CBSE Class 11 Chemistry Structure of Atom MCQs Set 07

Practice MCQs for Class 11 Chemistry Chapter 02 Structure of Atom

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Multiple Choice Questions

Question 1: The minimum frequency of light below which no photoelectrons are emitted from a metal surface is called:
(a) the critical frequency
(b) the threshold frequency
(c) the Rydberg frequency
(d) the resonance frequency
Show Answer & Explanation

Answer: (b) the threshold frequency

Explanation:
1. Each metal has a characteristic minimum frequency ν0.
2. If the incident frequency is below ν0, no electrons are ejected however intense the light is or however long it falls on the surface.

Teacher's Note:
1. The work function is W0 = hν0.
2. Different metals have different threshold frequencies.

Question 2: In the photoelectric effect, the kinetic energy of the emitted electrons does not depend on:
(a) the frequency of the incident light
(b) the work function of the metal
(c) the intensity of the incident light
(d) the wavelength of the incident light
Show Answer & Explanation

Answer: (c) the intensity of the incident light

Explanation:
1. K.E. = hν − hν0, so it depends on the frequency (or wavelength) and on the metal.
2. Intensity only changes the number of photons, and so the number of electrons ejected, not their kinetic energy.

Teacher's Note:
1. Brighter light gives more electrons, not faster ones.
2. Higher frequency gives faster electrons.

Question 3: The threshold frequency of a metal is 5 × 1014 s−1. The kinetic energy of electrons emitted by light of frequency 8 × 1014 s−1 is (h = 6.626 × 10−34 J s):
(a) 1.99 × 10−19 J
(b) 5.30 × 10−19 J
(c) 3.31 × 10−19 J
(d) 0 J
Show Answer & Explanation

Answer: (a) 1.99 × 10−19 J

Explanation:
1. K.E. = h(ν − ν0) = 6.626 × 10−34 × (8 − 5) × 1014.
2. K.E. = 6.626 × 10−34 × 3 × 1014 = 1.99 × 10−19 J.

Teacher's Note:
1. Subtract the threshold frequency from the incident frequency.
2. K.E. is zero only when ν equals ν0.

Question 4: A 100 W bulb emits light of wavelength 400 nm. The approximate number of photons emitted per second is:
(a) 2 × 1018
(b) 2 × 1015
(c) 2 × 1025
(d) 2 × 1020
Show Answer & Explanation

Answer: (d) 2 × 1020

Explanation:
1. Energy of one photon = hc/λ = 6.626 × 10−34 × 3 × 108 ÷ 4 × 10−7 = 4.97 × 10−19 J.
2. Photons per second = 100 ÷ 4.97 × 10−19 = 2.0 × 1020.

Teacher's Note:
1. Power is energy per second (1 W = 1 J s−1).
2. Divide the total energy per second by the energy of one photon.

Question 5: The number of photons of wavelength 4000 Å needed to supply 1 J of energy is about:
(a) 2 × 1018
(b) 2 × 109
(c) 2 × 10−18
(d) 2 × 1027
Show Answer & Explanation

Answer: (a) 2 × 1018

Explanation:
1. Energy of one photon = hc/λ = 6.626 × 10−34 × 3 × 108 ÷ 4 × 10−7 = 4.97 × 10−19 J.
2. Number = 1 ÷ 4.97 × 10−19 = 2.0 × 1018.

Teacher's Note:
1. Convert 4000 Å to 4 × 10−7 m.
2. Number of photons = total energy ÷ energy per photon.

Question 6: The spectrum obtained when light from a hot gaseous atom is analysed directly is:
(a) a continuous spectrum
(b) a line absorption spectrum only
(c) a line emission spectrum
(d) a band absorption spectrum
Show Answer & Explanation

Answer: (c) a line emission spectrum

Explanation:
1. Excited atoms return to lower levels and emit radiation of definite wavelengths only.
2. This gives bright lines on a dark background, a line emission spectrum, which is a fingerprint of the element.

Teacher's Note:
1. White light from a bulb gives a continuous spectrum.
2. Dark lines on a bright background form an absorption spectrum.

