Structure of atom practice · Mixed difficulty

Option B (Stark effect) is correct because the Stark effect describes the splitting of spectral lines due to an external electric field.

Option A (Momentum of H atom when electrons return from n=2 to n=1: $\frac{{3Rh}}{{4}}$) is correct because the momentum of the emitted photon equals the momentum of the recoiling H atom.

Option B ($\mathrm{\;}_{33}^{77}\mathrm{As}$) is correct because isotones have the same number of neutrons. $^{76}_{32}Ge$ has 76-32 = 44 neutrons. $^{77}_{33}As$ has 77-33 = 44 neutrons.

Option C ($\sqrt {\left( {\frac{{2eV}}{m}} \right)}$) is correct because it is derived from equating the kinetic energy gained by the electron to the work done by the electric field.

Option D (10.2) is correct because it represents the energy difference between the n=1 and n=2 energy levels in a hydrogen atom.

Option C (Both S and E are correct and E is correct explanation of S) is correct because the Humphrey series, with $n_1 = 6$, involves transitions to a higher energy level, resulting in lower energy radiations.

Option B (Wavelength and frequency) is correct because the product of wavelength and frequency equals the speed of light ($c = \lambda v$).

Option A ($6.02 \times {10^{23}}$) is correct because it represents the number of water molecules in 1.8 mL of water multiplied by the number of electrons in each water molecule.

Option B is correct because the mass of a nucleus is always less than the sum of the masses of its individual nucleons due to the binding energy.

Option B (Isobars and Isotones) is correct because $_{32}Ge^{76}$ and $_{34}Se^{76}$ have the same mass number (isobars), and $_{14}Si^{30}$ and $_{16}S^{32}$ have the same number of neutrons (isotones).

Option A (1∶1) is correct because the radius of the orbit depends on the principal quantum number and atomic number, not the mass of the nucleus.

The correct answer is D (11 or 12) because the quantum numbers given can correspond to either sodium (Na) or magnesium (Mg).

Option B (Wavelength of an electron and its momentum) is correct because the de Broglie equation directly relates wavelength and momentum.

Option B (${E_1} = 2{E_2}$) is correct because energy is inversely proportional to wavelength, so doubling the wavelength halves the energy.

Option D ($2.64 \times {10^{ - 19}}\mathrm{\;J}$) is correct because the minimum energy is the product of Planck's constant and the threshold frequency.

Option C ($5.79 \times {10^6}\mathrm{\;m\;}{\mathrm{s}^{ - 1}}$) is correct because it represents the uncertainty in velocity calculated using the Heisenberg uncertainty principle.

Option B (A line spectrum in emission) is correct because an incandescent source of hydrogen emits light at specific wavelengths, resulting in a line spectrum.

Option C (Cathode rays) is correct because cathode rays are a stream of electrons, which are matter particles, while the other options are electromagnetic radiation.

Option A (Atoms) is correct because line spectra are produced by the transitions of electrons within individual atoms.

The question is flawed. The correct sequence for filling electrons when n=6 is based on (n+l) rule and Aufbau principle, which gives the order: 6s → 4f → 5d → 6p. Therefore, option B is the closest to the correct answer.

Option B ($1 \times {10^{15}}\mathrm{\;}{\mathrm{s}^{ - 1}}$) is correct because it represents the threshold frequency calculated from the given maximum kinetic energy and frequency of the incident radiation.

Option C ($4.03 \times 10^{-6}$) is correct because it is calculated by determining the number of moles of C-14, then the number of atoms, and finally multiplying by the number of neutrons per atom and the mass of a neutron.

Option D (II,III,IV) is correct because isoelectronic species have the same number of electrons. H3O+, NH3, and CH3- all have 10 electrons.

The correct answer is D (An identical set of quantum numbers) because, according to the Pauli Exclusion Principle, no two electrons in an atom can have the same set of all four quantum numbers.

Option B is not possible because if n=2, l can only be 0 or 1. l cannot be equal to n.

Option A (3) is correct because the M-shell corresponds to n=3, and for each n, there are n possible values of l (0 to n-1), representing the number of elliptical orbits, including the circular one.

Option C (Several discrete series of lines with both intensity and spacings between lines decreasing as the wave number increase within each series) is correct because it accurately describes the characteristics of the hydrogen emission spectrum.

Option D is correct because both the assertion and the reason are incorrect regarding nuclear binding energy and its relationship with nucleon numbers.

Option C ($\mathrm{C}{\mathrm{o}^{2 + }}$) is correct because it has the fewest unpaired d-electrons compared to the other options.

Option B (eV) is correct because the kinetic energy gained by an electron accelerated through a potential difference V is equal to the charge of the electron (e) multiplied by the potential difference (V).