JEE Chemistry Structure Of Atom MCQs Set 15

Answer: (a) 333 m

Explanation:
1. \( \lambda = c/\nu = 3\times10^{8}/(9\times10^{5}) \) m.
2. Wavelength = 333 m.

Teacher's Note:
1. Convert kHz to Hz (1 kHz = 1000 Hz) before dividing.
2. Radio waves have very long wavelengths.

Answer: (b) \( 3.62\times10^{-4 \)} J

Explanation:
1. Energy of one photon = hν = 6.626 x 10^-34 x 9.09 x 10⁵ = 6.02 x 10^-28 J.
2. For 1 mole: 6.02 x 10^-28 x 6.022 x 10²³ = \( 3.62\times10^{-4 \)} J.

Teacher's Note:
1. The first option is the energy of one photon, not one mole.
2. Multiply by Avogadro's number for a mole of photons.

Answer: (a) 178 pm

Explanation:
1. KE = kT = 1.38 x 10^-23 x 300 = 4.14 x 10^-21 J.
2. \( \lambda = h/\sqrt{2m\,KE \)} with m = 1.675 x 10^-27 kg gives 1.78 x 10^-10 m.
3. Wavelength = 178 pm.

Teacher's Note:
1. Use the de Broglie relation with momentum written in terms of kinetic energy.
2. 1 pm = 10^-12 m.

Answer: (c) \( \dfrac{h^2}{32\pi^2 m a_0^2 \)}

Explanation:
1. For orbit n, r = n²a₀ (Z = 1), so r = 4a₀ for n = 2.
2. Bohr quantisation: mvr = nh/2π, so v = 2h/(2π m x 4a₀) = h/(4π m a₀).
3. KE = (1/2)mv² = \( h^2/(32\pi^2 m a_0^2) \).

Teacher's Note:
1. Combine r = n²a₀ with mvr = nh/2π to get v.
2. Substitute into (1/2)mv².

Answer: (b) -27.2 eV

Explanation:
1. Total energy = -13.6 x Z²/n² = -13.6 x 4/4 = -13.6 eV.
2. In a Bohr orbit PE = 2 x (total energy) = -27.2 eV.

Teacher's Note:
1. Remember: KE = -E, PE = 2E for the Bohr model.
2. Total energy is the sum of KE and PE.

Answer: (b) \( 6\pi X \)

Explanation:
1. Radius of third orbit = 9X.
2. Wavelength = circumference / n = 2π(9X)/3 = 6πX.

Teacher's Note:
1. An electron orbit holds n complete de Broglie waves.
2. Circumference = n x wavelength.

Answer: (d) \( 2.5\dfrac{h}{\pi \)}

Explanation:
1. Angular momentum = nh/2π with n = 5.
2. = 5h/2π = 2.5 h/π.

Teacher's Note:
1. Angular momentum in Bohr's model is quantised in units of h/2π.
2. Do not confuse with orbital angular momentum \( \sqrt{l(l+1)}\,h/2\pi \).

Answer: (d) 6

Explanation:
1. Fe: [Ar] 3d⁶ 4s².
2. Fe²⁺ loses the two 4s electrons, leaving 3d⁶.
3. Number of d electrons = 6.

Teacher's Note:
1. Transition metals lose 4s electrons first when forming cations.
2. Fe3+ would have 5 d electrons.

Answer: (a) Neutron particle emission

Explanation:
1. Isotopes have the same atomic number but different mass number.
2. Neutron emission changes only the mass number. Positron emission lowers Z by 1, \( \alpha \) emission lowers Z by 2, \( \beta \) emission raises Z by 1.
3. Isotope is produced by neutron emission.

Teacher's Note:
1. Only a change in A with Z fixed gives an isotope.
2. A beta decay gives an isobar.

Answer: (b) 0.400 nm

Explanation:
1. \( \lambda = h/mv = 6.626\times10^{-34}/(1.67\times10^{-27}\times10^{3}) \).
2. = 3.97 x 10^-10 m = 0.397 nm, i.e. 0.400 nm.

Teacher's Note:
1. Use SI units: kg and m/s.
2. 1 nm = 10^-9 m.

Answer: (c) 5

Explanation:
1. For a given \( \ell \) the number of orbitals is 2\( \ell \) + 1.
2. For \( \ell \) = 2 this is 5 (the 3d orbitals).

Teacher's Note:
1. m takes values from -\( \ell \) to +\( \ell \).
2. n only fixes the shell; the count depends on \( \ell \).

Answer: (a) \( -5.44\times10^{-19 \)} J

Explanation:
1. E₂ = -13.6/4 = -3.4 eV.
2. -3.4 x 1.602 x 10^-19 = \( -5.44\times10^{-19 \)} J per atom.

Teacher's Note:
1. Energy per atom is of the order of 10^-19 J.
2. 1 eV = 1.602 x 10^-19 J.

Answer: (b) \( 5\times10^{-27 \)}

Explanation:
1. \( \Delta v = h/(4\pi m\,\Delta x) \) with m = 5 x 10^-4 kg.
2. = 6.626 x 10^-34 / (4π x 5 x 10^-4 x 2 x 10^-5) = 5 x 10^-27 m/s.

Teacher's Note:
1. Convert grams to kilograms first.
2. The uncertainty is negligible for large objects.

Answer: (c) He

Explanation:
1. \( \ce{Li+ \)} has 3 - 1 = 2 electrons.
2. H has 1, \( \ce{Na^{10+ \)}} has 1, He has 2, \( \ce{He+ \)} has 1 electron.
3. The two-electron species is He.

Teacher's Note:
1. Isoelectronic species have the same number of electrons.
2. Count electrons = Z - charge.

Answer: (d) 4.77 A

Explanation:
1. r_n = n² x 0.53 A for hydrogen.
2. r₃ = 9 x 0.53 = 4.77 A.

Teacher's Note:
1. Radius grows as n².
2. For hydrogen-like ions divide by Z.

Answer: (d) All of the above

Explanation:
1. \( \ell \) runs from 0 to n - 1.
2. For n = 3 the values are 0, 1 and 2: all of the above.

Teacher's Note:
1. Maximum \( \ell \) is n - 1.
2. These correspond to 3s, 3p and 3d.

Answer: (a) 1 : 2

Explanation:
1. \( \lambda = h/\sqrt{2meV \)}, so \( \lambda \propto 1/\sqrt V \).
2. \( \lambda_{400}:\lambda_{100} = \sqrt{100}:\sqrt{400} = \) 1 : 2.

Teacher's Note:
1. Wavelength falls as the accelerating voltage rises.
2. Take square roots of the voltage ratio.

Answer: (a) 6

Explanation:
1. Configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s².
2. Each s subshell gives one such electron (4 in total: 1s, 2s, 3s, 4s); 2p and 3p each give one with m = 0 and s = +1/2 (2 more).
3. Total = 6.

Teacher's Note:
1. Within a p subshell only one orbital has m = 0.
2. Each orbital holds one electron with s = +1/2.