JEE Chemistry Structure of the Atom MCQs Set 06

Answer: (b) \(\gamma\)-rays

Explanation:
1. Electronic transitions involve energies of a few eV to keV, giving infrared, visible, UV (and X-rays for inner shells).
2. \(\gamma\)-rays come from nuclear transitions, not electronic ones.

Teacher's Note:
1. Different regions of the spectrum come from different processes.
2. \(\gamma\)-rays are of nuclear origin.

Answer: (c) \(\ce{He+}\)

Explanation:
1. Second Balmer transition: 4 to 2. \(\Delta E = 13.6 Z^2\left(\frac14 - \frac1{16}\right) = 2.55 Z^2\) eV.
2. Setting this equal to 10.2 gives \(Z^2 = 4\), so Z = 2.
3. The one-electron species with Z = 2 is \(\ce{He+}\).

Teacher's Note:
1. Only one-electron species follow the Bohr formula.
2. \(\ce{Li+}\) has two electrons, so the formula does not apply to it.

Answer: (c) \(8.22 \times 10^{14}\) s\(^{-1}\)

Explanation:
1. The limiting line has \(n_2 = \infty\) and \(n_1 = 2\).
2. \(\nu = Rc\left(\frac14 - 0\right) = \frac{3.29 \times 10^{15}}{4} = 8.22 \times 10^{14}\) s\(^{-1}\).

Teacher's Note:
1. Rc equals \(3.29 \times 10^{15}\) s\(^{-1}\).
2. Frequency cannot be negative.

Answer: (b) \(\frac{36x}{5}\)

Explanation:
1. The shortest wavelength of Lyman: \(\frac1x = R\), so \(x = \frac1R\).
2. First Balmer line: \(\frac1\lambda = R\left(\frac14 - \frac19\right) = \frac{5R}{36}\), so \(\lambda = \frac{36}{5R}\).
3. \(\lambda = \frac{36x}{5}\).

Teacher's Note:
1. Express the answer in terms of 1/R.
2. The Lyman series limit gives x = 1/R.

Answer: (a) \(\frac{5R}{36}\) cm\(^{-1}\)

Explanation:
1. The first Balmer line is 3 to 2.
2. \(\bar\nu = R\left(\frac1{2^2} - \frac1{3^2}\right) = R\left(\frac14 - \frac19\right) = \frac{5R}{36}\).

Teacher's Note:
1. Wave number is 1/wavelength.
2. 3R/4 is the Lyman first line and 7R/144 the Paschen first line.

Answer: (d) 4 to 2

Explanation:
1. H Lyman first line: \(\frac1\lambda = R\left(1 - \frac14\right) = \frac{3R}{4}\).
2. For \(\ce{He+}\) (Z = 2): \(\frac1\lambda = 4R\left(\frac1{n_1^2} - \frac1{n_2^2}\right)\), so \(\frac1{n_1^2} - \frac1{n_2^2} = \frac3{16}\).
3. n = 4 to 2: \(\frac14 - \frac1{16} = \frac3{16}\). This matches.

Teacher's Note:
1. Equate the wave numbers of the two transitions.
2. Check that 5 to 3 gives \(\frac19 - \frac1{25} = 0.071\), which is not 3/16.

Answer: (a) 6

Explanation:
1. Levels involved: n = 6, 5, 4, 3, 2, so the total number of possible lines is \(\frac{5 \times 4}{2} = 10\).
2. Lines ending at n = 2 (Balmer) are 6 to 2, 5 to 2, 4 to 2 and 3 to 2, which is 4 lines.
3. Remaining lines = 10 - 4 = 6.

Teacher's Note:
1. Total lines between levels = \(\frac{N(N-1)}{2}\) for N levels.
2. Subtract the lines that end at n = 2.

Answer: (c) \(1 + 2 + 3 + \dots + (x - 1)\)

Explanation:
1. Number of lines between x levels = \(\frac{x(x-1)}{2}\).
2. This equals \(1 + 2 + 3 + \dots + (x - 1)\).

Teacher's Note:
1. Each pair of levels gives one line.
2. The sum of the first (x - 1) natural numbers is \(\frac{x(x-1)}{2}\).

