Structure Of Atom – JEE Main ChemistryPractice Questions & PYQs

The photoelectric effect requires incident photons to carry at least enough energy to overcome the work function of a metal, which is the minimum energy needed to liberate an electron from its surface. This threshold value is determined by the product of the Planck constant and the metal's threshold frequency as shown here:
$E = h \nu_0$
Using the provided Planck constant of 6.6 x 10^-34 J s and the threshold frequency for platinum of 1.3 x 10^15 s^-1, we multiply these values to find the required energy. This calculation results in 8.58 x 10^-19 J as the minimum photon energy necessary to trigger the emission.

The quantum mechanical description of an atom relies on three specific quantum numbers to define an atomic orbital uniquely. Although the principal quantum number ($n$) determines size and the azimuthal quantum number ($l$) indicates shape, the magnetic quantum number ($m$) is required to specify the spatial orientation of the orbital. Because the full identity of an orbital is incomplete without the magnetic quantum number, the assertion that atomic orbitals are characterized by only $n$ and $l$ is scientifically inaccurate.

The total number of nodes in an orbital characterizes its spatial distribution, with angular nodes representing planar regions of zero electron probability and radial nodes occurring as spherical surfaces where the wavefunction vanishes. The number of angular nodes is given by $l$, while the number of radial nodes is determined by the formula $n - l - 1$. For a $4d$ orbital, we identify the principal quantum number as $n = 4$ and the azimuthal quantum number as $l = 2$, which results in 2 angular nodes and $4 - 2 - 1 = 1$ radial node.

The distribution of electron probability density within an orbital is characterized by nodal surfaces where the wave function vanishes. The total number of nodes is defined by the principal quantum number $n$ as $n - 1$, encompassing both angular and radial regions. Radial nodes, which represent spherical shells of zero electron density, are determined by the specific relation $n - l - 1$, where $l$ is the azimuthal quantum number. For a $4p$ orbital, where $n = 4$ and $l = 1$, the total number of nodes is calculated as $4 - 1 = 3$. Simultaneously, the number of radial nodes is determined as $4 - 1 - 1 = 2$.

Determining the percentage difference between subatomic particles requires finding the exact count of electrons and neutrons for the specific ionic state. For the species ${}_{22}^{48}X^{3-}$, the atomic number 22 indicates 22 protons, meaning the neutral atom would have 22 electrons. Because the ion carries a 3- charge, it has gained 3 electrons, resulting in a total of 25 electrons. The number of neutrons is determined by subtracting the atomic number from the mass number, which is 26 neutrons.

The variable 'a' represents the percentage by which the neutron count exceeds the electron count, defined by the relationship $a = \left( \frac{\text{neutrons} - \text{electrons}}{\text{electrons}} \right) \times 100$ Substituting the derived values results \in the equation $a = \left( \frac{26 - 25}{25} \right) \times 100$ which simplifies to a value of 4.

The emission of light in an atom occurs as an electron drops from a higher energy level to a lower one, with the total number of spectral lines defined by all possible intermediate transitions. When an electron transitions from an initial shell $n_2$ to a final shell $n_1$, the total number of emission lines is calculated using the formula $\frac{(n_2-n_1)(n_2-n_1+1)}{2}$. By substituting the starting level of $n_2=5$ and the ground state of $n_1=1$ into this expression, we obtain $\frac{(5-1)(5-1+1)}{2}$, which simplifies to $\frac{4 \times 5}{2}$ and results in 10 observed emission lines.

Chromium is a notable exception to standard electron filling rules because the extra stability gained from a half-filled $d$-subshell results in the electronic configuration $[\mathrm{Ar}]\ 3 d^5 4 s^1$. Magnetic quantum numbers define orbital orientation and span values from $-l$ to $+l$ including zero, confirming that they can indeed be negative, while the Aufbau principle correctly describes the ground state filling process as electrons occupy orbitals in order of increasing energy. However, the calculation provided for total nodes is inaccurate; since the total number of nodes is the sum of radial nodes ($n - l - 1$) and angular nodes ($l$), the resulting value is $n - 1$ rather than $n - 2$.

