Structure Of Atom – JEE Main ChemistryPractice Questions & PYQs

The de Broglie wavelength of a particle is defined by the inverse relationship with its momentum, specifically $\lambda = \frac{h}{p}$. Since the kinetic energy of a particle is related to its momentum by $K.E. = \frac{p^2}{2m}$, the wavelength can be expressed in terms of energy as
$\lambda = \frac{h}{\sqrt{2m K.E.}}$
According to the photoelectric effect, the energy of an ejected electron is determined by the incident radiation frequency $v$ and the threshold frequency $v_0$, expressed as $K.E. = hv - hv_0 = h(v - v_0)$. Substituting this energy expression into the wavelength formula yields
$\lambda = \frac{h}{\sqrt{2m h(v - v_0)}}$
Because the Planck constant $h$ and the electron mass $m$ are constants, the wavelength depends on the inverse square root of the difference between the incident and threshold frequencies, establishing the proportionality $\lambda \propto \frac{1}{(v - v_0)^{1/2}}$.

An element with an atomic number of 15 possesses an electronic configuration of $1s^2 2s^2 2p^6 3s^2 3p^3$, which simplifies to a valence shell structure of ${ns}^{2} {np}^{3}$. This arrangement shows 5 electrons in the outermost shell, placing the element in group 15 of the periodic table. Furthermore, since the element requires 3 additional electrons to achieve a stable octet, its valency is 3.

Einstein's photoelectric equation is expressed as $E = \Phi + K_{max}$, where $E$ is the incident photon energy, $\Phi$ is the work function, and $K_{max} = eV_s$ represents the maximum kinetic energy of the photoelectrons corresponding to the stopping potential $V_s$. The energy of the incident radiation is calculated using the relation $E = \frac{hc}{\lambda}$, which equates to approximately $\frac{1240 \text{ eV nm}}{250 \text{ nm}} = 4.96 \text{ eV}$. By substituting the given stopping potential of $0.5 \text{ V}$ into the equation, the work function is determined as $\Phi = 4.96 \text{ eV} - 0.5 \text{ eV} = 4.46 \text{ eV}$. Rounding this value to the nearest provided option yields $4.5 \text{ eV}$.

Wien's displacement law states that the wavelength corresponding to the maximum intensity of emission from a black body ($\lambda_{max}$) is inversely proportional to the absolute temperature $T$, given by the relation $\lambda_{max} T = b$. As the temperature of a black body increases, the peak intensity shifts toward shorter wavelengths, which corresponds to higher frequencies rather than higher wavelengths. Therefore, the assertion that the radiation shifts toward higher wavelengths as temperature increases is physically incorrect.

Bohr's postulate states that the circumference of an electron's orbit must be an integral multiple of its de Broglie wavelength, defined by the relation $2\pi r = n\lambda$. Rearranging this expression to solve for the wavelength yields $\lambda = \frac{2\pi r}{n}$. For the first Bohr orbit where the principal quantum number $n = 1$ and the radius $r = 0.529$ Å, the wavelength is calculated as $\lambda = 2\pi \times 0.529$ Å.

The Rydberg formula, which relates the wavelength of emitted radiation to electron transitions in hydrogen-like atoms, is expressed as:
$\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
In the Lyman series of a hydrogen atom where the atomic number is $Z=1$, the shortest wavelength occurs at the series limit, representing a transition from $n_2=\infty$ to the ground state $n_1=1$. Substituting these values gives $\frac{1}{A} = R (1)^2 (\frac{1}{1^2} - 0)$, which simplifies to the relationship $R = \frac{1}{A}$.

For the Paschen series of the helium ion $He^+$ with $Z=2$, the electron transitions to the energy level $n_1=3$. The longest wavelength corresponds to the first line of the series, which involves the smallest energy jump between adjacent levels, specifically from $n_2=4$ to $n_1=3$. Applying the Rydberg formula here yields:
$\frac{1}{\lambda} = R (2)^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{9} - \frac{1}{16} \right) = 4R \left( \frac{16-9}{144} \right) = \frac{7R}{36}$
Substituting the previously determined value of $R = \frac{1}{A}$ into this expression results in $\frac{1}{\lambda} = \frac{7}{36A}$, which leads to the wavelength being $\lambda = \frac{36A}{7}$.

The radius of a Bohr orbit is determined by the balance of electrostatic and centripetal forces, given by the expression $r = \frac{n^2 h^2}{4 \pi^2 m e^2 k}$ where $n$ is the principal quantum number. For the second orbit, we use $n = 2$ and the values $h = 6.6262 × 10^{-34}$ Js, $m = 9.1091 × 10^{-31}$ kg, $e = 1.60210 × 10^{-19}$ C, and $k = 9 × 10^9$. Substituting these into the formula results \in $r = \frac{2^2 \times (6.6262 \times 10^{-34})^2}{4 \times (3.1416)^2 \times 9.1091 \times 10^{-31} \times (1.60210 \times 10^{-19})^2 \times 9 \times 10^9}$ which calculates to 2.12 Å.

