Structure Of Atom – JEE Main ChemistryPractice Questions & PYQs

Germanium ($Z=32$) possesses the electronic configuration $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^2$. Every $s$-orbital is defined by an azimuthal quantum number $l=0$, and since $m_l$ is restricted to $m_l=0$ for these orbitals, every filled $s$-subshell contributes exactly one orbital that satisfies the condition. Because the $1s, 2s, 3s$, and $4s$ subshells are all completely filled, they provide four distinct orbitals with $m_l=0$.

Within the $p$-subshells ($l=1$), the magnetic quantum number can take values of $-1, 0,$ and $+1$, meaning each set of $p$-orbitals contains exactly one orbital where $m_l=0$. With both the $2p^6$ and $3p^6$ subshells completely filled, each contributes one orbital to the count, adding two more. Finally, the $3d$ subshell ($l=2$) consists of five orbitals, one of which corresponds to $m_l=0$; since the $3d^{10}$ subshell is fully occupied, this adds one final orbital. Summing these contributions-four from $s$-orbitals, two from $p$-orbitals, and one from $d$-orbitals-yields a total of seven completely filled orbitals where $m_l=0$.

Kinetic energy is proportional to the square of the electron's velocity, and with $v \propto \frac{Z}{n}$, it follows that $\text{KE} \propto \frac{Z^2}{n^2}$, confirming the first statement is correct. Since $v \propto \frac{Z}{n}$, the product $v \cdot n$ is proportional to $Z$ rather than $Z^2$, which makes the second statement incorrect. The frequency of revolution, defined as $f = \frac{v}{2\pi r}$, incorporates the proportionalities $v \propto \frac{Z}{n}$ and $r \propto \frac{n^2}{Z}$ to yield $f \propto \frac{Z^2}{n^3}$, proving the third statement inaccurate. Finally, the Coulombic force is governed by $F \propto \frac{Z}{r^2}$, and substituting the Bohr radius $r \propto \frac{n^2}{Z}$ results in $F \propto \frac{Z^3}{n^4}$, verifying that the fourth statement is correct.

Heisenberg's Uncertainty Principle states that there is a fundamental limit to the precision with which a particle's position and momentum can be known simultaneously, expressed by the relationship $\Delta x \cdot m \Delta v \geq \frac{h}{4\pi}$. This principle implies that the product of the uncertainty in position and the uncertainty in momentum must be at least equal to this constant threshold.

To determine the uncertainty in position for the ball, first convert the mass from $10\ \text{g}$ to $10 \times 10^{-3}\ \text{kg}$ to maintain standard SI units. Next, calculate the uncertainty in velocity by taking $5\%$ of the velocity, which yields $90 \times \frac{5}{100} = 4.5\ \text{ms}^{-1}$. Rearranging the uncertainty formula to solve for $\Delta x$ gives $\Delta x = \frac{h}{4 \pi m \Delta v}$. Substituting the given values results in $\Delta x = \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times (10 \times 10^{-3}) \times 4.5}$, which simplifies to approximately $1.17 \times 10^{-33}\ \text{m}$. Rounding this value to the nearest integer provides $1 \times 10^{-33}\ \text{m}$.

The photoelectric effect dictates that the energy of an incident photon must first satisfy the metal's threshold requirement, with any remaining energy converting into the kinetic energy of the ejected electron, which subsequently exhibits a wave-like de-Broglie wavelength. To begin, convert the incident light's wavelength of $248\;\text{nm}$ to $2480\;\text{\AA}$. Using the photon energy formula $E = \frac{12400}{\lambda (\text{\in } \text{\AA})}$, the incident energy is found to be $\frac{12400}{2480} = 5\;\text{eV}$.

