Structure Of Atom – JEE Main ChemistryPractice Questions & PYQs

The radial probability density function, represented by $\psi^2(r)$, depicts the likelihood of finding an electron at a specific distance from the nucleus. The number of radial nodes is determined by the formula $n - l - 1$, and for a $2s$ orbital with $n=2$ and $l=0$, this calculation yields $2 - 0 - 1 = 1$ node. Because the density is a squared value, $\psi^2(r)$ is always positive, resulting in a graph that touches the horizontal axis at the radial node before rising again, consistently remaining above or at the axis.

The Bohr model is strictly applicable only to single-electron species, meaning it cannot be used for a neutral lithium atom, but it is valid for the hydrogen-like $Li^{2+}$ ion. For a hydrogen-like species with atomic number $Z$, the energy of an electron in the $n$-th shell is given by the formula $E_n = E_1(H) \times \frac{Z^2}{n^2}$, where $E_1(H) = -2.2 \times 10^{-18} \mathrm{J}$ is the energy of the electron in the first shell of the hydrogen atom. Applying this for the $Li^{2+}$ ion ($Z=3$) for the $n=2$ state, the energy becomes $E = -2.2 \times 10^{-18} \times \frac{3^2}{2^2} = -4.95 \times 10^{-18} \mathrm{J}$. The energy required for ionization is the magnitude of this value, $4.95 \times 10^{-18} \mathrm{J}$, and substituting this into the relationship $\lambda = \frac{hc}{E}$ yields $\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{4.95 \times 10^{-18}} \approx 4.018 \times 10^{-8} \mathrm{m}$. Comparing this to the given expression $x \times 10^{-8} \mathrm{m}$, the nearest integer for $x$ is 4.

Quantum mechanics dictates that the azimuthal quantum number $l$ must be less than the principal quantum number $n$, and the magnetic quantum number $m_l$ must fall within the range $-l$ to $+l$. Looking at set A, $n=3$ and $l=3$ violates the requirement $l < n$ because the value of $l$ must always be strictly less than $n$. For set B, $n=3$ and $l=2$ satisfies the condition $l < n$, and since $m_l=2$ falls within the allowable range from -2 to +2, this is a valid set. Set C provides $n=2$ and $l=1$, which also follows the rule $l < n$, and with $m_l=1$ falling within the allowed range from -1 to +1, this set is valid. Finally, set D with $n=2$ and $l=2$ fails the required $l < n$ constraint and is therefore invalid, which means there are two valid sets of quantum numbers in total.

The energy of electromagnetic radiation is quantized, with the energy of a single photon defined by the relation $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength. To determine the energy content for one mole of such photons, we multiply the energy of a single photon by Avogadro's number, $N_A$, resulting \in the formula $E_{\text{mole}} = \frac{N_A h c}{\lambda}$. Substituting the provided values of $h = 6.63 × 10^{-34}\ \mathrm{J}\ \mathrm{s}$, $c = 3 × 10^8\ \mathrm{m}\ \mathrm{s}^{-1}$, $N_A = 6.02 × 10^{23}\ \mathrm{mol}^{-1}$, and a wavelength of $\lambda = 300 × 10^{-9}\ \mathrm{m}$, the total energy becomes $\frac{(6.02 \times 10^{23}) \times (6.63 \times 10^{-34}) \times (3 \times 10^8)}{300 \times 10^{-9}} \approx 399,126\ \mathrm{J}\ \mathrm{mol}^{-1}$. Converting this figure into kiloJoules per mole by dividing by $1000$ provides a final value of $399\ \mathrm{kJ}\ \mathrm{mol}^{-1}$.

The de Broglie hypothesis defines the wavelength $\lambda$ of an electron in terms of its momentum $p$, expressed as $\lambda = \frac{h}{p}$. When an electron of charge $e$ is accelerated from rest through a potential difference $V$, it gains kinetic energy equal to $eV$. Since kinetic energy relates to momentum through the equation $KE = \frac{p^2}{2m}$, the electron's momentum is $p = \sqrt{2meV}$, which transforms the wavelength formula into:

$\lambda = \frac{h}{\sqrt{2meV}}$

Substituting the known values-Planck's constant $h$ is 6.63 × 10⁻³⁴ J s, mass $m$ is 9.1 × 10⁻³¹ kg, charge $e$ is 1.6 × 10⁻¹⁹ C, and potential difference $V$ is 40 × 10³ V-the calculation becomes:

$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times 40 \times 10^3}} \approx 6.14 \times 10^{-12} \text{ m}$

This result corresponds to the form $x \times 10^{-12} \text{ m}$, meaning $x$ is 6.14, which rounds to the nearest integer of 6.