Isoelectronic species are defined by having an identical number of electrons, regardless of their atomic identity. Calculating the electron count for each ion reveals that oxygen with an atomic number of 8 gains 2 electrons to reach 10, fluorine with an atomic number of 9 gains 1 to reach 10, sodium with an atomic number of 11 loses 1 to reach 10, and magnesium with an atomic number of 12 loses 2 to reach 10. Because $\mathrm{O}^{2-}$, $\mathrm{F}^{-}$, $\mathrm{Na}^{+}$, and $\mathrm{Mg}^{2+}$ all share this total of 10 electrons, they constitute an isoelectronic series.

The radius of the $n$-th Bohr orbit of a hydrogen atom is determined by the formula $r_n = 0.529 \times n^2 \, \text{Å}$. By converting the given orbital radius of $211.6 \, \text{pm}$ to $2.116 \, \text{Å}$, we can calculate the principal quantum number $n$ as follows:

$2.116 = 0.529 \times n^2$
$n^2 = \frac{2.116}{0.529} = 4$
$n = 2$

Since the electron transitions to the $n=2$ orbit, this process signifies the emission of energy corresponding to the Balmer series.

The total number of orbitals in any principal energy level is determined by $n^2$. For the principal quantum number $n=5$, the allowed subshells are 5s, 5p, 5d, 5f, and 5g. Summing the individual orbitals found within these subshells requires adding 1 from the 5s, 3 from the 5p, 5 from the 5d, 7 from the 5f, and 9 from the 5g subshell. Combining these values gives the final count of $1 + 3 + 5 + 7 + 9 = 25$ orbitals.

The de Broglie relation defines the wavelength of a matter wave as $\lambda = \frac{h}{p}$, implying that the quantity $\frac{h}{\lambda}$ is equivalent to the momentum $p$ of the electron. When an electron is accelerated by a potential difference $V$, its kinetic energy increases to $K = eV$. Because the kinetic energy of a particle is related to its momentum by $K = \frac{p^2}{2m}$, we can isolate momentum as $p = \sqrt{2mK}$. Substituting the kinetic energy $eV$ into this expression yields $p = \sqrt{2meV}$, which confirms that $\frac{h}{\lambda}$ is equal to $\sqrt{2meV}$.

The discrete energy levels of a hydrogen atom are governed by the Bohr model, where the energy of the electron at any integer principal quantum number $n$ is given by the expression
$E_n = -13.6 \frac{Z^2}{n^2}$
Given that the atomic number $Z$ for hydrogen is 1, this formula simplifies to $E_n = -13.6/n^2$ in units of eV. Substituting $n=2$ for the first excited state, the calculation $-13.6/4$ yields $-3.4$ eV, confirming this as a valid energy level for the atom.

Rubidium possesses an atomic number of 37, which results in an electronic configuration where the single valence electron resides in the $5s$ orbital. Because this electron is located in the fifth principal shell, its principal quantum number is $n=5$, and its occupancy of an $s$ subshell dictates that the azimuthal quantum number is $l=0$. Consequently, the magnetic quantum number is constrained to $m=0$, and the spin quantum number is conventionally assigned the value $s=+\frac{1}{2}$.

Exciting an electron from the $n=1$ shell to the $n=2$ shell requires an input of energy equivalent to the difference between these two levels. For a hydrogen atom where the atomic number is 1, this energy gap is determined by the following formula:

$\Delta E = 2.178 \times 10^{-18} \text{ J} \times Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

Substituting $n_1 = 1$ and $n_2 = 2$ into this expression yields the energy difference, which is related to the wavelength of the required light by $\Delta E = \frac{hc}{\lambda}$. By rearranging this to $\lambda = \frac{hc}{\Delta E}$ and applying the provided constants $h = 6.62 \times 10^{-34}$ J s and $c = 3.0 \times 10^8$ m/s, the calculated wavelength is $1.214 \times 10^{-7}$ m.

To determine the relative energy levels of orbitals, we apply the $(n+l)$ rule, which states that the energy of an orbital increases as the sum of the principal quantum number $n$ and the azimuthal quantum number $l$ increases. In cases where this sum is identical, the orbital with the lower principal quantum number $n$ sits at a lower energy level. Evaluating the provided options, $3p$ ($n=3, l=1$) and $4s$ ($n=4, l=0$) both yield a sum of $4$, while $3d$ ($n=3, l=2$) and $4p$ ($n=4, l=1$) both result in a sum of $5$. Since the tie-breaker rule gives preference to the lower $n$ value, the final increasing energy sequence is $(d) < (b) < (c) < (a)$.