Question 7: The energy of the electron in the third orbit (n = 3) of a hydrogen atom is (RH = 2.18 × 10−18 J):
(a) −6.54 × 10−18 J
(b) −2.42 × 10−19 J
(c) −7.26 × 10−19 J
(d) +2.42 × 10−19 J
Show Answer & Explanation

Answer: (b) −2.42 × 10−19 J

Explanation:
1. En = −2.18 × 10−18 ÷ n2 J.
2. For n = 3, E = −2.18 × 10−18 ÷ 9 = −2.42 × 10−19 J.

Teacher's Note:
1. The energy is always negative for a bound electron.
2. Divide by n squared, not by n.

Question 8: The energy of the electron in the second Bohr orbit of He+ (Z = 2) is:
(a) −5.45 × 10−19 J
(b) −8.72 × 10−18 J
(c) −1.09 × 10−18 J
(d) −2.18 × 10−18 J
Show Answer & Explanation

Answer: (d) −2.18 × 10−18 J

Explanation:
1. En = −2.18 × 10−18 × Z2/n2 J.
2. For Z = 2 and n = 2: E = −2.18 × 10−18 × 4/4 = −2.18 × 10−18 J.

Teacher's Note:
1. Z2/n2 = 1 here, so the energy equals that of H in n = 1.
2. Always substitute both Z and n.

Question 9: The radius of the third Bohr orbit of hydrogen is (a0 = 52.9 pm):
(a) 159 pm
(b) 17.6 pm
(c) 476 pm
(d) 52.9 pm
Show Answer & Explanation

Answer: (c) 476 pm

Explanation:
1. rn = 52.9 × n2/Z pm.
2. For n = 3 and Z = 1: r = 52.9 × 9 = 476 pm (0.476 nm).

Teacher's Note:
1. The radius grows as n2.
2. r1 = 52.9 pm is the Bohr radius.

Question 10: The radius of the second Bohr orbit of Li2+ is:
(a) 70.5 pm
(b) 211.6 pm
(c) 17.6 pm
(d) 158.7 pm
Show Answer & Explanation

Answer: (a) 70.5 pm

Explanation:
1. rn = 52.9 × n2/Z pm.
2. For n = 2 and Z = 3: r = 52.9 × 4/3 = 70.5 pm.

Teacher's Note:
1. A larger Z shrinks the orbit.
2. Do not forget to divide by Z.

Question 11: According to Bohr's model, the angular momentum of an electron in the third orbit is:
(a) h/2π
(b) 9h/2π
(c) h/6π
(d) 3h/2π
Show Answer & Explanation

Answer: (d) 3h/2π

Explanation:
1. Angular momentum is quantised: mvr = nh/2π.
2. For n = 3, it is 3h/2π.

Teacher's Note:
1. The angular momentum is an integral multiple of h/2π.
2. It is proportional to n, not n2.

Question 12: Which series of the hydrogen spectrum lies in the visible region?
(a) Balmer series
(b) Lyman series
(c) Paschen series
(d) Brackett series
Show Answer & Explanation

Answer: (a) Balmer series

Explanation:
1. Balmer lines arise when electrons fall to n = 2 from higher levels (n = 3, 4, 5 ...).
2. Lyman lines are in the ultraviolet; Paschen, Brackett and Pfund lines are in the infrared.

Teacher's Note:
1. Lyman: n = 1, Balmer: n = 2, Paschen: n = 3, Brackett: n = 4, Pfund: n = 5.
2. Only Balmer is visible.

Question 13: The maximum number of different spectral lines that can be emitted when an electron of a hydrogen atom falls from n = 5 to the ground state is:
(a) 4
(b) 10
(c) 5
(d) 15
Show Answer & Explanation

Answer: (b) 10

Explanation:
1. Number of lines = n(n − 1)/2 for a fall from level n to the ground state.
2. For n = 5: 5 × 4/2 = 10.

Teacher's Note:
1. Count all possible jumps, not only those to n = 1.
2. For n = 6 the number would be 15.

Question 14: The wavelength of the first line of the Balmer series of hydrogen (n = 3 to n = 2) is about (R = 1.097 × 107 m−1):
(a) 486 nm
(b) 122 nm
(c) 656 nm
(d) 1875 nm
Show Answer & Explanation

Answer: (c) 656 nm

Explanation:
1. 1/λ = R(1/22 − 1/32) = 1.097 × 107 × 5/36 = 1.524 × 106 m−1.
2. λ = 6.56 × 10−7 m = 656 nm.

