3}K^2 \)
(c) \( \dfrac{64\pi^3 m e^4}{9h^3}K^2 \)
(d) \( \dfrac{9\pi^3 m e^4}{h^3}K^2 \)
Show Answer & Explanation

Answer: (d) \( \dfrac{9\pi^3 m e^4}{h^3}K^2 \)

Explanation:
1. In Bohr's model \( \omega = \dfrac{v}{r} = \dfrac{8\pi^3 m e^4 Z^2 K^2}{n^3 h^3} \).
2. For Z = 3 and n = 2: \( \omega = \dfrac{8\pi^3 m e^4 \times 9 K^2}{8h^3} = \dfrac{9\pi^3 m e^4 K^2}{h^3} \).

Teacher's Note:
1. Angular velocity depends on Z squared and falls as n cubed.
2. Substitute Z = 3 and n = 2 carefully.

Question 2: The value of Planck's constant is 6.63 × 10⁻³⁴ J s and the velocity of light is 3.0 × 10⁸ m s⁻¹. Which value is closest to the wavelength in nanometres of a quantum of light with frequency 8 × 10¹⁵ s⁻¹?
(a) 5 × 10⁻¹⁸
(b) 4 × 10¹
(c) 3 × 10⁷
(d) 2 × 10⁻²⁵
Show Answer & Explanation

Answer: (b) 4 × 10¹

Explanation:
1. \( \lambda = c/\nu = \dfrac{3.0\times10^8}{8\times10^{15}} = 3.75\times10^{-8} \) m.
2. In nanometres this is 37.5 nm, which is closest to 4 × 10¹ nm.

Teacher's Note:
1. Wavelength = c divided by frequency.
2. Convert metres to nanometres by multiplying by 10⁹.

Question 3: Which of the following is not a permissible arrangement of electrons in an atom?
(a) n = 5, l = 3, m = 0, s = +1/2
(b) n = 3, l = 2, m = -3, s = -1/2
(c) n = 3, l = 2, m = -2, s = -1/2
(d) n = 4, l = 0, m = 0, s = -1/2
Show Answer & Explanation

Answer: (b) n = 3, l = 2, m = -3, s = -1/2

Explanation:
1. For l = 2 the magnetic quantum number can only take the values -2, -1, 0, +1, +2.
2. m = -3 is therefore not allowed.

Teacher's Note:
1. m ranges from -l to +l.
2. Check l against n as well.

Question 4: Excited hydrogen atom emits light in the ultraviolet region at 2.47 × 10¹⁵ Hz. With this frequency, the energy of a single photon is (h = 6.63 × 10⁻³⁴ J s):
(a) 8.041 × 10⁻⁴⁰ J
(b) 2.680 × 10⁻¹⁹ J
(c) 1.640 × 10⁻¹⁸ J
(d) 6.111 × 10⁻¹⁷ J
Show Answer & Explanation

Answer: (c) 1.640 × 10⁻¹⁸ J

Explanation:
1. E = h\( \nu \) = 6.63 × 10⁻³⁴ × 2.47 × 10¹⁵ J.
2. This equals 1.64 × 10⁻¹⁸ J.

Teacher's Note:
1. Photon energy = Planck's constant times frequency.

Question 5: Calculate the velocity of the ejected electron when light of frequency 2 × 10¹⁵ Hz falls on a metal surface whose threshold frequency is 7 × 10¹⁴ Hz.
(a) 1.37 × 10⁶ m/s
(b) 1.26 × 10⁶ m/s
(c) 1.45 × 10⁷ m/s
(d) 1.37 × 10⁷ m/s
Show Answer & Explanation

Answer: (a) 1.37 × 10⁶ m/s

Explanation:
1. Kinetic energy = h(\( \nu-\nu_0 \)) = 6.626 × 10⁻³⁴ × 1.3 × 10¹⁵ = 8.61 × 10⁻¹⁹ J.
2. \( v = \sqrt{2KE/m} = \sqrt{2\times8.61\times10^{-19}/9.11\times10^{-31}} = 1.37\times10^{6} \) m/s.

