Structure Of Atom – JEE Main ChemistryPractice Questions & PYQs

Louis de Broglie proposed that particles like electrons possess wave-like properties, mathematically defined by the relationship $\lambda = \frac{h}{p}$, where $h$ denotes Planck's constant. This equation indicates that wavelength and momentum are inversely proportional to one another, represented as $\lambda \propto \frac{1}{p}$. As a result of this inverse dependence, where the product of the two variables remains a constant value, the corresponding graph takes the form of a rectangular hyperbola.

Bohr's atomic model provides the relationship for the velocity of an electron revolving in the $n$-th orbit of a hydrogen-like atom as $V = 2.18 \times 10^6 \times \frac{Z}{n}$ m/s, where $Z$ is the atomic number and $n$ is the principal quantum number. For a hydrogen atom, $Z = 1$, and for the first orbit, $n = 1$. Substituting these values into the formula yields a velocity of $2.18 \times 10^6$ m/s. Converting this value to be expressed in terms of $10^5$ m/s involves shifting the decimal point, resulting in $21.8 \times 10^5$ m/s. Rounding this figure to the nearest integer provides 22.

The electronic configuration of an atom is determined by filling atomic orbitals based on increasing energy levels, which results in the notation $\mathrm{[Kr]}5s^1$ for Rubidium with an atomic number of 37. This configuration indicates that the valence electron occupies the $5s$ subshell, meaning the principal quantum number $n=5$ and the azimuthal quantum number $l=0$ for an $s$-orbital. Given that $l=0$ restricts the magnetic quantum number $m=0$, and applying the standard convention $s=+\frac{1}{2}$ for the spin, the set of quantum numbers is $(5, 0, 0, +\frac{1}{2})$.

To determine the quantum numbers for the outermost electron of potassium, we distribute its 19 electrons according to the Aufbau principle by filling orbitals in increasing order of energy. The electron configuration is $1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$, which identifies the 4s orbital as the location of the outermost electron. Since this electron resides in the fourth shell, the principal quantum number $n=4$, and because it occupies an s orbital, the azimuthal quantum number $l=0$, which dictates that the magnetic quantum number $m=0$. Assigning the spin quantum number $s=+\frac{1}{2}$ for this single electron, we arrive at the set $n=4, l=0, m=0, s=+\frac{1}{2}$.

The candela is the SI base unit of luminous intensity, defined for a source emitting monochromatic radiation at a frequency of 540 x 10^12 Hz with a radiant intensity of 1/683 watts per steradian. By comparing the provided variable parameters to this physical definition, we see that $A = 540$. Since the radiant intensity in the problem is expressed as 1/B, the matching value indicates that B is 683. This standard ensures a consistent measurement of light perceived by the human eye across the visible spectrum.

In the $n=4$ principal shell, the set of orbitals characterized by a magnetic quantum number magnitude $|m_l|=1$ encompasses specific states within the $4p$, $4d$, and $4f$ subshells. Each of these three subshells contains two orbitals satisfying the condition $m_l=1$ or $m_l=-1$, which results in a total of 6 distinct orbitals. Since every individual orbital accommodates exactly one electron with a spin quantum number $m_s=-\frac{1}{2}$, there are 6 electrons in total that meet these specific quantum number requirements.

The fundamental relationship governing electromagnetic radiation is that the product of frequency and wavelength equals the speed of light, expressed by the equation $c = v \lambda$. Because electromagnetic waves propagate at a constant speed of 3 x 10^8 m/s, calculating the frequency involves dividing this speed by the given wavelength.

With the wavelength given as 6 x 10^-12 m, we use the rearranged expression $v = c / \lambda$. Performing this division results in a frequency of 0.5 x 10^20 Hz. To match the requested format of x x 10^19 Hz, we convert this value to 5 x 10^19 Hz, which confirms that $x = 5$.

