Structure Of Atom – NEET UG ChemistryPractice Questions & PYQs

The law of photochemical equivalence dictates that one mole of reactant molecules absorbs one mole of photons. Calculating the energy associated with this absorption requires the Planck-Einstein relation, expressed for a mole of particles as $E = \frac{h c N_A}{\lambda}$ Substituting the constant values, where $h = 6.62 × 10^{-27}$, $c = 3 × 10^{10}$, and $N_A = 6.02 × 10^{23}$, the expression simplifies by multiplying these constants together. The product of the numerator is approximately $1.196 × 10^8$, which leads to the final energy expression of $E = \frac{1.196 \times 10^8}{\lambda}$ ergs mol^-1.

Stability in electron configurations is driven by the exchange energy associated with half-filled or fully filled subshells, which significantly lowers the overall energy of the atom. Gadolinium, with an atomic number of 64, follows this principle because populating the $4f$ subshell to its half-filled $4f^{7}$ state is more energetically favorable than placing an additional electron into that same subshell. Consequently, after the $6s$ orbital is filled with two electrons, the remaining electron occupies the $5d$ orbital to preserve the high stability of the half-filled $4f^{7}$ shell, resulting in the configuration $[Xe] 4f^{7} 5d^{1} 6s^{2}$.

The relationship between the speed of light, its frequency, and its wavelength is governed by the fundamental equation $c = \upsilon \lambda$. To find the wavelength of the light quantum, we rearrange this formula to isolate $\lambda$ as $\lambda = \frac{c}{\upsilon}$. By substituting the given speed of light, $3 \times 10^{17} \text{ nm s}^{-1}$, and the frequency, $6 \times 10^{15} \text{ s}^{-1}$, we perform the division $\frac{3 \times 10^{17}}{6 \times 10^{15}}$. This calculation simplifies to $0.5 \times 10^2$, resulting in a final wavelength of 50 nm.

The energy of an electron in an atom is determined by $E = -2.178 \times 10^{-18} \text{J} \left( \frac{Z^2}{n^2} \right)$, where the negative sign indicates that the electron is more stable and at a lower potential energy compared to a free electron at an infinite distance from the nucleus. As the principal quantum number $n$ decreases, the energy value becomes increasingly negative, which reflects a stronger electrostatic attraction and a more tightly bound state for the electron. Therefore, the assertion that the electron is more loosely bound in the smallest allowed orbit is contradictory to physical reality, as smaller orbits actually represent regions of higher stability and tighter binding to the nucleus.

The electronic configuration of an atom provides the foundation for identifying the specific quantum numbers that define the state of its valence electron. For rubidium, which has an atomic number of 37, the electrons fill the orbitals according to the Aufbau principle, resulting in the configuration $1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{6} 4s^{2} 3d^{10} 4p^{6} 5s^{1}$. The single valence electron occupies the outermost 5s orbital.

The principal quantum number $n = 5$ is determined directly by the coefficient of the 5s orbital, indicating the electron's main energy level. For any electron within an s subshell, the azimuthal quantum number $l = 0$, which restricts the magnetic quantum number $m = 0$ as it must fall within the range of $-l$ to $+l$. The spin quantum number $s$ accounts for the intrinsic angular momentum of the electron and is conventionally assigned a value of $s = +1/2$. Consequently, the four quantum numbers describing this valence electron are $n=5, l=0, m=0, s=+1/2$.

The orbital angular momentum of an electron is determined by its azimuthal quantum number, $l$, expressed by the formula $\sqrt{l(l+1)} \frac{h}{2\pi}$. Since a p-electron corresponds to an azimuthal quantum number of $l = 1$, we substitute this value into the equation to obtain $\sqrt{1(1+1)} \frac{h}{2\pi}$, which simplifies to $\sqrt{2} \frac{h}{2\pi}$. Final algebraic reduction, recognizing that $\frac{\sqrt{2}}{2}$ is equivalent to $\frac{1}{\sqrt{2}}$, leads to the value $\frac{h}{\sqrt{2}\pi}$.

