Structure of atom practice · Mixed difficulty

Option A (Conservation of energy) is correct because the photoelectric effect demonstrates the conversion of light energy into kinetic energy of electrons.

Only B is correct because degenerate orbitals have the same n and l values. In B, both electrons have n=3 and l=2.

Option A (11,11,12) is correct because in a neutral atom, the number of protons equals the number of electrons, and the number of neutrons is the mass number minus the atomic number.

Option C ($1.5 a_{\mathrm{0}}, a_{\mathrm{0}}$) is correct because the average distance is 1.5 times the Bohr radius, while the most probable distance is equal to the Bohr radius.

Option A (6) is correct because the electrons ejected from He+ excite the H-atom to n=4, and the number of spectral lines is calculated using the formula n(n-1)/2.

Option C ($\lambda = \frac{{0.101}}{{\sqrt V }}{A^ \circ }$) is correct because it is the correct formula for calculating the de Broglie wavelength of alpha particles in terms of applied voltage.

Option A ((ii), (iii), (iv)) is correct because statements (ii), (iii), and (iv) are correct, while (i) is only applicable to single electron species and (v) Paschen series belongs to the infrared region.

Option A (Particle-like behaviour of light) is correct because the photoelectric effect demonstrates that light can behave as discrete packets of energy (photons), which is a particle-like property.

Option D (${\mathrm{\;}_{17}}\mathrm{C}{\mathrm{l}^ + }$) is correct because $_{16}S^{32}$ has 16 electrons, and $_{17}Cl^+$ also has 16 electrons (17 - 1 = 16).

Option B ($\frac{1}{{100}}\mathrm{th}$) is correct because the velocity of an electron in the first Bohr orbit is approximately $c/137$, which is about 1/100th of the speed of light.

Option D (Reduced by 25%) is correct because the mass of the neutrons is halved, and the mass of the electrons is doubled, resulting in a net decrease in the atomic mass.

Option D (494 nm) is correct because it represents the longest wavelength calculated using the energy required to break one mole of Cl-Cl bonds.

Option D (β-particle) is correct because the de Broglie wavelength is inversely proportional to mass ($λ = \frac{{h}}{{mv}}$), and the beta particle (electron) has the smallest mass among the given options, resulting in the longest wavelength for a given speed.

Option D ($\lambda = \sqrt {\frac{h}{m}}$) is correct because if wavelength equals the distance traveled in one second, then $\lambda = v$, and using de Broglie's equation $\lambda = \frac{h}{mv}$, we can derive the given relation.

Option D ($s$) is correct because two electrons in the same orbital must have opposite spins, thus differing in their spin quantum number ($s$).

Option A (Independent of its wavelength) is correct because the speed of a photon in a vacuum is a constant, denoted by 'c', and is independent of its wavelength.

Option A (Square of the orbital wave function i.e., $\psi^2$) is correct because the probability of finding an electron at a point is proportional to the square of the wave function at that point.

Option B (${10^{ - 15}}$) is correct because the nucleus is much smaller than the atom, and the ratio of their volumes is approximately ${10^{ - 15}}$.

Option A (6) is correct because the number of spectral lines is calculated using the formula n(n-1)/2 where n is the orbit number.

Option B ($L{i^ + }$) is correct because both $B{e^{^{ + 2}}}$ and $L{i^ + }$ have 2 electrons, making them isoelectronic.

Option D (7, 4) is correct because the change in mass number is 235-207 = 28, and each alpha particle emission reduces the mass number by 4, so 28/4 = 7 alpha particles are emitted. The change in atomic number is 92-82 = 10, and each alpha particle reduces the atomic number by 2, and each beta particle increases the atomic number by 1. Thus, 10 = 7*2 - x*(-1) => x = 4.

Option B (Mg) is correct because Magnesium (Mg) has the electronic configuration 1s² 2s² 2p⁶ 3s², which has 6 s-electrons and 6 p-electrons.

Option A ($4.21 \times 10^{-14} \mathrm{~m}$) is correct because it is calculated using the conservation of energy, equating the initial kinetic energy to the potential energy at the distance of closest approach.

Option B (Electron) is correct because electrons have a very small mass and a significant charge, resulting in a high charge-to-mass ratio compared to protons, neutrons, and alpha particles.

Option A (2) is correct because the potential energy (PE) is -2 times the total energy (TE) in a Bohr orbit, so PE/TE = 2.

Option D ($n=1$ to $n=\infty$) is correct because it represents ionization, which requires the greatest amount of energy to completely remove the electron from the atom.

Option B (52.9 pm) is correct because the radius of the nth Bohr orbit is given by $r_n = 0.529 \times \frac{n^2}{Z}$ Å, where n is the orbit number and Z is the atomic number.

Option A (Half times) is correct because the de-Broglie wavelength is inversely proportional to the square root of the kinetic energy.

Option A ($\sqrt{6} \hbar$) is correct because the orbital angular momentum for a d-electron (l=2) is given by $\sqrt{l(l+1)}\hbar$, which equals $\sqrt{6} \hbar$.

Option B ($6\pi a$) is correct because the circumference of the 3rd Bohr orbit is $2\pi r = n\lambda$, and $r = n^2a$, so $2\pi(9a) = 3\lambda$, which gives $\lambda = 6\pi a$.