Structure Of Atom – JEE Main ChemistryPractice Questions & PYQs

The quantization of electronic energy levels in gas-phase atoms mandates that emission spectra consist of discrete, well-defined lines corresponding to specific energy transitions, rather than a continuous spread of wavelengths. This characteristic line structure acts as a unique fingerprint for each element, enabling the study of electronic configurations, the observation of absorption spectra as negative matches to emission patterns, and the identification of extraterrestrial elements like helium within the sun. Since the claim that atomic emission spectra are continuous is scientifically inaccurate, whereas the remaining three statements are factual, there is only one incorrect statement in the set.

The Rydberg formula, defined as $\frac{1}{\lambda} = R_H Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$, determines the wavelengths of electronic transitions within a hydrogen atom. For the Paschen series, transitions end at the energy level $n_1 = 3$. The first spectral line involves a transition from $n_2 = 4$ to $n_1 = 3$, and the second spectral line corresponds to a transition from $n_2 = 5$ to $n_1 = 3$.

Applying this formula to both transitions provides the necessary relationship between their wavelengths. For the first line, the expression is $\frac{1}{720} = R_H Z^2 \left(\frac{1}{3^2} - \frac{1}{4^2}\right)$, which simplifies to $\frac{1}{720} = R_H Z^2 \left(\frac{7}{144}\right)$. For the second line, the expression is $\frac{1}{\lambda_2} = R_H Z^2 \left(\frac{1}{3^2} - \frac{1}{5^2}\right)$, which simplifies to $\frac{1}{\lambda_2} = R_H Z^2 \left(\frac{16}{225}\right)$. Taking the ratio of these equations isolates $\lambda_2$ by eliminating the Rydberg constant and the nuclear charge term, yielding $\frac{720}{\lambda_2} = \frac{16 \times 144}{225 \times 7}$. Solving this equation leads to $\lambda_2 = 720 \times \frac{25 \times 7}{16 \times 16}$, which calculates to 492 nm when rounded to the nearest integer.

To determine the energy required for an electronic transition in a hydrogen-like species such as $\mathrm{Li}^{2+}$, we apply the Rydberg equation, which accounts for the energy difference between two distinct orbits. For these systems, the transition energy is governed by the Rydberg constant, the square of the atomic number, and the inverse squares of the initial and final quantum levels, with Lithium having an atomic number of 3.

The energy change follows the relationship:
$\Delta E = R_{\mathrm{H}} Z^2 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$
Substituting the given values, $1.47 \times 10^{-17} = 2.18 \times 10^{-18} \times 3^2 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$. Simplifying the constant terms leads to the equation $1.47 \times 10^{-17} = 1.96 \times 10^{-17} \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$. Dividing the energy by this factor yields a ratio of $3/4$. Because the difference $\frac{1}{n^2} - \frac{1}{(n+1)^2}$ equates to $0.75$ only when $n=1$, the final value is 1.

The probability distribution of electrons in atoms is described by the wave function, where the square of this function determines the probability density of finding an electron at a given point. For a 1s orbital, this probability density is highest at the nucleus, whereas for p-orbitals, the presence of an angular node at the nucleus results in a probability density of zero at that specific location. Regarding orbital nodes, the number of angular nodes is determined by the azimuthal quantum number $l$, meaning p-orbitals with $l=1$ possess one angular node and d-orbitals with $l=2$ possess two. Conversely, boundary surface diagrams are drawn to encompass regions where an electron is most likely to be found, typically representing roughly 90 to 95 percent of the total probability rather than a complete 100 percent, and for a 2s orbital, the probability density behavior is characterized by an initial decrease followed by an increase rather than the reverse.

The energy of a single photon is determined by the product of Planck's constant and the frequency of the radiation, defined by the relationship $E = h\nu$. To determine the total energy contained within one mole of these photons, we multiply the energy of a single photon by Avogadro's number, yielding the formula $E_{mole} = N_A \times h \times \nu$.