The ionization of an atom occurs when an incoming photon carries sufficient energy to overcome the binding force holding the electron. We can determine the energy of the photon using the relationship $E = \frac{1240}{\lambda(\text{nm})}$, where the energy is obtained directly in electron volts. Using the provided wavelength of 242 nm, the photon energy is 5.12 eV.

The energy in Joules per atom is determined by the equation $E = 5.12 \times 1.6 \times 10^{-19} = 8.198 \times 10^{-19}$. Scaling this energy to a molar basis by multiplying with Avogadro's constant and converting the resulting units from Joules to kilojoules gives a final molar ionization energy of 494 kJ mol⁻¹.

When the principal quantum number is $n=4$, the shell contains 4s, 4p, 4d, and 4f subshells, which hold a maximum of 2, 6, 10, and 14 electrons respectively. Since each completely filled orbital contains one electron with $s=+\frac{1}{2}$ and one with $s=-\frac{1}{2}$, exactly half of the total electrons in each subshell possess an upward spin. The number of electrons with this spin in each subshell is calculated as $2/2=1$ for 4s, $6/2=3$ for 4p, $10/2=5$ for 4d, and $14/2=7$ for 4f. Aggregating these contributions results in a total of $1 + 3 + 5 + 7 = 16$ electrons.

The de Broglie relationship establishes that the wavelength of a moving particle relates to its mass and kinetic energy through the equation $\lambda = \frac{h}{\sqrt{2mK}}$. Using the given values of $h = 6.6 \times 10^{-34}$ J s, $m = 9 \times 10^{-31}$ kg, and $K = 4.5 \times 10^{-29}$ J, we substitute these into the formula to calculate the wavelength.

We first evaluate the expression under the square root in the denominator:
$\sqrt{2 \times (9 \times 10^{-31}) \times (4.5 \times 10^{-29})} = 9 \times 10^{-30}$
Dividing Planck's constant by this result yields the wavelength in meters:
$\lambda = \frac{6.6 \times 10^{-34}}{9 \times 10^{-30}} = 0.733 \times 10^{-4}$
Converting this result into the scientific notation format of $10^{-5}$ yields $7.3 \times 10^{-5}$ meters. Rounding to the nearest integer, we obtain the final value of 7.

An atom with 55 protons is Cesium, which in its neutral ground state has an electron configuration terminating in $6s^1$. Forming a unipositive ion, $\mathrm{Cs}^{+}$, involves removing this outer $6s$ electron, which results in the electronic structure of the noble gas Xenon. This configuration includes the s-subshells $1s^2$, $2s^2$, $3s^2$, $4s^2$, and $5s^2$, each containing two electrons, resulting in a total of 10 s-electrons.

The energy of an atomic orbital is determined by the sum of its principal quantum number $n$ and azimuthal quantum number $l$, where a higher $(n+l)$ value signifies greater energy. When multiple orbitals possess the same $(n+l)$ sum, the orbital with the lower principal quantum number $n$ is assigned a lower energy state.

Evaluating the orbitals provided, (A) has $n=3$ and $l=0$ for a sum of 3, (B) has $n=4$ and $l=0$ for a sum of 4, (C) has $n=3$ and $l=1$ for a sum of 4, and (D) has $n=3$ and $l=2$ for a sum of 5. Comparing these values, orbital (D) has the highest energy with a sum of 5, followed by the orbitals with a sum of 4. Between (B) and (C), (B) has a higher $n$ value, placing it at a higher energy than (C), while (A) has the lowest energy sum of 3. Arranging these in decreasing order results in $D > B > C > A$.

A black body is an ideal physical system that absorbs all incident electromagnetic radiation and emits thermal radiation characterized by Planck's Law. The spectral intensity distribution depends solely on temperature, where the peak frequency of emission shifts to higher values as the temperature increases, governed by the relation $\nu_{\text{max}} \propto T$. Because statements (A), (B), (C), and (D) accurately describe these fundamental physical behaviors, there are zero incorrect statements.

