Structure Of Atom – JEE Main ChemistryPractice Questions & PYQs

The Balmer series is defined by electronic transitions in a hydrogen atom terminating at the energy level $n_1 = 2$, where the starting level $n_2$ takes integer values starting from 3. Since the Rydberg equation $\overline{v}=R_H\left\{\frac{1}{2^2}-\frac{1}{n_2^2}\right\}$ indicates that the energy gap between levels diminishes as $n_2$ increases, the spectral lines inevitably crowd together at shorter wavelengths, confirming that the series converges as wavelength decreases. The longest wavelength is associated with the transition involving the smallest energy change, which is the transition from $n_2=3$ to $n_1=2$. Because ionization energy refers specifically to removing an electron from the ground state $n=1$ to the limit $n=\infty$, it cannot be calculated directly from the Balmer series transitions.

The radius of an electron orbit in a hydrogen-like species is determined by the formula $r_n = \frac{n^2 a_0}{Z}$, where $n$ denotes the principal quantum number, $Z$ is the atomic number, and $a_0$ is the Bohr radius. In the case of $\mathrm{Li}^{2+}$, the atomic number is $Z = 3$, and since the problem requires the radius for the second orbit, we set $n = 2$. By substituting these specific values into the orbital radius expression, we obtain $r_2 = \frac{2^2 a_0}{3}$, which simplifies to $\frac{4a_0}{3}$.

Einstein's photoelectric equation is based on energy conservation, stating that the energy of an incident photon is divided between the work function of the metal and the maximum kinetic energy of the ejected electron. We first determine the incident photon energy using the expression $E = \frac{hc}{\lambda}$, which yields $E = 6.63 \times 10^{-19} \, \text{J}$ when substituting the provided physical constants and wavelength. By applying the relation $K.E_{\max} = E - W$, we subtract the work function $W = 4.41 \times 10^{-19} \, \text{J}$ from the incident energy to calculate $K.E_{\max} = 2.22 \times 10^{-19} \, \text{J}$. Converting this result into the requested unit of $10^{-21} \, \text{J}$ leads to the final value of 222.00.

The Rydberg formula, $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, defines the wavelength $\lambda$ of radiation emitted during electronic transitions between energy levels $n_2$ and $n_1$, where $R_H = 1 \times 10^7 \text{ m}^{-1}$. For the Paschen series limit represented by the transition $\infty \to 3$, substituting $n_1 = 3$ and $n_2 = \infty$ yields $\frac{1}{\lambda} = 10^7 \left( \frac{1}{3^2} - 0 \right) = \frac{10^7}{9} \text{ m}^{-1}$. This calculation results in $\lambda = 9 \times 10^{-7} \text{ m}$, which simplifies to $900 \text{ nm}$.

The magnetic moment of a transition metal complex is determined by the number of unpaired electrons present, calculated using the relationship $\mu = \sqrt{n(n+2)}$ B.M., where $n$ is the number of unpaired electrons. Since a transition metal $d$-subshell contains exactly five orbitals, the highest possible value for $n$ is 5, a configuration found in ions like $\mathrm{Mn}^{2+}$.

Calculating for $n=5$ gives $\mu = \sqrt{5(5+2)} = \sqrt{35}$, which equals approximately $5.92$ B.M. Other choices like $3.87$ and $4.90$ relate to complexes with fewer unpaired electrons, specifically $n=3$ and $n=4$, while $6.93$ is not attainable in this context as it would require more unpaired electrons than there are $d$-orbitals available. Thus, $5.92$ B.M. stands as the maximum value achievable for a transition metal complex.

Einstein's photoelectric equation defines the relationship between the maximum kinetic energy of emitted electrons and the frequency of incident light through the expression $K.E. = h\nu - h\nu_0$, where $h$ is Planck's constant and $\nu_0$ is the threshold frequency. This formula corresponds to the linear equation $y = mx + c$, which dictates that the graph must be a straight line with a positive slope equal to $h$ and a y-intercept of $-h\nu_0$. Because the emission of electrons only occurs once the incident frequency exceeds $\nu_0$, the physical graph must intersect the y-axis at a negative value and only show kinetic energy for frequencies greater than $\nu_0$. Any graph that passes through the origin or fails to exhibit this negative intercept is an inaccurate representation of the energy balance described by the photoelectric effect.

The fundamental quantization condition for an electron in a Bohr orbit is given by $2\pi r_n = n\lambda$, where $r_n$ denotes the orbital radius. For a hydrogenic atom, this radius is expressed as $r_n = a_0 \frac{n^2}{z}$, so the condition becomes $2\pi a_0 \frac{n^2}{z} = n\lambda$. By substituting the given wavelength $\lambda = 1.5 \pi a_0$ into the equation, we obtain $2\pi a_0 \frac{n^2}{z} = n(1.5 \pi a_0)$. Canceling common terms on both sides simplifies the relationship to $2\frac{n}{z} = 1.5$, which results in the ratio $\frac{n}{z} = 0.75$.