Substituting the provided constants into this equation requires multiplying Avogadro's number, $N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$, by Planck's constant, $h = 6.626 \times 10^{-34}\ \text{J s}$, and the frequency of the radiation, $\nu = 2 \times 10^{12}\ \text{Hz}$. Executing the multiplication $E_{mole} = (6.022 \times 10^{23}) \times (6.626 \times 10^{-34}) \times (2 \times 10^{12})$ results in a value of $797.956\ \text{J mol}^{-1}$. Rounding this calculation to the nearest integer provides a final value of $798\ \text{J mol}^{-1}$.

A radial node identifies a distance $r_0$ where the electron probability density vanishes, which requires the wave function $\Psi_{2s}$ to be equal to zero. Within the expression for $\Psi_{2s}$, the constant coefficients and the exponential decay term $e^{-\frac{r}{2 a_0}}$ never reach zero, so the node is defined entirely by the root of the polynomial factor. Setting this term to zero results in $2 - \frac{r_0}{a_0} = 0$, and solving for the distance yields $r_0 = 2 a_0$.

The count of radial nodes in an atomic orbital follows the formula $n - \ell - 1$, utilizing the principal quantum number $n$ and the azimuthal quantum number $\ell$. Applying this to the provided orbitals 7s, 7p, 6s, 8p, 8d reveals which ones have 5 radial nodes. For 7s, $n=7$ and $\ell=0$ gives $7 - 0 - 1 = 6$. For 7p, $n=7$ and $\ell=1$ results in $7 - 1 - 1 = 5$. For 6s, $n=6$ and $\ell=0$ results in $6 - 0 - 1 = 5$. For 8p, $n=8$ and $\ell=1$ leads to $8 - 1 - 1 = 6$. Finally, for 8d, $n=8$ and $\ell=2$ results in $8 - 2 - 1 = 5$. Identifying the matches, we find that the 7p, 6s, and 8d orbitals meet the condition, providing a total of 3 orbitals.

The maximum electron capacity for any principal energy level is determined by the formula $2n^2$, where $n$ represents the principal quantum number. Substituting $n=4$ into this expression results in $2 \times (4)^2 = 32$ electrons. This total is further confirmed by summing the capacities of the individual subshells within the fourth shell, where the 4s subshell holds 2 electrons, the 4p subshell holds 6, the 4d subshell holds 10, and the 4f subshell holds 14, collectively amounting to 32 electrons.

A metal emits photoelectrons only when the energy of the incident photon is at least equal to its work function, $W_0$. We calculate the energy of the incident light with a wavelength of $400\ \text{nm}$ using the equation $E = \frac{hc}{\lambda}$. By incorporating the constants $h = 6.6 \times 10^{-34}\ \text{J s}$ and $c = 3 \times 10^8\ \text{m s}^{-1}$, and converting the result into electron volts by dividing by $e = 1.6 \times 10^{-19}\ \text{C}$, the incident photon energy evaluates to $E = \frac{1240}{400} = 3.1\ \text{eV}$.

Comparing this incident energy of $3.1\ \text{eV}$ against the work functions provided for each metal reveals which ones will undergo the photoelectric effect, as the condition $E \ge W_0$ must be met. Specifically, Li at $2.42\ \text{eV}$, Na at $2.3\ \text{eV}$, and K at $2.25\ \text{eV}$ all possess work functions lower than the incident light energy, allowing electron emission. Meanwhile, the work functions for Mg at $3.7\ \text{eV}$, Cu at $4.8\ \text{eV}$, and Ag at $4.3\ \text{eV}$ all exceed $3.1\ \text{eV}$, meaning these metals do not meet the requirement for the photoelectric effect. Therefore, exactly three of the listed metals are capable of emitting photoelectrons under these conditions.