Answer: (a) 1

Explanation:
1. Lines in the UV belong to the Lyman series (ending at n = 1). Three Lyman lines mean the electron started at n = 4 (4 to 1, 3 to 1, 2 to 1).
2. Other possible lines from n = 4: 4 to 3 (infrared), 4 to 2 and 3 to 2 (visible).
3. So only one infrared line.

Teacher's Note:
1. Series by region: Lyman is ultraviolet, Balmer is visible and Paschen is infrared.
2. List all lines from the starting level.

Answer: (a) 3

Explanation:
1. Absorption lines start from the ground state n = 1.
2. The possible lines are 1 to 2, 1 to 3 and 1 to 4, which is 3 lines.

Teacher's Note:
1. At ordinary temperature almost all atoms are in the ground state.
2. Excited-state absorptions are not normally observed.

Answer: (a) 0

Explanation:
1. The smallest excitation energy of hydrogen is 10.2 eV (n = 1 to n = 2).
2. An atom can absorb only an energy that exactly matches a gap.
3. 9.9 eV is not a gap, so no absorption and no emission happens.

Teacher's Note:
1. Atomic energy levels are quantised.
2. Energy below the first excitation energy cannot be absorbed by a photon.

Answer: (c) \(9.28 \times 10^{-8}\) m

Explanation:
1. \(\lambda = \frac{h}{\sqrt{2mK}}\).
2. \(2mK = 2 \times 9.1 \times 10^{-31} \times 2.8 \times 10^{-23} = 5.096 \times 10^{-53}\), whose square root is \(7.14 \times 10^{-27}\).
3. \(\lambda = \frac{6.626 \times 10^{-34}}{7.14 \times 10^{-27}} = 9.28 \times 10^{-8}\) m.

Teacher's Note:
1. Momentum is \(p = \sqrt{2mK}\).
2. Wavelength is h divided by momentum.

Answer: (a) equal to \(n\lambda\)

Explanation:
1. For a stable (standing wave) orbit the circumference must contain a whole number of wavelengths.
2. So \(2\pi r = n\lambda\).

Teacher's Note:
1. Waves that do not fit would interfere destructively.
2. This is de Broglie's explanation of Bohr's quantisation.

Answer: (c) 0.175 cm

Explanation:
1. \(\Delta v = 0.011\% \text{ of } 3 \times 10^{4} = 3.3\) cm/s.
2. \(\Delta x = \frac{h}{4\pi m \Delta v} = \frac{6.626 \times 10^{-27}}{4\pi \times 9.1 \times 10^{-28} \times 3.3}\).
3. The denominator is \(3.77 \times 10^{-26}\), so \(\Delta x = 0.176\) cm, about 0.175 cm.

Teacher's Note:
1. Use the uncertainty relation \(\Delta x\,\Delta p = \frac{h}{4\pi}\) with \(\Delta p = m\Delta v\).
2. Keep CGS units consistently.

Answer: (b) a motor car

Explanation:
1. The uncertainty \(\Delta x \Delta p \ge \frac{h}{4\pi}\) is tiny in magnitude.
2. For a heavy macroscopic object like a car, the uncertainties in position and velocity are far below anything measurable.
3. For electrons, protons and neutrons the effect is significant.

Teacher's Note:
1. The effect matters only for microscopic particles.
2. h is very small, so large masses show no noticeable uncertainty.

Answer: (a) \(6.6252 \times 10^{-19}\) g cm/s

Explanation:
1. \(p = \frac{h}{\lambda}\).
2. \(\lambda = 1\) Å \(= 10^{-8}\) cm.
3. \(p = \frac{6.6252 \times 10^{-27}}{10^{-8}} = 6.6252 \times 10^{-19}\) g cm/s.

Teacher's Note:
1. Convert 1 Å to \(10^{-8}\) cm for CGS units.
2. Momentum is inversely proportional to the wavelength.

