CBSE Class 11 Chemistry Structure of Atom MCQs Set 06

Answer: (b) electrons

Explanation:
1. Cathode rays are deflected towards the positive plate in an electric field, so they carry negative charge.
2. Their e/m value is the same whatever the gas or the electrode material, which identifies them as electrons.

Teacher's Note:
1. Cathode rays start at the cathode and travel to the anode.
2. Their behaviour does not depend on the gas in the tube.

Answer: (d) the nature of the gas in the discharge tube

Explanation:
1. Canal rays are positive ions formed when electrons knock electrons out of the gas atoms.
2. The mass of the ion depends on the gas, so e/m changes from gas to gas; it is highest for hydrogen.

Teacher's Note:
1. For cathode rays e/m is constant, but for canal rays it is not.
2. Hydrogen gives the largest e/m, which led to the idea of the proton.

Answer: (a) H⁺

Explanation:
1. Taking charge in units of e and mass in amu: H⁺ is 1/1 = 1, He²⁺ is 2/4 = 0.5 and He⁺ is 1/4 = 0.25.
2. The neutral H atom has no charge, so its ratio is zero.

Teacher's Note:
1. Divide the charge by the mass number for each species.
2. The lightest charged particle gives the highest ratio.

Answer: (c) Millikan, using the oil drop experiment

Explanation:
1. Millikan found that the charge on every oil drop was a whole-number multiple of 1.6 × 10⁻¹⁹ C.
2. Thomson measured only e/m; combining it with Millikan's charge gave the electron mass.

Teacher's Note:
1. Thomson gave e/m, Millikan gave e.
2. Mass = e ÷ (e/m) = 9.11 × 10⁻³¹ kg.

Answer: (b) James Chadwick

Explanation:
1. Chadwick bombarded a thin sheet of beryllium with α-particles in 1932 and obtained neutral particles.
2. The neutral particles had a mass almost equal to that of a proton.

Teacher's Note:
1. Thomson found the electron, Goldstein the canal rays and Rutherford the nucleus.
2. Remember: neutron, 1932, Chadwick.

Answer: (d) 1840

Explanation:
1. m(n) = 1.675 × 10⁻²⁷ kg and m(e) = 9.109 × 10⁻³¹ kg.
2. Ratio = 1.675 × 10⁻²⁷ ÷ 9.109 × 10⁻³¹ ≈ 1839, so about 1840.

Teacher's Note:
1. A proton or neutron is about 1840 times heavier than an electron.
2. This is why nearly all the mass of an atom is in the nucleus.

Answer: (a) 20

Explanation:
1. Number of neutrons = mass number − atomic number.
2. 37 − 17 = 20.

Teacher's Note:
1. Mass number A = protons + neutrons.
2. Atomic number Z = number of protons.

Answer: (c) 20, 20 and 18

Explanation:
1. Protons = Z = 20 and neutrons = 40 − 20 = 20.
2. The ion has lost 2 electrons, so electrons = 20 − 2 = 18.

Teacher's Note:
1. A positive charge means electrons have been lost.
2. The nucleus does not change when an ion forms.

Answer: (b) Na⁺

Explanation:
1. Ne has 10 electrons. Na⁺ has 11 − 1 = 10 electrons.
2. Cl⁻ and K⁺ have 18 electrons and Li⁺ has only 2.

Teacher's Note:
1. Isoelectronic species have the same number of electrons.
2. Count electrons, not protons.

Answer: (d) ⁴⁰Ar and ⁴⁰Ca

Explanation:
1. Isobars have the same mass number but different atomic numbers: ⁴⁰Ar (Z = 18) and ⁴⁰Ca (Z = 20).
2. The other three pairs have the same atomic number and different mass numbers, so they are isotopes.

Teacher's Note:
1. Isotopes: same Z. Isobars: same A.
2. Isobars are atoms of different elements.

Answer: (a) the plum pudding (watermelon) model

Explanation:
1. Thomson pictured a positive sphere with electrons embedded in it like plums in a pudding or seeds in a watermelon.
2. The planetary and nuclear model belongs to Rutherford.

Teacher's Note:
1. Thomson's model has no nucleus.
2. It could not explain the α-particle scattering results.

Answer: (c) most of the space inside an atom is empty

Explanation:
1. If the atom were a solid positive sphere, α-particles would be deflected much more often.
2. Passing straight through shows that most of the atom is empty space, and the positive charge is packed into a tiny nucleus.

Teacher's Note:
1. Very few particles bounced back, which shows a tiny but dense nucleus.
2. Undeflected means nothing was in the way.

Answer: (d) 10⁵

Explanation:
1. Ratio = 10⁻¹⁰ ÷ 10⁻¹⁵ = 10⁵.
2. So if the nucleus were a cricket ball, the atom would have a radius of several kilometres.

Teacher's Note:
1. Subtract the powers of ten when dividing.
2. The volume ratio would be (10⁵)³ = 10¹⁵.

Answer: (b) It could not explain the stability of the atom

Explanation:
1. An electron moving round the nucleus is accelerated and, by classical theory, should radiate energy continuously.
2. It would then spiral into the nucleus, but atoms are stable. The model also said nothing about how electrons are arranged or their energies.

Teacher's Note:
1. Stability and electronic arrangement are the two main drawbacks.
2. Maxwell's theory is the reason for the first one.

Answer: (a) 40

Explanation:
1. The neutral atom has 18 + 2 = 20 electrons, so Z = 20.
2. Mass number = 20 + 20 = 40.

Teacher's Note:
1. Add back the electrons lost to find Z.
2. A = Z + number of neutrons.

Answer: (d) 6 × 10¹⁴ Hz

Explanation:
1. λ = 5000 Å = 5 × 10⁻⁷ m.
2. ν = c/λ = 3 × 10⁸ ÷ 5 × 10⁻⁷ = 6 × 10¹⁴ Hz.

Teacher's Note:
1. Convert angstrom to metre first (1 Å = 10⁻¹⁰ m).
2. ν = c/λ.

Answer: (b) 2.5 × 10⁶ m⁻¹

Explanation:
1. Wave number = 1/λ = 1 ÷ 4 × 10⁻⁷ m.
2. = 2.5 × 10⁶ m⁻¹.

Teacher's Note:
1. Wave number is the number of waves per metre.
2. Do not confuse it with frequency, which is c/λ.

Answer: (c) radio waves < infrared < visible < X-rays

Explanation:
1. Energy E = hν increases with frequency.
2. Frequency increases in the order radio waves, infrared, visible, X-rays.

Teacher's Note:
1. Short wavelength means high frequency and high energy.
2. Gamma rays are at the high-energy end of the spectrum.

Answer: (a) 3.98 × 10⁻¹⁹ J

Explanation:
1. E = hν = 6.626 × 10⁻³⁴ × 6 × 10¹⁴.
2. E = 3.98 × 10⁻¹⁹ J.

Teacher's Note:
1. Use E = hν when frequency is given.
2. Use E = hc/λ when wavelength is given.

Answer: (d) in discrete packets, as whole-number multiples of hν

Explanation:
1. According to Planck, radiant energy is transferred in packets called quanta (photons for light).
2. The energy of one quantum is hν, so the total energy is nhν with n = 1, 2, 3 and so on.

Teacher's Note:
1. Quantisation means energy comes in steps, not a smooth flow.
2. h = 6.626 × 10⁻³⁴ J s.