The radius of an electron orbit in a hydrogen-like atom is governed by the Bohr model, which expresses the radius as $r \propto \frac{n^2}{Z}$, where $n$ is the principal quantum number and $Z$ is the atomic number. Given that the radius of the second orbit ($n=2$) for $\mathrm{Li}^{2+}$ ($Z=3$) is defined as $x = k \frac{2^2}{3}$, we can determine the radius for the third orbit ($n=3$) of $\mathrm{Be}^{3+}$ ($Z=4$) as $r = k \frac{3^2}{4}$. Calculating the ratio of these two radii, $\frac{r}{x} = \frac{3^2 / 4}{2^2 / 3}$, leads to $\frac{9}{4} \times \frac{3}{4}$, which simplifies to $\frac{27}{16}$. This results in the expected radius being $\frac{27}{16} x$.

Isoelectronic species are defined as atoms or ions that share an identical number of electrons despite having different atomic numbers. Determining the electron count involves adjusting the atomic number based on the ionic charge, where cations lose electrons and anions gain them. The group $(\mathrm{K}^{+}, \mathrm{Cl}^{-}, \mathrm{Ca}^{2+}, \mathrm{Sc}^{3+})$ consists of ions that each possess exactly 18 electrons, calculated from their respective atomic numbers of 19, 17, 20, and 21 after accounting for the loss or gain of electrons.

Orbital shapes are characterized by the spatial distribution of electron density, which is either oriented along the coordinate axes or between them. The three $p$ orbitals-$p_x, p_y,$ and $p_z$-consist of lobes lying directly along the Cartesian axes, and among the $d$ orbitals, $d_{z^2}$ and $d_{x^{2-}y^2}$ also feature electron density directed along the axes. Conversely, the $d_{xy}, d_{yz},$ and $d_{xz}$ orbitals have lobes situated in the planes between the axes, bringing the total count of axial orbitals in the given set to 5.

The radius of an electron's orbit in hydrogen-like species depends directly on the square of the orbit number $n$ and inversely on the atomic number $Z$, expressed by the proportionality $r_n \propto \frac{n^2}{Z}$. This fundamental relationship allows us to determine the dimensions of any Bohr orbit for single-electron systems by scaling the known radius of the first hydrogen orbit.

For the He$^+$ ion, which has an atomic number $Z = 2$, we seek the radius of the third orbit where $n = 3$. Using the provided baseline of 0.6 Å for the first hydrogen orbit, the calculation follows the formula $r = 0.6 \times \frac{n^2}{Z} = 0.6 \times \frac{3^2}{2}$. Simplifying this expression leads to 0.6 × 4.5, which results in 2.7 Å. Converting this measurement to picometers, where 1 Å is equal to 100 pm, yields a final radius of 270 pm.

The de Broglie wavelength is determined by the equation $\lambda = \frac{h}{mv}$, which relates wavelength to Planck's constant, particle mass, and velocity. Substituting the provided values of 6 x 10⁻³⁴ Js for h, 9 x 10⁻³¹ kg for m, and 1000 ms⁻¹ for v, the calculation yields:

$\lambda = \frac{6 \times 10^{-34}}{9 \times 10^{-31} \times 1000} = 666.67 \times 10^{-9} \text{ m} = 666.67 \text{ nm}$

This calculation confirms that the first statement is true. Cathode rays are composed of electrons, which are universal subatomic particles, meaning their characteristics are inherently independent of the material used for the electrodes or the nature of any gas present in the tube. This invalidates both the second and fourth statements. Finally, cathode rays consistently originate from the cathode and travel toward the anode, which confirms that the third statement is true. Since exactly two of the four provided statements are factually correct, the total number of true statements is two.

The Rydberg formula $\frac{1}{\lambda} = R Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$ determines the wavelength of spectral transitions based on the atomic number $Z$ and the principal quantum numbers $n_1$ and $n_2$. For the $\mathrm{He}^{+}$ ion where $Z=2$, the transition from $n=4$ to $n=2$ results in:
$\frac{1}{\lambda} = R(2)^2 \left(\frac{1}{2^2} - \frac{1}{4^2}\right) = R(4) \left(\frac{1}{4} - \frac{1}{16}\right) = R \left(\frac{3}{4}\right)$
For a hydrogen atom, $Z=1$, so the corresponding equation is $\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. Equating the two expressions requires finding integers $n_1$ and $n_2$ such that $\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4}$, which is satisfied when $n_1=1$ and $n_2=2$, corresponding to the transition from the second energy level to the first.