The maximum electron capacity of a subshell is governed by the azimuthal quantum number $l$ and is calculated using the formula $2(2l + 1)$. For a subshell where $l = 3$, we perform the following calculation:

$2(2 \times 3 + 1) = 14$

Although the principal quantum number $n = 4$ defines the energy level as the 4f subshell, the capacity is determined solely by $l$, confirming that the subshell holds 14 electrons.

The total number of atomic orbitals contained within any principal energy level is calculated using the square of the principal quantum number, $n^2$. Applying this principle to the fourth energy level means setting $n = 4$ in the expression. Consequently, $4^2 = 16$ provides the total number of orbitals found in that specific shell.

According to the Bohr theory, the energy of a photon emitted during an electronic transition is proportional to the difference between the inverse squares of the quantum numbers, expressed as $\Delta E \propto \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. Energy levels in a hydrogen atom become progressively closer together as the principal quantum number $n$ increases, which means transitions occurring between higher-energy shells involve significantly smaller changes in energy. Because the transition from $n=6$ to $n=5$ involves the largest principal quantum numbers among the given choices, the interval between these two states is the smallest, resulting in the release of the least energetic photon.

The Aufbau principle dictates that electrons populate atomic orbitals in order of increasing energy, which is determined by the $(n+l)$ rule. For the principal quantum number $n=6$, calculating these values reveals that $6s$ ($6+0=6$) has the lowest sum and is filled first, while the remaining orbitals $4f$ ($4+3=7$), $5d$ ($5+2=7$), and $6p$ ($6+1=7$) share a higher sum of $7$. Among orbitals with identical $(n+l)$ values, the one with the lowest principal quantum number takes priority, establishing a sequence of $4f$ ($n=4$), followed by $5d$ ($n=5$), and finally $6p$ ($n=6$), which matches the general pattern $ns \rightarrow (n-2)f \rightarrow (n-1)d \rightarrow np$.

The energy of a photon is inversely proportional to its wavelength, described by the relationship $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant and $c$ is the speed of light. Because both $h$ and $c$ are constant, the ratio of the two energies $E_1$ and $E_2$ corresponds to the inverse ratio of their wavelengths $\lambda_1$ and $\lambda_2$, written as $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$. By substituting the values 25 eV for $E_1$ and 50 eV for $E_2$ into this equation, we find that $\frac{25}{50} = \frac{\lambda_2}{\lambda_1}$, which simplifies to $\frac{1}{2} = \frac{\lambda_2}{\lambda_1}$, ultimately yielding $\lambda_1 = 2\lambda_2$.

The wave-particle duality proposed by de-Broglie states that every moving object possesses an associated wavelength determined by its momentum. This relationship is quantified by the formula $\lambda = \frac{h}{mv}$ where $\lambda$ is the wavelength, $h$ is Planck's constant, $m$ is the mass of the object, and $v$ is its velocity.

Substituting the provided values into this expression results in the equation $\lambda = \frac{6.6 \times 10^{-34}}{0.66 \times 100}$ Multiplying the mass and velocity \in the denominator yields a total momentum of 66 kg m/s. Dividing the Planck constant by this momentum value gives the final result, defined by the equation $\lambda = 1.0 \times 10^{-35} \text{ m}$

The allowed states for an electron are strictly governed by quantum numbers, where the magnetic quantum number $m_l$ must always fall within the range $-l \leq m_l \leq l$ for a given azimuthal quantum number $l$. For an orbital defined by $l=2$, the permissible values for $m_l$ are restricted to integers starting from $-2$ and ending at $2$. Because the arrangement $n=3, l=2, m=-3, s=-\frac{1}{2}$ specifies $m=-3$ while assigning $l=2$, this configuration contradicts the fundamental rules of quantum mechanics and is physically impossible.

The maximum number of electrons in a subshell is governed by the azimuthal quantum number $l$, which defines the total number of orbitals available as $2l+1$. According to the Pauli exclusion principle, every individual orbital can accommodate a maximum of two electrons with opposite spins. Multiplying the number of orbitals by this capacity per orbital yields the final value of $2(2l+1)$, which simplifies to the expression $4l+2$.