The wave number of an emission line is derived from the difference in energy between an initial state $n_i = 8$ and a final state $n_f = n$. Following the established energy relationship for this transition, the expression for the wave number is $\overline{v} = -R_H \left( \frac{1}{n^2} - \frac{1}{8^2} \right)$. Rearranging this formula to match the standard linear equation $y = mx + c$ yields $\overline{v} = -\frac{R_H}{n^2} + \frac{R_H}{64}$, where $\overline{v}$ serves as the vertical axis $y$ and $\frac{1}{n^2}$ serves as the horizontal axis $x$. This comparison reveals that the coefficient multiplying the variable $\frac{1}{n^2}$ is $-R_H$, which denotes the slope of the plot.

The shortest wavelength for any hydrogen spectral series occurs when an electron transitions from an infinitely distant orbit to the specific base energy level $n_1$, a relationship described by the Rydberg formula $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{\infty } \right)$, which simplifies to $\lambda = \frac{n_1^2}{R_H}$. For the Lyman series, where the base level is $n_1 = 1$, the shortest wavelength is $\lambda = \frac{1}{R_H}$. In the case of the Paschen series, the base level is $n_1 = 3$, resulting in a shortest wavelength of $\lambda = \frac{3^2}{R_H} = \frac{9}{R_H}$. Comparing the Paschen wavelength to the Lyman wavelength yields a ratio of exactly 9, identifying these two series as the ones satisfying the given condition.

Isoelectronic species are defined as atoms or ions that possess an identical number of electrons, regardless of their atomic number. To identify the correct set, calculate the total electron count for each species by adjusting the atomic number based on the ionic charge. For the set containing $N^{3-}$, $O^{2-}$, $F^{-}$, and $Na^{+}$, nitrogen (atomic number 7) gains 3 electrons, oxygen (atomic number 8) gains 2, fluorine (atomic number 9) gains 1, and sodium (atomic number 11) loses 1, resulting in each ion containing exactly 10 electrons.

The spectral lines of atomic hydrogen follow the Rydberg formula, which relates the wavenumber $\bar{v}$ of emitted radiation to the transitions between energy levels defined by principal quantum numbers $n_1$ and $n_2$:
$\bar{v} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
In this formula, $R$ is the Rydberg constant, $n_1$ is the lower energy level, and $n_2$ is the higher energy level. For any specific series, the maximum frequency difference $\Delta \bar{v}$ is obtained by finding the difference between the limiting line ($n_2 = \infty$) and the first line of the series ($n_2 = n_1 + 1$).

For the Lyman series, where $n_1 = 1$, the maximum frequency corresponds to $n_2 = \infty$, yielding $\bar{v}_{\text{max}} = R(1 - 0) = R$. The minimum frequency occurs at the first possible transition, where $n_2 = 2$, giving $\bar{v}_{\text{min}} = R(1 - \frac{1}{4}) = \frac{3}{4}R$. The difference for the Lyman series is therefore $\Delta \bar{v}_{\text{Lyman}} = R - \frac{3}{4}R = \frac{1}{4}R$.

For the Balmer series, where $n_1 = 2$, the maximum frequency corresponds to $n_2 = \infty$, yielding $\bar{v}_{\text{max}} = R(\frac{1}{4} - 0) = \frac{1}{4}R$. The minimum frequency occurs at $n_2 = 3$, giving $\bar{v}_{\text{min}} = R(\frac{1}{4} - \frac{1}{9}) = R(\frac{9-4}{36}) = \frac{5}{36}R$. The difference for the Balmer series is $\Delta \bar{v}_{\text{Balmer}} = \frac{1}{4}R - \frac{5}{36}R = \frac{9-5}{36}R = \frac{4}{36}R = \frac{1}{9}R$.

Calculating the ratio of the frequency differences for the Lyman and Balmer series involves dividing these two values:
$\frac{\Delta \bar{v}_{\text{Lyman}}}{\Delta \bar{v}_{\text{Balmer}}} = \frac{\frac{1}{4}R}{\frac{1}{9}R} = \frac{9}{4}$

When the work function of a metal is negligible compared to the kinetic energy of an ejected photoelectron, the incident photon energy is effectively equivalent to the kinetic energy of the electron, represented by the following equation:

$\frac{hc}{\lambda} = \frac{p^2}{2m}$

This relationship establishes that the wavelength of the incident light is inversely proportional to the square of the electron's momentum. For the initial state with wavelength $\lambda$ and momentum $p$, the energy is proportional to $p^2$. When the momentum increases to $1.5p$, the new wavelength $\lambda'$ must satisfy the condition where the energy is proportional to $(1.5p)^2$, which equals $2.25p^2$. Dividing the initial energy relation by the new energy relation yields $\frac{\lambda'}{\lambda} = \frac{p^2}{2.25p^2}$, which simplifies to $\frac{1}{2.25}$. Therefore, the required wavelength is $\frac{4}{9} \lambda$.