Bohr's model is fundamentally built on the quantization of angular momentum, where an electron is restricted to stationary orbits with a value of $\frac{nh}{2\pi}$. While the Heisenberg uncertainty principle describes the nature of electrons in terms of probability distributions within orbitals, Bohr's theory of fixed orbits does not constitute a violation of this principle; rather, it represents a specific, quantized model of atomic structure that operates under its own distinct set of postulates.

Isoelectronic species are defined as chemical entities that possess an identical total number of electrons. Evaluating the provided set of ions, we determine the electron count for each by adjusting the atomic number according to the ionic charge: $\mathrm{O}^{2-}$ ($8+2=10$), $\mathrm{F}^-$ ($9+1=10$), $\mathrm{Na}^+$ ($11-1=10$), $\mathrm{Mg}^{2+}$ ($12-2=10$), $\mathrm{Al}^{3+}$ ($13-3=10$), $\mathrm{P}^{3-}$ ($15+3=18$), $\mathrm{S}^{2-}$ ($16+2=18$), and $\mathrm{Cl}^-$ ($17+1=18$). Because the five species $\mathrm{O}^{2-}, \mathrm{F}^-, \mathrm{Na}^+, \mathrm{Mg}^{2+},$ and $\mathrm{Al}^{3+}$ each contain 10 electrons, they form a group of five isoelectronic species, while the remaining ions contain 18 electrons.

The Rydberg formula, $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, defines the wavelength of emitted radiation based on the atomic number $Z$ and electron energy levels $n_1$ and $n_2$. Hydrogen's shortest Lyman series wavelength involves an electron transition from $n_2=\infty$ to $n_1=1$, which simplifies to the relationship $\frac{1}{\lambda} = R(1)^2(1) = R$. The longest wavelength in the Balmer series of the $\mathrm{He}^{+}$ ion corresponds to the transition from $n_2=3$ to $n_1=2$, with $Z=2$. Calculation using the Rydberg equation yields the following expression:
$\frac{1}{\lambda_{\mathrm{He}^{+}}} = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{5}{36} \right) = \frac{5R}{9}$
Substituting $R = \frac{1}{\lambda}$ into this equality results in $\frac{1}{\lambda_{\mathrm{He}^{+}}} = \frac{5}{9\lambda}$, which identifies the required wavelength as $\frac{9\lambda}{5}$.

The energy of orbitals is determined primarily by the $(n+l)$ rule, where a higher sum indicates greater energy, and secondarily by the value of $n$ when these sums are identical. Calculating these values, we find for (A) $n+l=3$, for (B) $n+l=4$, for (C) $n+l=4$, and for (D) $n+l=5$. Since (D) has the highest sum of 5, it holds the most energy, while (A) has the lowest sum of 3 and thus the lowest energy. Comparing (B) and (C), which both share a sum of 4, we prioritize the orbital with the higher principal quantum number, placing (B) above (C) because its $n=4$ exceeds the $n=3$ of (C), resulting in the final energy order $(D) > (B) > (C) > (A)$.

Heisenberg's uncertainty principle defines the fundamental limit of precision for simultaneous position and momentum measurements as
$\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$
Because momentum is defined as the product of mass and velocity, this principle translates to $\Delta x \cdot m \Delta v \geq \frac{h}{4\pi}$, which allows us to isolate the minimum uncertainty in velocity as
$\Delta v = \frac{h}{4\pi m \Delta x}$
Substituting the known values-where $\Delta x = 2 \times 52.9 \times 10^{-12}$ meters, mass $m = 9.1 \times 10^{-31}$ kilograms, and Planck's constant $h = 6.63 \times 10^{-34}$ Joules seconds-into this equation results in a calculation of
$\Delta v = \frac{6.63 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 2 \times 52.9 \times 10^{-12}}$
The result is 548273 meters per second, which converts to 548.273 kilometers per second or approximately 548 kilometers per second.