Structure of atom practice · Medium

Option A ($7.2 \times 10^4 \mathrm{~cm}^{-1}$) is correct because it is calculated using the Rydberg formula and the given wavenumber of the first line in the Balmer series of $\mathrm{Be}^{3+}$.

Option A ($1.988 \times 10^{-19} \mathrm{~J}$) is correct because it is the kinetic energy of the emitted electron calculated using Einstein's photoelectric equation.

Option B (If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.) is correct because both the assertion about the frequency of emitted radiation and the reason about energy levels are true, but the energy level difference is the direct cause of the frequency.

Option D (4.79 Å) is correct because it is the calculated radius of the third Bohr orbit of ${\mathrm{H}^ + }$ using the Bohr model formula.

Option A ($6.02 \times {10^{23}}$) is correct because it calculates the number of electrons in 1.8 g of water.

Option D (– 1.70 eV) is correct because the potential energy is twice the total energy, and the total energy in the nth orbit is given by $E_n = E_1/n^2$.

Option A ($8.2 \times {10^{ - 7}}\,{\text{erg}}$) is correct because it's the energy calculated using Einstein's mass-energy equivalence formula, $E = mc^2$, with the mass of an electron.

Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because the number of radial nodes is determined by n-l-1, which results in equal nodes for 3s and 4p, and the reason correctly explains this.

Option C (The energy of 2S orbital is less than the energy of 2P orbital in case of H-like atoms) is the wrong statement because for H-like atoms, the energy of 2S and 2P orbitals is the same.

Option D ($2.12\,\mathop \mathrm{A}\limits^ \circ$) is correct because it is calculated using the formula for the radius of the nth Bohr orbit for hydrogen atom.

Option A (Atoms can be created in a chemical reaction) is correct because Dalton's atomic theory states that atoms are indivisible and cannot be created or destroyed in a chemical reaction.

Option D (Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube) is correct because the characteristics of cathode rays do not depend on the gas present in the tube.

Option A ([Xe] 4f14 5d9 6s1) is correct because Platinum (Pt) has an atomic number of 78, and its electronic configuration follows Hund's rule and the Aufbau principle, with one electron promoted from the 5d to the 6s orbital for stability.

Option D (iii)<(iv)<(i)<(ii) is correct because electron affinity increases across a period and decreases down a group, with exceptions due to small size.

Option B ($1.46 \times 10^{-33} \mathrm{~m}$) is correct because it is the uncertainty in position calculated using Heisenberg's uncertainty principle.

Option B ($6.63 \times 10^{-29} \mathrm{~m}$) is correct because it is calculated by substituting the given values into the de Broglie wavelength formula.

Option C (0.5) is correct because the quantum efficiency is calculated by dividing the moles of B formed by the moles of photons absorbed.

Option A ($1.14 \times 10^{-6} \mathrm{~m}$) is correct because the wavelength is calculated using the Rydberg formula with the correct values for n1, n2, and Z.

Option C (Both (1) and (2)) is correct because both the dual character of electromagnetic radiation and experimental results regarding atomic spectra were responsible for Bohr's model.

Option B ($\mathrm{F}{\mathrm{e}^{2 + }}$) is correct because it has the most unpaired electrons (4) in its d-orbital configuration.

Option A (662) is correct because it is the result of applying the formula relating energy and wavelength of a photon.

Option C (4p) is correct because after filling the 3d orbitals, the next electron enters the 4p orbitals according to the Aufbau principle and Hund's rule.

Option D (Both statement I and statement II are correct) is correct because the kinetic energy of photoelectrons increases with the frequency of incident radiation, and the number of photoelectrons increases with the intensity of incident radiation.

Option C is correct because for a 4f orbital, n=4, l=3, m can be any value from -3 to +3, and s can be +1/2 or -1/2.

Option D (Reduced by 25%) is correct because halving the neutron mass reduces the total atomic mass from 12 to 9, which is a 25% reduction.

Option A ($\gamma$-rays) is correct because gamma rays have the highest frequency in the electromagnetic spectrum.

Option B is correct because the principal quantum number is derived from the Rydberg formula for hydrogen atom emission.

Option A is the incorrect statement because the increasing order of wavelength should be Radiowaves > Microwaves > IR waves > visible waves > UV waves.

Option B is correct. When electrons de-excite from the $n^{th}$ shell to the second shell ($n_f = 2$), the spectral lines fall in the visible region (Balmer series), and the number of lines is (n-2).

Option B ($2.52 \times 10^{24}$) is correct because it is the total number of protons in 8.4 g of MgCO₃, calculated by determining the number of moles of MgCO₃ and the number of protons per molecule.