This physical process is governed by the law of conservation of energy, where the energy of the absorbed photon is distributed between the two resulting emitted photons. Because photon energy is proportional to the inverse of its wavelength, the relationship follows the equation $\frac{1}{\lambda} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$. Substituting the known wavelengths yields $\frac{1}{355} = \frac{1}{680} + \frac{1}{\lambda_2}$. Rearranging to solve for the missing reciprocal wavelength gives $\frac{1}{\lambda_2} = \frac{1}{355} - \frac{1}{680}$, which results in a value of approximately 0.001346 per nanometer, identifying the second emission wavelength as 742.77 nm, or 743 nm.

The energy of an electron in a hydrogen-like species is governed by the relationship $E_n = -E_H \times \frac{Z^2}{n^2}$, where $E_H$ represents the ground state energy constant for hydrogen and $Z$ is the atomic number. Ionization energy represents the energy required to remove an electron from the ground state to infinity, where the potential energy is defined as zero, meaning the ionization energy is equivalent to the negative of the ground state energy. Given the ionization energy of $\mathrm{He}^{+}$ is $19.6 \times 10^{-18} \ \mathrm{J} \ \mathrm{atom}^{-1}$, its ground state energy $E_1$ satisfies the equation $E_1 = -19.6 \times 10^{-18} \ \mathrm{J} \ \mathrm{atom}^{-1}$.

Using this value, the ground state energy constant for hydrogen is derived by rearranging the equation $E_1 = -E_H \times Z^2$ for $\mathrm{He}^{+}$, where $Z = 2$, yielding $E_H = \frac{19.6 \times 10^{-18}}{4} \ \mathrm{J} \ \mathrm{atom}^{-1}$. For $\mathrm{Li}^{2+}$, which has an atomic number of $Z = 3$, the energy of the first stationary state is calculated as $E_1 = -E_H \times 3^2$. Substituting the derived value of $E_H$ into this expression gives $E_1 = -\frac{19.6 \times 10^{-18}}{4} \times 9$, which results in $-4.41 \times 10^{-17} \ \mathrm{J} \ \mathrm{atom}^{-1}$.

Breaking a single Cl-Cl bond requires a photon with energy at least equal to the bond dissociation energy per molecule. Because the energy provided is 242 kJ/mol, we first convert this to the energy required for one molecule by dividing 242,000 J by 6.02 × 10^23.

The relationship between energy and wavelength is expressed as $E = \frac{hc}{\lambda}$. Rearranging this to solve for the wavelength gives $\lambda = \frac{hc}{E}$, where $h$ is 6.626 × 10^-34 J s and $c$ is 3 × 10^8 m/s. Inserting these constants and the calculated energy per molecule into the equation results in 494 × 10^-9 m, which corresponds to 494 nm.

Heisenberg's Uncertainty Principle asserts that it is impossible to simultaneously measure the position and momentum of a subatomic particle with absolute precision, with the product of their uncertainties defined by the inequality $\Delta x \cdot \Delta p \geq \frac{h}{4\pi}$. Since momentum uncertainty is the product of the particle mass and the uncertainty in its velocity, we first determine $\Delta v$ by applying the percentage error to the given speed. A speed of 600 m/s with an accuracy of 0.005% results in $\Delta v = 600 \times \frac{0.005}{100} = 0.03\ \mathrm{m/s}$.

Using the rearranged principle $\Delta x = \frac{h}{4\pi m_e \Delta v}$ and substituting the provided constants $h = 6.6 \times 10^{-34}\ \mathrm{kg}\ \mathrm{m}^2\ \mathrm{s}^{-1}$ and $m_e = 9.1 \times 10^{-31}\ \mathrm{kg}$, we calculate the uncertainty in position. Performing the calculation leads to the following expression:

$\Delta x = \frac{6.6 \times 10^{-34}}{4 \times 3.14 \times 9.1 \times 10^{-31} \times 0.03} \approx 1.92 \times 10^{-3}\ \mathrm{m}$

To determine the wavelength of the moving proton, we use the de Broglie relation, which states that the wavelength is inversely proportional to momentum, given by $\lambda = \frac{h}{mv}$. We plug the provided values for Planck's constant, the proton's mass, and its velocity into this equation: $\lambda = \frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 10^3}$. Calculating this quotient yields a value of 0.4 x 10^-9 meters. Since 1 nanometer is equal to 10^-9 meters, the wavelength is 0.4 nm.