Structure Of Atom – JEE Main ChemistryPractice Questions & PYQs

Orbitals within the same subshell that share identical energy levels are termed degenerate orbitals, and the $2p_x$ and $2p_y$ orbitals are perfect examples of this property. A spectral line only appears when an electron shifts between states of different energy, as the energy difference determines the frequency of the emitted or absorbed radiation. Since there is no energy difference between these two orbitals, a $2p_x \longrightarrow 2p_y$ transition cannot generate a spectral line.

The photoelectric effect occurs if and only if the incident light frequency $\nu$ exceeds the material's threshold frequency $\nu_0$, where the resulting photocurrent magnitude is directly proportional to the incident light intensity. Yellow, blue, and white light frequencies exceed the threshold of $5.16 \times 10^{14} \, \text{Hz}$, thereby facilitating electron emission and generating current, whereas red light remains below this threshold and produces no current at all. Consequently, dimming the yellow light reduces the rate of photoemission and the associated current, while the higher frequencies of blue and white light successfully trigger the photoelectric effect to register current on the ammeter.

The orbital angular momentum of an electron is governed by the azimuthal quantum number $l$ according to the expression $\frac{h}{2\pi}\sqrt{l(l+1)}$. For an $s$ orbital, the azimuthal quantum number is 0, which makes the angular momentum calculation equal to 0. Since a $p$ orbital corresponds to an $l$ value of 1, substituting this into the formula results in $\frac{h}{2\pi}\sqrt{1(1+1)}$, which simplifies to $\sqrt{2}\frac{h}{2\pi}$.

The total number of orbitals in a principal shell defined by the quantum number $n$ is calculated by the relationship $n^2$, which accounts for the combined capacity of all subshells within that energy level. Furthermore, the spatial orientation of orbitals is determined by the magnetic quantum number $m_l$, which allows for all integer values ranging from $-l$ to $+l$ for any given subshell $l$, including zero.

The Rydberg formula provides the theoretical foundation for determining the wavelength of spectral lines emitted by a hydrogen atom, represented by the equation
$\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Given a target wavelength of 900 nm, which converts to $9 \times 10^{-5} \text{ cm}$, the reciprocal of the wavelength is $\frac{10^5}{9} \text{ cm}^{-1}$. Equating this to the Rydberg expression using $R_H = 10^5 \text{ cm}^{-1}$ simplifies the calculation to $\frac{1}{9} = \frac{1}{n_1^2} - \frac{1}{n_2^2}$. This specific relationship is satisfied when an electron transitions from an infinite energy level to the third orbit, which is characteristic of the $\infty \rightarrow 3$ transition within the Paschen series.

The energy of a hydrogen-like ion is defined by the relation $E = -13.6 \frac{Z^2}{n^2}$ where $E$ represents the energy \in eV, $Z$ is the atomic number, and $n$ is the principal quantum number. For beryllium, $Z = 4$, and the first excited state corresponds to the quantum level $n = 2$. Substituting these values gives $E = -13.6 \times \frac{4^2}{2^2} = -54.4$ in units of eV. The magnitude of this energy is 54.4, which rounds to the nearest integer of 54.

Isoelectronic species are defined by having the exact same electron configuration, which is $1s^{2} 2s^{2} 2p^{6}$ for $\mathrm{F}^{-}$, $\mathrm{Ne}$, and $\mathrm{Na}^{+}$. Since the number of electrons remains constant across these particles, any difference in their atomic or ionic size is driven by the specific nuclear charge ($z$) of each element. An increase in the number of protons creates a stronger pull on the shared electron cloud, effectively drawing the electrons inward and reducing the overall size of the species.

The total energy $E_n$ and kinetic energy $K_n$ of an electron in a hydrogen atom orbit are related by the expression $K_n = -E_n$. Substituting the given total energy $E_n = -3.4$ eV into this relation yields $K_n = -(-3.4)$ eV, which simplifies to $K_n = 3.4$ eV. Since the kinetic energy is defined as $x$ eV, setting $x = 3.4$ and expressing this value in the required format gives $x = 34 \times 10^{-1}$. Thus, the required value for the blank is 34.

The spectral series of the hydrogen atom are determined by the specific energy level to which an electron transitions during emission, with each jump producing a photon of a characteristic wavelength. Transitions terminating at the ground state, $n_1 = 1$, constitute the Lyman series, which releases high-energy photons in the ultraviolet region. The Balmer series involves transitions to the first excited state, $n_1 = 2$, which corresponds to the visible region of the electromagnetic spectrum. As electrons cascade to higher orbits, such as $n_1 = 3$ for the Paschen series and $n_1 = 5$ for the Pfund series, the emitted photons possess significantly lower energy, placing their wavelengths squarely within the infrared region. This relationship matches the Lyman series to the ultraviolet range, the Balmer series to the visible range, and both the Paschen and Pfund series to the infrared range.

Heisenberg's uncertainty principle establishes a fundamental limit on the precision with which the position and velocity of a particle can be simultaneously measured, governed by the equation $\Delta x \cdot m \Delta v = \frac{h}{4\pi}$. When applying this principle to an electron confined within a nucleus of diameter $10^{-15} \mathrm{m}$, the uncertainty in position $\Delta x$ is taken as $10^{-15} \mathrm{m}$. Substituting the mass $m = 9.1 \times 10^{-31} \mathrm{kg}$ and Planck's constant $h = 6.626 \times 10^{-34} \mathrm{Js}$ into the rearranged formula results in $\Delta v = \frac{6.626 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 10^{-15}}$. This yields a value of $57.97 \times 10^{9} \mathrm{ms}^{-1}$, which is 58 when rounded to the nearest integer.