Teacher's Note:
1. Use n1 = 2 and n2 = 3.
2. This red line is called H-alpha.

Question 15: The energy of the electron in a hydrogen atom is negative because:
(a) the electron carries a negative charge
(b) the energy is zero near the nucleus
(c) of repulsion between electrons
(d) the electron is bound; its energy is lower than that of a free electron at rest, taken as zero
Show Answer & Explanation

Answer: (d) the electron is bound; its energy is lower than that of a free electron at rest, taken as zero

Explanation:
1. A free electron at rest infinitely far away (n = ∞) is assigned zero energy.
2. As the electron comes closer, energy is released and its energy falls below zero, so the bound electron has negative energy. The most negative value is at n = 1.

Teacher's Note:
1. Negative energy means a stable, bound state.
2. The ground state is the lowest, most negative level.

Question 16: The ionisation energy of the He+ ion (in its ground state) is:
(a) 54.4 eV
(b) 13.6 eV
(c) 27.2 eV
(d) 122.4 eV
Show Answer & Explanation

Answer: (a) 54.4 eV

Explanation:
1. Ionisation energy = 13.6 × Z2 eV for a one-electron species in n = 1.
2. For He+, Z = 2: 13.6 × 4 = 54.4 eV.

Teacher's Note:
1. The energy needed is proportional to Z2.
2. Li2+ would need 13.6 × 9 = 122.4 eV.

Question 17: Which of the following is a limitation of Bohr's model?
(a) It cannot explain the line spectrum of hydrogen
(b) It cannot explain the spectra of atoms with more than one electron
(c) It does not use quantised energy levels
(d) It does not give the radius of the first orbit
Show Answer & Explanation

Answer: (b) It cannot explain the spectra of atoms with more than one electron

Explanation:
1. Bohr's model works well for hydrogen and hydrogen-like ions only.
2. It cannot explain multi-electron spectra, the splitting of lines in magnetic and electric fields, or chemical bonding, and it ignores the dual nature of matter and the uncertainty principle.

Teacher's Note:
1. It gave the correct hydrogen spectrum and orbit radius.
2. Its fixed circular orbits conflict with the uncertainty principle.

Question 18: The energy released when an electron in a hydrogen atom falls from n = 4 to n = 2 is:
(a) 2.04 × 10−19 J
(b) 1.36 × 10−18 J
(c) 4.09 × 10−19 J
(d) 4.09 × 10−18 J
Show Answer & Explanation

Answer: (c) 4.09 × 10−19 J

Explanation:
1. ΔE = 2.18 × 10−18 × (1/22 − 1/42) J.
2. ΔE = 2.18 × 10−18 × 3/16 = 4.09 × 10−19 J.

Teacher's Note:
1. Take 1/nf2 − 1/ni2, with nf the lower level.
2. The emitted photon has this energy.

Question 19: The speed of the electron in the first Bohr orbit of hydrogen is about:
(a) 3.0 × 108 m s−1
(b) 2.18 × 108 m s−1
(c) 6.6 × 103 m s−1
(d) 2.18 × 106 m s−1
Show Answer & Explanation

Answer: (d) 2.18 × 106 m s−1

Explanation:
1. The speed in the nth orbit is vn = 2.188 × 106 × Z/n m s−1.
2. For n = 1 and Z = 1 this is 2.188 × 106 m s−1, about 1/137 of the speed of light.

Teacher's Note:
1. Speed falls as n increases.
2. It is much less than the speed of light.

Question 20: Which transition in the He+ ion gives a line of the same wavelength as the n = 2 to n = 1 line of hydrogen?
(a) n = 4 to n = 2
(b) n = 3 to n = 1
(c) n = 3 to n = 2
(d) n = 2 to n = 1
Show Answer & Explanation

Answer: (a) n = 4 to n = 2

Explanation:
1. For He+, 1/λ = R × 4 × (1/n12 − 1/n22). For hydrogen n = 2 to n = 1: 1/λ = R × 3/4.
2. For n = 4 to n = 2 in He+: 4 × (1/4 − 1/16) = 4 × 3/16 = 3/4. The two match.

Teacher's Note:
1. The factor Z2 = 4 makes He+ levels four times as far apart.
2. Equate the two expressions and test the options.

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