When an electron is removed from a hydrogen atom, the total energy absorbed consists of the ionization energy required to release it and the kinetic energy it possesses after emission. We determine the electron's momentum $p$ using the de Broglie relation $p = \frac{h}{\lambda}$, which yields $p = \frac{6.626 \times 10^{-34}}{3.3 \times 10^{-10}} = 2 \times 10^{-24} \text{ kg m s}^{-1}$. Using this momentum, the kinetic energy is calculated as $KE = \frac{p^2}{2m} = \frac{(2 \times 10^{-24})^2}{2 \times 9.1 \times 10^{-31}} \approx 2.18 \times 10^{-18} \text{ J}$, or $21.8 \times 10^{-19} \text{ J}$. Adding this to the ionization energy of the ground-state hydrogen atom, which is $21.76 \times 10^{-19} \text{ J}$, gives a total absorbed energy of $43.56 \times 10^{-19} \text{ J}$. Comparing this total to the ionization energy reveals that the absorbed energy is approximately 2 times the minimum energy required for the electron's escape.

Species are defined as isoelectronic when they possess an identical total number of electrons regardless of their atomic identity. Since the aluminum atom has an atomic number of 13, the $\mathrm{Al}^{3+}$ ion contains $13 - 3 = 10$ electrons. The oxygen ion $\mathrm{O}^{2-}$ and the magnesium ion $\mathrm{Mg}^{2+}$ also reach this same electron count, as the oxygen atom gains two electrons and the magnesium atom loses two, confirming they are isoelectronic with $\mathrm{Al}^{3+}$.

Heisenberg's Uncertainty Principle establishes that there is a fundamental limit to the precision with which the position and momentum of a particle can be simultaneously defined, expressed by the relationship $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$. Since momentum is the product of mass and velocity, this principle rearranges to $\Delta x \cdot m \cdot \Delta v = \frac{h}{4\pi}$. Substituting the provided values of $\Delta x = 10^{-7}\,\mathrm{m}$, $\Delta v = 2.4 \times 10^{-26}\,\mathrm{m/s}$, and Planck's constant $h = 6.626 \times 10^{-34}\,\mathrm{J\,s}$ into the equation results in $10^{-7} \times m \times 2.4 \times 10^{-26} = \frac{6.626 \times 10^{-34}}{4\pi}$. Simplifying the expression yields $m \times 2.4 \times 10^{-33} = \frac{6.626 \times 10^{-34}}{12.566}$, and calculating for the mass gives $0.022\,\mathrm{kg}$. Converting this value into grams, the mass is $22\,\mathrm{g}$.

The principal quantum number $n$ represents the energy shell and is defined by positive integers such as $1, 2, 3, \dots$, making the first statement valid, whereas the azimuthal quantum number $l$ ranges from $0$ to $n-1$, meaning it cannot equal $n$ as suggested. The magnetic quantum number $m_l$ identifies the orbital's orientation by spanning from $-l$ to $+l$ inclusive, yielding $2l+1$ total values, and electron spin is characterized by two potential orientations of $\pm \frac{1}{2}$. Finally, applying the capacity formula $2l+1$ for $l=5$ results in 11 orbitals rather than 9.

Degenerate orbitals possess the same energy, which requires identical values for the principal quantum number $n$ and the azimuthal quantum number $l$ as these determine the orbital's energy level. In pair (A), the values $l=1$ and $l=2$ distinguish the electrons into different subshells, meaning they are not degenerate. Pair (B) contains electrons with matching $n=3$ and $l=2$ values for both, confirming that they occupy the same $3d$ subshell where they are degenerate regardless of their $m_l$ or $m_s$ values. In pair (C), the differing principal quantum numbers of $n=4$ and $n=3$ establish that these electrons reside in orbitals with unequal energy levels.

The de Broglie hypothesis establishes that the wavelength of a moving particle is inversely proportional to its momentum, expressed by the equation $\lambda = \frac{h}{mv}$, where $h$ is Planck's constant. Setting the wavelength of an electron equal to that of a neutron implies that their momenta are also equivalent, leading to the equality $m_e v_e = m_n v_n$.

By isolating the ratio of the electron's velocity to the neutron's velocity, denoted as $x$, we derive the following relationship:
$x = \frac{v_e}{v_n} = \frac{m_n}{m_e}$
Substituting the provided mass values of $m_n = 1.6 \times 10^{-27}$ kg and $m_e = 9.1 \times 10^{-31}$ kg into this ratio results in $\frac{1.6 \times 10^{-27}}{9.1 \times 10^{-31}}$. Calculating this expression yields approximately $1758.24$, which rounds to a final integer value of $1758$.