By applying the principle of energy conservation, $E = \phi + K_{\max}$, and subtracting the threshold energy of $3\;\text{eV}$ from the photon energy of $5\;\text{eV}$, the kinetic energy of the emitted electrons is determined to be $2\;\text{eV}$. The de-Broglie wavelength $\lambda$ for these electrons can then be computed using the relation $\lambda = \sqrt{\frac{150}{V}}\;\text{\AA}$, where $V$ represents the kinetic energy in electron-volts. Inserting the value $V = 2$, the result is $\sqrt{75}\;\text{\AA}$, which simplifies to $5\sqrt{3}\;\text{\AA}$. Given the approximation $\sqrt{3} \approx 1.73$, this equals $8.65\;\text{\AA}$, yielding a final rounded integer value of $9$.

The azimuthal quantum number $l$ defines the shape of an orbital, with $l=0$ assigned specifically to $s$-subshells. Gallium has an atomic number of 31, which corresponds to an electron configuration of $[Ar] 3d^{10} 4s^2 4p^1$. Removing the outermost electron to form the $Ga^{+}$ ion leaves the electron configuration as $[Ar] 3d^{10} 4s^2$. Since the valence electrons now occupy the $4s$ subshell, their azimuthal quantum number is 0.

The total power emitted by a light source represents the energy radiated per unit of time, which equals the product of the number of photons emitted per second and the energy carried by a single photon. Given that the energy of one photon is defined as $E = \frac{hc}{\lambda}$, we can establish the relationship $P = n \frac{hc}{\lambda}$, where $P$ is the power, $n$ is the number of photons per second, $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. Rearranging this equation to solve for the number of photons yields $n = \frac{P \lambda}{hc}$. Substituting the given values where $P = 50 \text{ W}$, $\lambda = 795 \times 10^{-9} \text{ m}$, $h = 6.63 \times 10^{-34} \text{ Js}$, and $c = 3.0 \times 10^{8} \text{ ms}^{-1}$, we find $n = \frac{50 \times 795 \times 10^{-9}}{6.63 \times 10^{-34} \times 3.0 \times 10^{8}}$. Simplifying the denominator gives $19.89 \times 10^{-26}$, and performing the division results in approximately $1.998 \times 10^{20}$. Rounding this value to the nearest integer, we determine that $x = 2$.

The radial distribution function for an orbital illustrates the probability density at distance $r$ from the nucleus, where the total number of peaks is defined by the formula $n - l$. For the 3s-orbital, the principal quantum number is $n = 3$ and the azimuthal quantum number is $l = 0$, resulting in $3 - 0 = 3$ peaks. Graph D is the only plot that displays exactly three peaks, identifying it as the correct representation for this orbital.

The kinetic energy of an electron in a hydrogen atom is intrinsically linked to the quantization of its angular momentum, which is defined by the relationship $mvr = \frac{nh}{2\pi}$. By rearranging this equation to isolate the velocity $v = \frac{nh}{2\pi mr}$ and substituting it into the standard classical kinetic energy formula $KE = \frac{1}{2}mv^{2}$, we can derive the energy expression as $KE = \frac{n^{2}h^{2}}{8\pi^{2}mr^{2}}$.

For an electron in the second Bohr orbit, the principal quantum number $n$ is 2, and the orbital radius is given by $r = n^{2}a_{0}$, which results in $r = 4a_{0}$. Substituting these specific values into the energy formula yields $KE = \frac{2^{2}h^{2}}{8\pi^{2}m(4a_{0})^{2}}$. Upon simplifying the denominator, the expression becomes $KE = \frac{4h^{2}}{8\pi^{2}m(16a_{0}^{2})}$, which reduces to $KE = \frac{h^{2}}{32\pi^{2}ma_{0}^{2}}$.

Matching this result to the provided form $\frac{h^{2}}{xma_{0}^{2}}$ reveals that $x$ is equal to $32\pi^{2}$. Using the approximation $\pi = 3.14$, we find that $x = 32 \times (3.14)^{2}$, which equals 315.5072. Consequently, the value of $10x$ is 3155.072, making the nearest integer 3155.