Teacher's Note:
1. Use the photoelectric equation.
2. Then convert KE to speed.

Question 6: An electron e₁ is moving in the fifth stationary state and another electron e₂ in the fourth stationary state. The radius of the orbit of e₁ is five times the radius of the orbit of e₂. The ratio of the velocity of e₁ to the velocity of e₂ is:
(a) 5 : 1
(b) 4 : 1
(c) 1 : 5
(d) 1 : 4
Show Answer & Explanation

Answer: (d) 1 : 4

Explanation:
1. Radius is proportional to n²/Z: 25/Z₁ = 5 × 16/Z₂, so Z₂/Z₁ = 80/25 = 3.2.
2. Velocity is proportional to Z/n: v₁/v₂ = (Z₁/5)/(Z₂/4) = 0.8 × (Z₁/Z₂) = 0.8/3.2 = 1/4.

Teacher's Note:
1. Use both r and v proportionalities.
2. The two electrons belong to different species.

Question 7: Among the following groupings, which represents the collection of isoelectronic species?
(a) \( \ce{NO+, C2^2-, O2^-, CO} \)
(b) \( \ce{N2, C2^2-, CO, NO} \)
(c) \( \ce{CO, NO+, CN-, C2^2-} \)
(d) \( \ce{NO, CN-, N2, O2^-} \)
Show Answer & Explanation

Answer: (c) \( \ce{CO, NO+, CN-, C2^2-} \)

Explanation:
1. Isoelectronic species have the same number of electrons.
2. CO, NO⁺, CN⁻ and C₂²⁻ each have 14 electrons.
3. The other groups mix species with 14, 15 and 17 electrons.

Teacher's Note:
1. Count electrons: add atomic numbers and adjust for charge.

Question 8: At temperature T, the average kinetic energy of any particle is \( \tfrac32 kT \). The de Broglie wavelength follows the order:
(a) visible photon > thermal neutron > thermal electron
(b) thermal proton > thermal electron > visible photon
(c) thermal proton > visible photon > thermal electron
(d) visible photon > thermal electron > thermal neutron
Show Answer & Explanation

Answer: (d) visible photon > thermal electron > thermal neutron

Explanation:
1. For a particle of fixed kinetic energy, \( \lambda = \dfrac{h}{\sqrt{2mKE}} \), so heavier particles have shorter wavelengths.
2. Visible light has a wavelength of about 500 nm, while thermal electrons are about 6 nm and thermal neutrons only about 0.1 nm, so the order is visible photon, thermal electron, thermal neutron.

Teacher's Note:
1. Wavelength falls as mass rises at equal kinetic energy.
2. The electron is far lighter than the neutron.

Question 9: What is the maximum number of electrons that can be accommodated in an atom in which the highest principal quantum number value is 4?
(a) 10
(b) 18
(c) 36
(d) 54
Show Answer & Explanation

Answer: (c) 36

Explanation:
1. With n = 4 as the highest shell, the subshells that can be occupied are those up to 4p, which are 1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p.
2. These hold 2 + 2 + 6 + 2 + 6 + 10 + 2 + 6 = 36 electrons (krypton).

Teacher's Note:
1. Fill subshells in Aufbau order until 4p is full.
2. The next subshell is 5s, which is outside n = 4.

Question 10: In an atom, how many orbitals have the quantum numbers n = 3, l = 2 and ml = +2?
(a) 5
(b) 3
(c) 1
(d) 7
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. For l = 2 there are five d orbitals, each with a different m value.
2. Only one orbital has m = +2.

Teacher's Note:
1. Each (n, l, m) combination specifies exactly one orbital.