The number of angular nodes in an orbital is determined by the azimuthal quantum number $l$, with values assigned as $s=0$, $p=1$, $d=2$, and $f=3$, while the number of radial nodes is calculated using the formula $n - l - 1$. Since the problem requires an orbital with two angular nodes, the azimuthal quantum number must be $l=2$, which identifies the orbital as a $d$-orbital. By substituting $l=2$ and the requirement of two radial nodes into the node formula, we obtain the equation $2 = n - 2 - 1$, which simplifies to $2 = n - 3$. Solving for the principal quantum number gives $n=5$, confirming that the 5d orbital satisfies the given conditions, whereas the other options like $3p$, $4f$, and $4d$ result in different node counts.

To identify the number of orbitals with $n=5$ and $m_l = +2$, we must consider the allowable values for the azimuthal quantum number $l$. For a principal quantum number $n=5$, the possible values for $l$ range from $0$ to $n-1$, resulting in the set $0, 1, 2, 3, 4$. The magnetic quantum number $m_l$ for any subshell is restricted to the range $-l \leq m_l \leq +l$. Consequently, an orbital with $m_l = +2$ can only exist in subshells where $l \geq |m_l|$, which requires $l \geq 2$. Among the available values for $l$ at this energy level, only $2, 3,$ and $4$ satisfy this requirement, corresponding to the $5d, 5f,$ and $5g$ subshells. Each of these three subshells contributes exactly one orbital where $m_l = +2$, yielding a total of 3 orbitals.

The photoelectric effect follows the principle of energy conservation, where the energy of an incident photon equals the sum of the metal's work function and the kinetic energy of the ejected electron. This is expressed as:

$\frac{hc}{\lambda} = h\nu_0 + \frac{1}{2} m_e v^2$

The incident photon energy, determined by the expression $\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}}$, is $3.978 \times 10^{-19} \text{ J}$. The work function of the metal is $6.63 \times 10^{-34} \times 4.3 \times 10^{14}$, which evaluates to $2.851 \times 10^{-19} \text{ J}$. Subtracting the work function from the incident photon energy leaves a kinetic energy of $1.127 \times 10^{-19} \text{ J}$.

Using the relation for kinetic energy with an electron mass of $9.0 \times 10^{-31} \text{ kg}$, the velocity squared becomes $v^2 = \frac{2 \times 1.127 \times 10^{-19}}{9.0 \times 10^{-31}}$. Solving this yields $v^2 \approx 25 \times 10^{10} \text{ m}^2\text{s}^{-2}$, resulting in an electron velocity of $5 \times 10^5 \text{ ms}^{-1}$.

For a metal to function as a photoelectric electrode, it must possess a sufficiently low work function, $W$, such that incident light can trigger the emission of electrons. Within Group 1, ionization energy decreases as the atomic size increases down the column, indicating that the outermost electrons become progressively easier to remove. Since Caesium exhibits the lowest ionization energy among the provided metals, it holds the minimum threshold energy required for this effect, distinguishing it as the only element on the list typically utilized for this purpose.

The photoelectric effect relies on the principle that each photon striking a metal surface has the potential to eject one electron, meaning the rate of electron emission is equal to the number of photons arriving at the surface per second. First, calculate the power delivered by the radiation source by dividing the total energy by the time elapsed, which yields 100 J/s. Next, the energy of a single photon is determined by the expression $\varepsilon = \frac{hc}{\lambda}$, using the values $h = 6.626 \times 10^{-34}$ Js, $c = 3 \times 10^8$ m/s, and wavelength $\lambda = 400 \times 10^{-9}$ m.

The total number of electrons ejected per second, denoted by $N$, is found by dividing the power $P$ by the energy per photon $\varepsilon$:
$N = \frac{P}{\left( \frac{hc}{\lambda} \right)}$
Substituting the known values into the equation, the calculation becomes:
$N = \frac{100}{\left( \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} \right)}$
Evaluating this expression results in approximately $2.01 \times 10^{20}$ electrons per second. Comparing this result to the given form $x \times 10^{20}$, the value of x is 2.