The orbital angular momentum of an electron is determined by the formula $\sqrt{l(l+1)}\frac{h}{2\pi}$, where $l$ is the azimuthal quantum number. For any $s$ orbital, including the 3s orbital specified, the azimuthal quantum number is $l=0$. Substituting this value into the equation yields $\sqrt{0(0+1)}\frac{h}{2\pi}$, which reduces to 0. Since the problem states the angular momentum equals $\frac{xh}{2\pi}$, it follows that $x=0$.

The spin-only magnetic moment of a transition metal ion is calculated using the formula $\mu = \sqrt{n(n+2)}$ B.M., where $n$ represents the number of unpaired electrons. A magnetic moment of 3.87 B.M. equates to $\sqrt{15}$ B.M., which means $n(n+2) = 15$, solving for $n = 3$ unpaired electrons. Comparing the electronic configurations, $\text{Ti}^{2+}$ has 2 unpaired electrons, $\text{Cr}^{2+}$ has 4, $\text{Mn}^{2+}$ has 5, and $\text{V}^{2+}$ has a $3d^3$ configuration, which contains the required 3 unpaired electrons.

Moseley's law provides the fundamental relationship in this context, establishing that the square root of the frequency of emitted characteristic X-rays is directly proportional to the atomic number, represented by $\sqrt{v} \propto (Z - b)$. Since the square root of a value is identical to raising that value to the power of one-half, the relationship can be rewritten as $v^{1/2} \propto (Z - b)$. By comparing this expression with the provided graph of $v^n$ against atomic number $Z$, it is clear that $n = 1/2$.

The photoelectric effect is characterized by the instantaneous emission of electrons, occurring without a time lag as long as the incident light frequency exceeds the material's specific threshold frequency ($\nu_0$). Although this physical phenomenon aligns with the assertion, the provided reason is incorrect because it falsely claims that any photon, regardless of its energy, will transfer energy to an electron to cause ejection. In truth, a photon must possess an energy $E = h\nu$ that is at least equal to the work function of the metal ($\Phi$) to trigger the emission of an electron, meaning a photon with insufficient energy cannot facilitate this process.

The behavior of an electron in a hydrogen atom is governed by the Bohr quantization condition, which dictates that the circumference of its orbit must equal an integer multiple of its de Broglie wavelength, expressed as $2 \pi r_n = n \lambda$. Since the radius of the $n$-th orbit is given by $r_n = a_0 n^2$, we can determine the radius of the third orbit by setting $n = 3$, resulting in $r_3 = 9 a_0$. Substituting this radius into the quantization condition, the equation becomes $2 \pi (9 a_0) = 3 \lambda$. Solving for the wavelength, we divide $18 \pi a_0$ by 3 to obtain $\lambda = 6 \pi a_0$.

Moseley's law establishes a fundamental relationship between the atomic number of an element and the characteristic X-ray frequencies it emits. Through his experimental observations, Henry Moseley determined that the square root of the X-ray frequency, $\sqrt{v}$, is directly proportional to the atomic number $Z$. This relationship, mathematically expressed as $\sqrt{v} = a(Z - b)$, describes a linear function where $a$ and $b$ are constants. Consequently, a graph plotting $\sqrt{v}$ on the vertical axis against $Z$ on the horizontal axis yields a straight line, reflecting that the square root of the frequency increases linearly with the atomic number.

The energy of an electron in a hydrogen atom is inversely proportional to the square of the principal quantum number $n$, represented by the relation $E_n = \frac{E_1}{n^2}$. With the first Bohr orbit energy defined as $E_1 = -2.18 \times 10^{-18} \mathrm{J}$, the energy for the third orbit requires substituting $n=3$ into the denominator of this formula. This calculation results in $E_3 = \frac{E_1}{3^2}$, which simplifies to $E_3 = \frac{1}{9} E_1$. Therefore, the energy in the third Bohr orbit is exactly $\frac{1}{9}$th of the energy of the first Bohr orbit.