Question 11: If the de Broglie wavelength of a particle of mass m is 100 times its velocity, then its value in terms of its mass m and Planck's constant h is:
(a) \( \dfrac{1}{10}\sqrt{\dfrac{m}{h}} \)
(b) \( 10\sqrt{\dfrac{h}{m}} \)
(c) \( \dfrac{1}{10}\sqrt{\dfrac{h}{m}} \)
(d) \( 10\sqrt{\dfrac{m}{h}} \)
Show Answer & Explanation

Answer: (b) \( 10\sqrt{\dfrac{h}{m}} \)

Explanation:
1. Given \( \lambda = 100v \), so \( v = \lambda/100 \).
2. Substituting in \( \lambda = h/(mv) \) gives \( \lambda = \dfrac{100h}{m\lambda} \), so \( \lambda^2 = \dfrac{100h}{m} \) and \( \lambda = 10\sqrt{h/m} \).

Teacher's Note:
1. Eliminate velocity using the given relation.

Question 12: An electron in an atom jumps in such a way that its kinetic energy changes from x to x/4. The change in potential energy will be:
(a) \( +\dfrac32 x \)
(b) \( -\dfrac38 x \)
(c) \( +\dfrac34 x \)
(d) \( -\dfrac34 x \)
Show Answer & Explanation

Answer: (a) \( +\dfrac32 x \)

Explanation:
1. In Bohr orbits PE = -2KE.
2. Initial PE = -2x and final PE = -x/2.
3. Change in PE = -x/2 - (-2x) = +3x/2.

Teacher's Note:
1. Use potential energy = -2 times kinetic energy.

Question 13: What atomic number would an element X have to be so that its 4th orbit would fit inside the first Bohr orbit of the H atom (that is, have the same radius)?
(a) 3
(b) 4
(c) 16
(d) 25
Show Answer & Explanation

Answer: (c) 16

Explanation:
1. Radius is proportional to n²/Z.
2. Setting 4²/Z equal to 1²/1 gives Z = 16.

Teacher's Note:
1. Equate the radius expressions.

Question 14: Which combination of quantum numbers n, l, m and s for the electron in an atom does not provide a permissible solution of the wave equation?
(a) 3, 2, 1, +1/2
(b) 3, 1, 1, -1/2
(c) 3, 3, 1, -1/2
(d) 3, 2, -2, +1/2
Show Answer & Explanation

Answer: (c) 3, 3, 1, -1/2

Explanation:
1. For n = 3, l can only be 0, 1 or 2.
2. l = 3 is not allowed, so the combination (3, 3, 1, -1/2) is impossible.

Teacher's Note:
1. l must be less than n.

Question 15: Two fast moving particles X and Y are associated with de Broglie wavelengths 1 nm and 4 nm respectively. If the mass of X is nine times the mass of Y, the ratio of kinetic energies of X and Y would be:
(a) 3 : 1
(b) 9 : 1
(c) 5 : 12
(d) 16 : 9
Show Answer & Explanation

Answer: (d) 16 : 9

Explanation:
1. \( KE = \dfrac{h^2}{2m\lambda^2} \).
2. \( \dfrac{KE_X}{KE_Y} = \dfrac{m_Y}{m_X}\times\dfrac{\lambda_Y^2}{\lambda_X^2} = \dfrac19\times16 = \dfrac{16}{9} \).

Teacher's Note:
1. Express KE using the wavelength and mass.

Question 16: In Bohr's model, for a unielectronic species the time period of revolution is represented by Tn,Z, where n is the shell number and Z the atomic number. The value of T₁,₂ : T₂,₁ will be:
(a) 8 : 1
(b) 1 : 8
(c) 1 : 1
(d) 1 : 32
Show Answer & Explanation

Answer: (d) 1 : 32

Explanation:
1. Time period is proportional to n³/Z².
2. T₁,₂ is proportional to 1/4 and T₂,₁ to 8.
3. The ratio is (1/4) : 8 = 1 : 32.

Teacher's Note:
1. Use T proportional to n cubed over Z squared.