Rutherford's gold foil experiment was instrumental in showing that electrons are bound to the nucleus by electrostatic forces and travel in high-speed circular paths, yet this model failed to provide any basis for the discrete line spectrum observed in hydrogen. Moving to the atomic structure proposed by Bohr, it relies on the calculation of exact speed and distance of an electron from the nucleus, which creates a fundamental conflict with Heisenberg's uncertainty principle. Since this principle dictates that it is impossible to simultaneously determine the exact position and momentum of an electron, governed by the relation $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$, Bohr's assumption of fixed, calculable orbits is conceptually incompatible with the inherent uncertainty of quantum systems.

Heisenberg's Uncertainty Principle states that the product of the uncertainties in position and momentum for a subatomic particle is at least $\frac{h}{4 \pi}$, providing the mathematical foundation $\Delta x \cdot m \cdot \Delta v = \frac{h}{4 \pi}$. With the electron's speed at $5 \times 10^{6} \text{ ms}^{-1}$ and an uncertainty of $0.02\%$, the uncertainty in velocity is $\Delta v = \frac{0.02}{100} \times 5 \times 10^{6} = 10^{3} \text{ ms}^{-1}$.

Substituting the known values of mass $9.1 \times 10^{-31} \text{ kg}$, Planck's constant $6.63 \times 10^{-34} \text{ Js}$, and velocity uncertainty $10^{3} \text{ ms}^{-1}$ into the rearranged principle $\Delta x = \frac{h}{4 \pi m \Delta v}$ results in $\Delta x = \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{3}}$.

Simplifying the calculation yields $\Delta x \approx 0.058 \times 10^{-6} \text{ m}$. Because the location uncertainty is given in the form $x \times 10^{-9} \text{ m}$, converting the value to $58 \times 10^{-9} \text{ m}$ shows that the integer value of $x$ is $58$.

The radius of an orbit in a hydrogen-like species is governed by the Bohr model, where the orbital radius scales as $r_n \propto \frac{n^2}{Z}$. To calculate the difference in radii for $\mathrm{Li}^{2+}$ ($Z=3$), the value is proportional to $\frac{4^2}{3} - \frac{3^2}{3} = \frac{7}{3}$. Similarly, for $\mathrm{He}^{+}$ ($Z=2$), the difference in radii follows the proportionality $\frac{4^2}{2} - \frac{3^2}{2} = \frac{7}{2}$. Comparing these differences as the ratio $\frac{\Delta R_1}{\Delta R_2}$ results in $\frac{7/3}{7/2}$, which simplifies to $2 : 3$.

Isotopes are variants of the same chemical element that share an identical number of protons but possess a different number of neutrons. Hydrogen occurs in three primary isotopic forms: protium ($^{1}_{1}\mathrm{H}$), which has zero neutrons; deuterium ($^{2}_{1}\mathrm{H}$), which contains one neutron; and tritium ($^{3}_{1}\mathrm{H}$), which contains two neutrons. To find the sum of these values, we simply add the neutron counts for each isotope together, which results in $0 + 1 + 2 = 3$.

Bohr's condition for stable orbits states that the circumference of the electron path must be an integer multiple of its de Broglie wavelength, expressed as $2\pi r = n\lambda$. With the radius of the $n^{th}$ orbit defined as $r = n^2 a_0$, the radius for the $4^{th}$ orbit becomes $16 a_0$. Substituting these values into the quantization equation yields $2\pi(16 a_0) = 4\lambda$, resulting \in a wavelength of $8\pi a_0$.

Probability density is defined as the square of the wavefunction, $|\psi|^2$, which is mathematically non-negative and represents the likelihood of locating an electron at a specific point. Nodes are defined as regions in space where the wavefunction $\psi$ equals zero, causing the probability density $|\psi|^2$ to also vanish at those locations. A 3p orbital possesses one radial node, calculated by the formula $n - l - 1 = 3 - 1 - 1 = 1$, where the probability density is zero, unlike the 1s orbital which lacks nodes at finite distances and the 2s orbital which does possess a node.

