Structure of atom practice · Mixed difficulty

Option D (Pauli's exclusion principle) is correct because it states that no two electrons in an atom can have the same set of all four quantum numbers.

Option A (2, 0) is correct because the number of radial nodes is calculated using the formula n - l - 1, where n is the principal quantum number and l is the azimuthal quantum number.

Option C ($17.65\,x$ unit) is correct because the e/m ratio is inversely proportional to the mass, and the mass of the meson is given relative to the electron.

Option B (Energy is absorbed) is correct because electrons absorb energy when transitioning from a lower to a higher energy orbit.

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 ($2.3 \mathrm{~eV}$) is correct because it is calculated by using the photoelectric equation $E = \phi + eV_s$, where E is the energy of the photon, $\phi$ is the work function, and $V_s$ is the stopping potential.

Option A (Nucleus) is correct because Rutherford's alpha-ray scattering experiment demonstrated the existence of a small, dense, positively charged nucleus within the atom.

Option C ($Y$ will be halved and $Z$ will remain the same) is correct because intensity is directly proportional to the number of electrons emitted ($Y$), while the maximum kinetic energy ($Z$) depends on the wavelength (energy) of the light, not its intensity.

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 D ($n=4, l=1, m=+2, s=-\frac{1}{2}$) is correct because the value of the magnetic quantum number ($m$) must be within the range of $-l$ to $+l$, and when $l=1$, $m$ can only be -1, 0, or +1.

The correct answer is A because S-I accurately describes Heisenberg's Uncertainty Principle, and S-II correctly states the wave-particle duality of electrons.

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 C ($n=3,l=2$) is correct because if the Aufbau rule is not followed, the 19th electron would fill the 3d orbital, corresponding to n=3 and l=2.

Option A (m) is correct because the mass number (m) represents the total number of nucleons (protons and neutrons) in the nucleus of an atom.

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 B (Radio wave) is correct because radio waves have the maximum wavelength among the given options.

Option C (Cesium) is correct because Cesium has the lowest work function among the given options, making it the most efficient for photoemission.

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

The correct answer is C (Excited state of Mg) because the ground state of Mg is $1s^22s^22p^63s^2$, and the given configuration has one electron promoted from 3s to 3p.

The correct answer is Decreases with increase of speed of $\mathrm{e}^{-}$(electron) because the wavelength associated with a moving electron is inversely proportional to its speed, as described by de Broglie's equation.

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 (Higher atomic number) is correct because an alpha particle consists of 2 protons and 2 neutrons, so combining it with a nuclide increases the atomic number by 2.

Option A (K.E of first orbit) is correct because the ionization potential of singly ionized Helium is equivalent to the kinetic energy of an electron in the first orbit of Hydrogen multiplied by Z².

Option C ($\frac{1}{{1800}}$) is correct because the mass of an electron is approximately $\frac{1}{{1837}}$ of the mass of the lightest nuclei (hydrogen).

Option D (Atomic number) is correct because the atomic number represents the number of protons in an atom's nucleus, which must be a whole number.

Option D is correct because both the assertion and the reason are false regarding nuclear binding energy and its relationship with neutron-proton difference.

Option D (Atoms can be created as well as destroyed in a chemical reaction) is correct because Dalton's atomic theory states that atoms are neither created nor destroyed in a chemical reaction, only rearranged.