Question 17: The energy of the electron in the first orbit of He⁺ is -871.6 × 10⁻²⁰ J. The energy of the electron in the first orbit of H is:
(a) -871.6 × 10⁻²⁰ J
(b) -435.8 × 10⁻²⁰ J
(c) -217.9 × 10⁻²⁰ J
(d) -108.9 × 10⁻²⁰ J
Show Answer & Explanation

Answer: (c) -217.9 × 10⁻²⁰ J

Explanation:
1. Energy is proportional to Z².
2. He⁺ has Z = 2 and H has Z = 1, so E(H) = -871.6 × 10⁻²⁰/4 = -217.9 × 10⁻²⁰ J.

Teacher's Note:
1. Same orbit, so the ratio is the ratio of Z squared.

Question 18: The velocity of particle A is 0.1 m s⁻¹ and that of particle B is 0.05 m s⁻¹. If the mass of particle B is five times that of particle A, the ratio of the de Broglie wavelengths associated with the particles A and B is:
(a) 2 : 5
(b) 3 : 4
(c) 6 : 4
(d) 5 : 2
Show Answer & Explanation

Answer: (d) 5 : 2

Explanation:
1. \( \lambda = h/mv \).
2. \( \dfrac{\lambda_A}{\lambda_B} = \dfrac{m_Bv_B}{m_Av_A} = \dfrac{5\times0.05}{1\times0.1} = 2.5 \), which is 5 : 2.

Teacher's Note:
1. Take the ratio of momenta in reverse.

Question 19: An electron travels with a velocity of x m s⁻¹. For a proton to have the same de Broglie wavelength, the velocity will be approximately:
(a) \( \dfrac{1840}{x} \)
(b) \( \dfrac{x}{1840} \)
(c) 1840x
(d) x
Show Answer & Explanation

Answer: (b) \( \dfrac{x}{1840} \)

Explanation:
1. Same wavelength means mv is the same for both.
2. A proton is about 1840 times heavier than an electron, so its velocity is x/1840.

Teacher's Note:
1. Momentum must be equal.

Question 20: The total spin and magnetic moment for the atom with atomic number 24 are:
(a) \( \pm3, \sqrt{48} \) BM
(b) \( \pm3, \sqrt{35} \) BM
(c) \( \pm\tfrac32, \sqrt{48} \) BM
(d) \( \pm\tfrac32, \sqrt{35} \) BM
Show Answer & Explanation

Answer: (a) \( \pm3, \sqrt{48} \) BM

Explanation:
1. Chromium (Z = 24) has the configuration [Ar] 3d⁵ 4s¹ with six unpaired electrons.
2. Total spin = 6 × 1/2 = ±3.
3. Magnetic moment = \( \sqrt{n(n+2)} = \sqrt{6\times8} = \sqrt{48} \) BM.

Teacher's Note:
1. Chromium has an exceptional half-filled configuration.
2. Count unpaired electrons first.

Question 21: The ionisation potential of the hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by monochromatic light of energy 12.1 eV. The number of spectral lines emitted according to Bohr's theory will be:
(a) one
(b) two
(c) three
(d) four
Show Answer & Explanation

Answer: (c) three

Explanation:
1. Energy of level n = 3 is -1.51 eV, so the excitation from the ground state is 13.6 - 1.51 = 12.09 eV, which is about 12.1 eV.
2. The atoms are excited to n = 3, and the lines possible are 3 to 1, 3 to 2 and 2 to 1, which are three lines.

Teacher's Note:
1. 12.1 eV takes the electron to n = 3.
2. Number of lines = n(n - 1)/2.

Question 22: If the wavelength of electromagnetic radiation is increased to three times its original value, what is the percentage change in the frequency of the radiation?
(a) increases by 33%
(b) decreases by 33%
(c) increases by 66%
(d) decreases by 66%
Show Answer & Explanation

Answer: (d) decreases by 66%

Explanation:
1. Frequency is inversely proportional to wavelength, so the new frequency is 1/3 of the original.
2. The decrease is 100 × (1 - 1/3) = 66.7 percent, which is about 66%.

Teacher's Note:
1. Frequency = c / wavelength.

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