The magnetic quantum number $m$ is determined by the azimuthal quantum number $l$, where $m$ can take any integer value in the range from $-l$ to $+l$. For the principal quantum number $n=4$, the allowed values for $l$ are 0, 1, 2, and 3. Since the existence of an $m=-2$ orbital requires $l \geq |m|$, only the values $l=2$ and $l=3$ satisfy this condition. These azimuthal quantum numbers correspond to the 4d and 4f subshells, resulting in exactly 2 subshells that accommodate the given quantum numbers.

The total number of orbitals available within any principal energy level is determined by the square of the principal quantum number, expressed as $n^2$. Since each distinct orbital is capable of holding exactly one electron with a specific spin orientation of $m_s = +1/2$, the number of such electrons corresponds directly to the total number of orbitals present in that shell. For the shell defined by $n = 5$, calculating the number of orbitals involves squaring the value to get $5^2 = 25$.

The behavior of matter at the atomic level is fundamentally defined by the quantization of energy, where states are restricted to discrete levels rather than a continuous range. While phenomena such as the photoelectric effect, atomic spectra, and black body radiation are direct consequences of these quantized energy levels, classical physics assumes that internal energy can take on any continuous value as temperature increases. Consequently, a depiction showing that all values of internal energy are accessible corresponds to a classical interpretation rather than a manifestation of atomic quantum mechanics.

In this hypothetical chemical system, the electronic configuration of elements is governed by the $(n+l)$ rule, where the energy of an orbital increases as the sum of the principal quantum number $n$ and the azimuthal quantum number $l$ increases. When two orbitals share the same $(n+l)$ value, the orbital with the lower $n$ is filled first. Given that the azimuthal quantum number $l$ extends up to $n+1$, the orbitals are filled in the energy sequence $1s < 1p < 2s < 1d < 2p < 3s < 2d$, and so on. Since an $s$ subshell accommodates 2 electrons and a $p$ subshell accommodates 6 electrons, we determine the configuration by filling these orbitals according to this specific hierarchy.

For an element with an atomic number of 8, the first two electrons fill the $1s$ subshell, and the remaining six electrons fill the $1p$ subshell, resulting in a total configuration of $1s^2 1p^6$. At this stage, the $1p$ subshell reaches its maximum capacity, signifying a completely filled shell. Because any additional electrons would be forced to enter the $2s$ orbital, thereby starting a new period, this state of full occupancy identifies the atom with atomic number 8 as the first noble gas in this hypothetical model.

The emission spectrum of a hydrogen atom is produced when electrons transition from higher energy states to lower ones, resulting in discrete lines of radiation categorized into specific series. When an electron drops to the energy level $n=2$, the energy released by the transition corresponds precisely to the range of visible light. Consequently, the spectral lines associated with this specific transition pattern, known as the Balmer series, are found within the visible portion of the electromagnetic spectrum.

The relationship between the wavelength of light emitted during an electronic transition and the energy levels involved is expressed by the Rydberg equation:
$\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
For the shortest wavelength in the Lyman series of the hydrogen atom, the transition occurs from the highest energy level, $n = \infty$, to the ground state, $n = 1$, where the atomic number $Z = 1$. Substituting these values into the formula yields $\frac{1}{\lambda_1} = R(1)^2(1 - 0)$, which simplifies to the expression $\lambda_1 = \frac{1}{R}$.

The longest wavelength in the Balmer series of the helium ion, $\mathrm{He}^{+}$, corresponds to the smallest energy gap, which occurs when an electron transitions between adjacent levels from $n = 3$ to $n = 2$ with $Z = 2$. Applying the Rydberg equation results in:
$\frac{1}{\lambda_2} = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$
This calculation simplifies to $4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{5}{36} \right) = \frac{5R}{9}$. By inverting this relationship, we find that $\lambda_2 = \frac{9}{5R}$, and since $\lambda_1 = \frac{1}{R}$, the final wavelength is expressed as $\lambda_2 = \frac{9\lambda_1}{5}$.