The uncertainty in momentum is fundamentally defined as the product of an object's mass and its uncertainty in velocity, which can be expressed as $\Delta p = m \Delta v$. To determine the uncertainty in velocity, we rearrange this relationship to isolate $\Delta v$, giving the equation $\Delta v = \frac{\Delta p}{m}$. By substituting the provided values into this formula, we divide the momentum uncertainty of $1 \times 10^{-18} \text{ g cm s}^{-1}$ by the electron mass of $9 \times 10^{-28} \text{ g}$. Performing this calculation results in $\frac{1}{9} \times 10^{10} \text{ cm s}^{-1}$, which yields approximately $1.1 \times 10^9 \text{ cm s}^{-1}$.

Heisenberg's uncertainty principle dictates that the product of the uncertainty in position and the uncertainty in momentum must be greater than or equal to $\frac{h}{4\pi}$. When the uncertainty in position is set equal to the uncertainty in momentum, the relationship is expressed as $(\Delta p)^2 \ge \frac{h}{4\pi}$. Substituting the definition of momentum as the product of mass and velocity, where $\Delta p = m \Delta v$, transforms the inequality into $(m \Delta v)^2 \ge \frac{h}{4\pi}$. Taking the square root of both sides, the equation simplifies to show that the uncertainty in velocity is $\Delta v \ge \frac{1}{2m} \sqrt{\frac{h}{\pi}}$.

The validity of quantum number sets relies on two primary rules: the azimuthal quantum number $l$ must strictly follow the condition $0 \le l < n$, and the magnetic quantum number $m$ must reside within the range $-l \le m \le +l$. Applying these constraints, set (ii) is impossible because $l$ cannot equal $n$ when $n=2$, and set (iv) fails because $m=-1$ falls outside the permissible range of $0$ allowed when $l=0$. Furthermore, set (v) is invalid because the magnetic quantum number $m=3$ exceeds the maximum value of $l=2$ allowed for that specific subshell.

Heisenberg's uncertainty principle states that there is a fundamental limit to the precision with which the position and momentum of a particle can be known simultaneously. This principle is mathematically expressed as:

$\Delta x \cdot m \Delta v = \frac{h}{4\pi}$

In this calculation, we are given the position uncertainty $\Delta x$ as $0.1 \times 10^{-10} \text{ m}$, the mass of the electron $m = 9.11 \times 10^{-31} \text{ kg}$, and Planck's constant $h = 6.626 \times 10^{-34} \text{ Js}$. Rearranging the principle to solve for the velocity uncertainty $\Delta v$ yields the formula $\Delta v = \frac{h}{4 \pi m \Delta x}$. Inserting the known values and using $3.143$ for the value of $\pi$, we divide $6.626 \times 10^{-34}$ by the product of $4$, $3.143$, $9.11 \times 10^{-31}$, and $0.1 \times 10^{-10}$, which results in a velocity uncertainty of $5.79 \times 10^6 \text{ ms}^{-1}$.

Atomic orbitals occupy distinct regions in space, and the specific way these orbitals are oriented relative to one another is determined by the magnetic quantum number. While the principal quantum number clarifies the energy and scale of the shell and the azimuthal quantum number indicates the shape of the orbital, the magnetic quantum number explicitly defines how the orbital is situated within the three-dimensional space of the atom.

The energy levels within a hydrogen atom are determined by Bohr's model, where the energy of an electron in a given orbit is inversely proportional to the square of the principal quantum number $n$, represented by the relation $E_n \propto \frac{1}{n^2}$. This means the energy of any orbit can be expressed as $E_n = -\frac{K}{n^2}$ for some constant $K$, allowing us to relate the energy of the fourth orbit to that of the second orbit using the squared inverse ratio of their quantum numbers. Given that the second orbit corresponds to $n=2$ and the fourth to $n=4$, the ratio of their energy values is $\frac{E_4}{E_2} = \frac{2^2}{4^2} = \frac{4}{16} = \frac{1}{4}$. Since the energy of the second orbit is -328 kJ mol⁻¹, the energy of the fourth orbit is calculated as -328 kJ mol⁻¹ multiplied by $\frac{1}{4}$, which results in -82 kJ mol⁻¹.