Orbitals sharing identical energy levels are defined as degenerate orbitals, which confirms the accuracy of the first statement. This principle serves as a fundamental classification in quantum mechanics for atomic states.

In the specific case of single-electron species like the hydrogen atom, the energy of an orbital depends exclusively on the principal quantum number $n$. Since both 3p and 3d orbitals share the same value of $n=3$, they reside at the same energy level and are therefore degenerate. Because the second statement incorrectly asserts that these orbitals are not degenerate, it is false.

Dalton's Atomic Theory establishes that when atoms of different elements combine to form a compound, they must do so in a precise, fixed ratio by mass. Because this law of definite proportions is a cornerstone of the theory, suggesting that elements combine in any arbitrary ratio is factually wrong. While the other statements correctly describe atoms as indivisible building blocks, attribute identical properties to atoms of the same element, and explain chemical reactions as the reorganization of these atoms, this claim about variable ratios contradicts the theory's foundational requirement.

The energy of an orbital in a multielectron system is determined by the $(n+\ell)$ rule, where a larger sum of the principal quantum number $n$ and the azimuthal quantum number $\ell$ indicates higher energy. In cases where the $(n+\ell)$ sums are identical, the orbital with the higher principal quantum number $n$ is at a higher energy level.

Calculating the sums for the provided sets gives $(n+\ell)=5$ for (A), $(n+\ell)=6$ for (B), $(n+\ell)=4$ for (C), $(n+\ell)=5$ for (D), and $(n+\ell)=4$ for (E). Analyzing the orbitals with a sum of 4, (C) has $n=3$ while (E) has $n=4$, so (C) has less energy than (E). For the orbitals with a sum of 5, (D) has $n=3$ while (A) has $n=4$, meaning (D) has less energy than (A). Since (B) has the highest sum of 6, it possesses the highest energy among all the sets. This leads to an increasing energy order of $(C) < (E) < (D) < (A) < (B)$.

The frequency of the de Broglie wave associated with an electron is fundamentally determined by the ratio of its velocity to its wavelength, represented as $f = \frac{v}{\lambda}$. Incorporating the de Broglie wavelength formula $\lambda = \frac{h}{mv}$, we can derive the expression:
$f = \frac{mv^2}{h}$
Since the kinetic energy of the electron is given as $\frac{1}{2} mv^2 = 2.18 \times 10^{-18} \mathrm{J}$, the term $mv^2$ is simply double that value, totaling $4.36 \times 10^{-18} \mathrm{J}$. By dividing this energy product by Planck's constant $h = 6.6 \times 10^{-34} \mathrm{J \cdot s}$, we calculate the frequency as $\frac{4.36 \times 10^{-18}}{6.6 \times 10^{-34}} \mathrm{Hz}$. This division yields approximately $0.6606 \times 10^{16} \mathrm{Hz}$, which is equivalent to $660.6 \times 10^{13} \mathrm{Hz}$, rounding to 661 when expressed in the specified units.

Radial nodes define the spherical regions around the nucleus where the probability of finding an electron is zero, with the total number of these nodes depending on the specific quantum numbers of an orbital. For a 3p orbital, the principal quantum number $n$ is 3 and the azimuthal quantum number $\ell$ is 1. Applying the formula for radial nodes, $n - \ell - 1$, we substitute these values to calculate $3 - 1 - 1$, which equals 1.

When an electron falls from a higher energy state to a lower one, it can follow multiple pathways, producing a specific number of spectral lines that depend on the principal quantum numbers of the initial and final energy levels. To identify these levels, note that the fifth excited state is $n = 6$ and the first excited state is $n = 2$, which gives a difference of $\Delta n = 4$. By applying the formula $N = \frac{\Delta n(\Delta n + 1)}{2}$ for the maximum number of spectral lines, we perform the calculation $\frac{4(4 + 1)}{2}$, resulting in 10.

Wavenumber is defined as the reciprocal of wavelength, expressed as $\bar{\nu} = 1/\lambda$. Converting the wavelength of 5800 angstroms into centimeters requires multiplying by $10^{-8}$, yielding a wavelength of $5800 \times 10^{-8}$ cm. The reciprocal of this wavelength results in approximately 17241 cm⁻¹. Expressing this quantity as $1724 \times 10$ cm⁻¹ aligns it with the required format, identifying the value of x as 1724.

The principal quantum number $n=4$ defines the fourth energy shell, which contains four distinct subshells characterized by azimuthal quantum numbers $l=0, 1, 2, \text{ and } 3$. For any given subshell, the magnetic quantum number $m_l$ ranges from $-l$ to $+l$, ensuring that each of these subshells possesses exactly one orbital where $m_l=0$. Since each of the four subshells contributes one specific orbital with this magnetic quantum number, the total number of such orbitals in the $n=4$ level is 4.

Dalton's atomic theory originally postulated that atoms are the indivisible building blocks of matter and that all atoms of a specific element are identical in all respects, including mass. These claims are considered incorrect today because atoms are now known to be divisible into subatomic particles like protons, neutrons, and electrons, and the existence of isotopes demonstrates that atoms of the same element can possess different masses.

Bohr's quantization condition states that the circumference of an electron's orbit must equal an integer multiple of its de-Broglie wavelength, expressed as $2 \pi r_n = n \lambda$ For an electron \in the fourth orbit, the radius is defined by the formula $r_n = a_0 n^2$, which becomes $r_4 = 16 a_0$ when $n=4$. Substituting this orbital radius into the quantization condition produces the equation $2 \pi (16 a_0) = 4 \lambda$ Solving this expression for the wavelength confirms that $\lambda = 8 \pi a_0$