The energy levels of a hydrogen-like ion are determined by the relationship $E_n = -13.6 \frac{Z^2}{n^2}$ eV, where $Z$ is the atomic number and $n$ is the principal quantum number. For a helium ion, the atomic number is 2, and the second excited state corresponds to the third energy level, meaning $n = 3$. Substituting these values into the energy formula identifies the specific energy for this state.

$E = -13.6 \times \frac{2^2}{3^2} = -6.04$ eV

The number of radial nodes in an atomic orbital is determined by the expression $n - l - 1$. A 2s orbital corresponds to the quantum numbers $n=2$ and $l=0$, which yields $2 - 0 - 1 = 1$ radial node, manifesting as the radial probability density function $\Psi^2$ touching the $r$-axis exactly once at a non-zero distance. Since both 1s and 2p orbitals possess zero radial nodes and do not intersect the $r$-axis, a graph exhibiting one radial node uniquely identifies the 2s orbital.

The ozone layer located in the upper stratosphere acts as a critical planetary shield by absorbing high-energy solar radiation that would otherwise be detrimental to life on Earth. Its primary function involves filtering out ultraviolet light in the 200 nm to 315 nm range. By effectively blocking these specific wavelengths, the ozone layer prevents significant environmental and biological damage that this type of radiation would naturally cause upon reaching the surface.

The probability density of finding an electron in space is mathematically represented by the square of the wavefunction, $\Psi^2$. Based on the relationship $\Psi^{2} = a \cdot c$, the likelihood of an electron's presence is linked specifically to these designated areas. Consequently, the particle possesses the highest probability of being located within both region $a$ and region $c$.

Einstein's photoelectric equation represents the conservation of energy in this process, established by the relation $h\nu = \Phi + K$. This principle dictates that the total energy of an incident photon $h\nu$ is distributed between the metal's work function $\Phi$ and the maximum kinetic energy of the emitted photoelectron, denoted as $K$.

The incident photon energy is calculated using $E = \frac{hc}{\lambda}$. By substituting 6.626 x 10^-34 Js for Planck's constant, 3 x 10^8 ms^-1 for the velocity of light, and 4000 x 10^-10 m for the wavelength, the photon energy is determined. The kinetic energy of the electron is then found using the expression $K = \frac{1}{2}mv^2$, substituting the mass of 9 x 10^-31 kg and the velocity of 6 x 10^5 ms^-1. Subtracting this kinetic energy from the incident photon energy provides the work function in Joules. Finally, dividing this result by the electron charge of 1.6 x 10^-19 J/eV converts the work function into the final value of approximately 2.1 eV.

Energy levels of electrons are determined by the sum of their principal quantum number $n$ and azimuthal quantum number $l$, where a lower $(n+l)$ value indicates lower energy. When two orbitals share the same $(n+l)$ sum, the one with the smaller principal quantum number $n$ is lower in energy. Applying this to the given values, electron I ($n=4, l=2$) has a sum of $n+l=6$, electron II ($n=3, l=2$) results in $n+l=5$, electron III ($n=4, l=1$) also results in $n+l=5$, and electron IV ($n=3, l=1$) results in $n+l=4$. Because electron IV has the smallest sum, it occupies the lowest energy state, while electron I has the highest energy. Between II and III, both have equal sums, but because electron II has a lower $n$ value than electron III, it has lower energy, establishing the final descending energy order as $I > III > II > IV$.

The energy of an electron occupying the 1s orbital in a hydrogen atom is governed by the ground state configuration, where the system settles into its most stable arrangement. According to physical principles, the total energy of this electron reaches its minimum value specifically at the Bohr radius, $a_0$, due to the negative potential energy associated with the electrostatic attraction to the nucleus. Stating that the total energy is at a maximum at this distance is therefore incorrect, as the 1s orbital represents the lowest possible energy state of the atom rather than a peak in the energy profile.

Quantum mechanics dictates that an electron's radial distribution is intimately linked to its angular momentum, which is defined by the azimuthal quantum number $l$. For a fixed principal quantum number $n$, an electron occupying an orbital with a higher angular momentum is subjected to an increased centrifugal potential, which effectively keeps it at a greater distance from the nucleus compared to an electron in a lower angular momentum orbital. Additionally, the assertion that the angular momentum of the ground state is equal to $\frac{h}{2\pi}$ aligns with the quantization condition from the Bohr model for the first principal energy level where $n=1$. These principles together explain the fundamental interpretations of atomic orbital behavior described in the statements.