The azimuthal quantum number $l$ must always be less than the principal quantum number $n$, restricted to integer values ranging from $0$ to $n-1$. For a principal quantum number of $n=3$, the allowed values for $l$ are confined to $0, 1,$ and $2$. Consequently, the set provided in option C is invalid because it assigns $l=3$ to a shell where $n=3$, which violates the fundamental rule of quantum mechanics that $l$ cannot equal or exceed $n$.

The radius of an electron's orbit in a hydrogen atom follows the proportionality $r_n \propto n^2$, where $n$ is the principal quantum number. Because the third orbit is defined by $n=3$ and the fourth by $n=4$, the ratio between their radii is determined by the squares of these quantum numbers. This relationship is expressed as $\frac{r_4}{r_3} = \frac{4^2}{3^2}$, which simplifies to $\frac{16}{9}$. Consequently, the mathematical link between these two radii is $r_4 = \frac{16}{9} r_3$.

Effective covalent bonding requires atomic orbitals to maintain specific spatial alignment and symmetry to allow for net overlap, whereas zero orbital overlap represents an out of phase interaction where these conditions are not met. This result arises directly because the orbitals possess mismatched symmetry or occupy different planes, preventing any constructive interaction between the electron clouds. Consequently, the distinct orientation and geometric configuration described in the reason act as the primary factors that preclude orbital overlap, making the reason a valid explanation for the assertion.

The energy of an orbital in a hydrogen-like system is defined by the formula
$E_{\text{total}} = -13.6 \frac{Z^2}{n^2} \text{ eV}$
where $Z$ is the atomic number and $n$ is the principal quantum number. Because this value is inherently negative, an increase in the atomic number $Z$ makes the total energy more negative, which signifies a decrease in the orbital energy. This relationship confirms that for a fixed subshell, the energy decreases as the atomic number increases, which is why the 2s orbital of hydrogen possesses a higher energy than the 2s orbital of lithium.

Ionization enthalpy generally increases across a period due to increasing effective nuclear charge, yet significant deviations occur at fully filled or half-filled subshells which enhance stability. Comparing the configurations, element (B) $3s^2 3p^1$ is the easiest to ionize due to its outer electron placement, followed by (A) $3s^2$, which benefits from the extra stability of a fully filled s-orbital. Between the remaining elements, (C) $3s^2 3p^3$ possesses a highly stable half-filled p-orbital, making it harder to ionize than (D) $3s^2 3p^4$. Consequently, the correct order of ionization enthalpy is $(B) < (A) < (D) < (C)$.

The energy of an electron is governed by the $(n+l)$ rule, which states that subshells with a lower sum of the principal quantum number $n$ and the azimuthal quantum number $l$ have lower energy. If two subshells possess the same $(n+l)$ value, the one with the smaller principal quantum number $n$ holds lower energy.

Calculating the sum for each electron, we find values of $n+l=4$ for $D$, $n+l=3+2=5$ for $A$, $n+l=4+1=5$ for $B$, and $n+l=4+2=6$ for $C$. Comparing these totals, $D$ sits at the lowest energy level, while $C$ is at the highest. Because $A$ and $B$ share an identical $(n+l)$ sum of $5$, we look to their principal quantum numbers, where $A$ ($n=3$) is lower than $B$ ($n=4$), making $A$ lower in energy than $B$. This logic results in an increasing energy order of $D < A < B < C$.

The work function of a metal defines the minimum energy required to liberate an electron from its surface, which corresponds to the photon energy at the threshold wavelength. We use the relationship $\Phi = \frac{hc}{\lambda}$, where $\Phi$ is the work function, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the threshold wavelength. Rearranging this expression to isolate the wavelength gives the equation $\lambda = \frac{hc}{\Phi}$.

Substituting the provided constants, we calculate $\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6.63 \times 10^{-19}}$. This calculation simplifies to $\lambda = 3 \times 10^{-7}$ meters, and since 1 meter is equal to 1,000,000,000 nanometers, the final value is 300 nm.