Answer: (d) \(3.746 \times 10^{-36}\) kg

Explanation:
1. \(m = \frac{h}{\lambda c}\).
2. \(\lambda = 5894 \times 10^{-10} = 5.894 \times 10^{-7}\) m.
3. \(m = \frac{6.6252 \times 10^{-34}}{5.894 \times 10^{-7} \times 3 \times 10^{8}} = 3.746 \times 10^{-36}\) kg.

Teacher's Note:
1. The equivalent mass of a photon comes from \(E = mc^2 = \frac{hc}{\lambda}\).
2. A photon has no rest mass.

Answer: (d) \(\ce{SO2}\)

Explanation:
1. \(\lambda = \frac{h}{mv}\), so at the same velocity the heaviest particle has the least wavelength.
2. Masses: electron is smallest, proton is 1 u, \(\ce{CO2}\) is 44 u and \(\ce{SO2}\) is 64 u.
3. So \(\ce{SO2}\) has the least wavelength.

Teacher's Note:
1. At equal speed, wavelength is inversely proportional to mass.
2. The condition of equal velocity must be stated for a valid comparison.

Answer: (a) 1

Explanation:
1. n = 3 and \(\ell\) = 2 denote the 3d subshell.
2. m = +2 selects one particular d orbital.
3. So only 1 orbital fits these quantum numbers.

Teacher's Note:
1. n, \(\ell\) and m together fix a single orbital.
2. Spin quantum number distinguishes electrons within it.

Answer: (d) 9

Explanation:
1. The number of orbitals in a shell is \(n^2\).
2. For n = 3 it is 9 (one 3s, three 3p and five 3d).

Teacher's Note:
1. Shell n holds \(n^2\) orbitals and \(2n^2\) electrons.
2. List the subshells to check: 1 + 3 + 5 = 9.

Answer: (c) n = 2, \(\ell\) = 1

Explanation:
1. The number of angular (planar) nodes equals \(\ell\).
2. Only \(\ell\) = 1 (p orbital) has a nodal plane among the options.
3. s orbitals (\(\ell\) = 0) have no nodal plane.

Teacher's Note:
1. Angular nodes = \(\ell\) and radial nodes = n - \(\ell\) - 1.
2. A p orbital has one nodal plane through the nucleus.

Answer: (a) 11

Explanation:
1. The subshells s, p, d, f, g, h correspond to \(\ell\) = 0, 1, 2, 3, 4, 5.
2. The h subshell has \(\ell\) = 5, so \(2\ell + 1 = 11\) orbitals.

Teacher's Note:
1. Number of orbitals in a subshell is \(2\ell + 1\).
2. After f the letters continue g, h, i alphabetically.

Answer: (c) 3

Explanation:
1. An orbital is defined by n (size/energy), \(\ell\) (shape) and m (orientation).
2. The spin quantum number is needed only to describe the electron itself, not the orbital.
3. So 3 quantum numbers are required.

Teacher's Note:
1. Four quantum numbers describe an electron, three describe an orbital.
2. The spin quantum number distinguishes the two electrons in one orbital.

Answer: (a) s < p < d < f

Explanation:
1. Within a shell, penetration of the nucleus decreases from s to f.
2. Greater penetration means lower energy, so s < p < d < f.

Teacher's Note:
1. This ordering applies to multi-electron atoms.
2. In hydrogen all subshells of a shell have equal energy.

Answer: (a) \(\sqrt{s(s+1)}\,\frac{h}{2\pi}\)

Explanation:
1. Spin angular momentum \(= \sqrt{s(s+1)}\,\frac{h}{2\pi}\).
2. With s = 1/2 this is \(\frac{\sqrt3}{2}\frac{h}{2\pi}\).

Teacher's Note:
1. Orbital angular momentum is \(\sqrt{\ell(\ell+1)}\frac{h}{2\pi}\).
2. The spin formula has the same form with s in place of \(\ell\).

Answer: (c) 2

Explanation:
1. Specifying n = 6, \(\ell\) = 1 and m = 0 identifies a single 6p orbital.
2. One orbital holds at most 2 electrons (opposite spins).

Teacher's Note:
1. Every orbital holds a maximum of two electrons.
2. Pauli exclusion principle limits it.