The de-Broglie wavelength of a charged particle accelerated through a potential difference $V$ is given by $\lambda = \frac{h}{\sqrt{2mqV}}$, where $h$ is Planck's constant, $m$ is the particle mass, and $q$ is its charge. For two particles accelerated by the same potential, the ratio of their wavelengths simplifies to $\frac{\lambda_{\mathrm{Li}}}{\lambda_{\mathrm{P}}} = \sqrt{\frac{m_{\mathrm{p}} \cdot q_{\mathrm{p}}}{m_{\mathrm{Li}} \cdot q_{\mathrm{Li}}}}$.

Substituting the relationships $m_{\mathrm{Li}} = 8.3 m_{\mathrm{p}}$, $q_{\mathrm{p}} = e$, and $q_{\mathrm{Li}} = 3e$ yields $\frac{\lambda_{\mathrm{Li}}}{\lambda_{\mathrm{P}}} = \sqrt{\frac{m_{\mathrm{p}} \cdot e}{8.3 m_{\mathrm{p}} \cdot 3e}}$. This expression simplifies to $\sqrt{\frac{1}{24.9}}$, which is approximately $\sqrt{\frac{1}{25}}$, or $0.2$. Since $0.2 = 2 \times 10^{-1}$, the value of $x$ satisfies the equation $x = 2$.

In the Thomson model of the atom, the positive charge is considered to be distributed uniformly throughout the entire volume of the atom rather than being concentrated at the center. As $\alpha$-particles travel through this diffused positive charge, they experience continuous, weak repulsive forces from these distributed charges. Consequently, the particles would undergo only minor deflections from their original path while their kinetic energy is gradually sapped, leading to a decrease in their speed.

Radial nodes are spherical regions around a nucleus where the probability of finding an electron is exactly zero, and their quantity is determined by the principal quantum number $n$ and the azimuthal quantum number $l$. Because the magnetic quantum number $m_l$ must fall within the range of $-l$ to $l$, a value of $m_l = -3$ restricts the azimuthal quantum number to be at least 3. Given the principal quantum number $n=4$, the azimuthal quantum number is fixed at $l=3$, which characterizes the orbital as $4f$. Applying the calculation $n - l - 1$ by substituting the known values of $n=4$ and $l=3$ results in $4 - 3 - 1$, confirming there are 0 radial nodes.

Bohr's atomic model is strictly limited to hydrogen-like species, which are atoms or ions possessing only a single electron. While this theory successfully explains the spectral behavior of ions like $Li^{2+}$, it is incapable of describing systems with more than one electron, such as $Li^+$, meaning the first statement is invalid. Additionally, the model is unable to account for the Zeeman effect, which describes the splitting of spectral lines that occurs when an atom is placed within an external magnetic field, confirming that the second statement correctly identifies a known limitation of the theory.

An orbital with no angular nodes corresponds to an azimuthal quantum number $l = 0$, which characterizes it as an $s$ orbital. The number of radial nodes is determined by the formula $n - l - 1$, and setting this value to two for an $s$ orbital leads to the equation $n - 0 - 1 = 2$. Solving this for the principal quantum number yields $n = 3$, identifying the orbital as 3s.

The total energy emitted by a source over a given time interval is equal to the product of the number of photons and the energy of a single photon, where the energy of an individual photon is determined by its wavelength according to the relation $E = \frac{hc}{\lambda}$. First, we calculate the energy of a single photon using the wavelength of 1000 nm, which is equivalent to $10^{-6}$ m, Planck's constant $h = 6.63 \times 10^{-34}$ Js, and the speed of light $c = 3.00 \times 10^{8}$ m/s. Substituting these values, the energy per photon is $E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{10^{-6}}$ J, which equals $1.989 \times 10^{-19}$ J.

Since the source operates at a power of 1 mW, the total energy emitted in 0.1 s is the product of power and time, resulting in $10^{-3} \times 0.1 = 10^{-4}$ J. Dividing this total energy by the energy of a single photon yields the total number of photons, calculated as $N = \frac{10^{-4}}{1.989 \times 10^{-19}}$. This calculation results in approximately $5.027 \times 10^{14}$, which is equivalent to $50.27 \times 10^{13}$. Rounding to the nearest integer, we find $x = 50$.

An element featuring an outermost configuration of $4s^{2} 4p^{1}$ is Gallium, which resides in Group 13 of the fourth period. A diagonal relationship connects an element to the one located one period down and one group to the right, meaning the required element must belong to the fifth period and Group 14. This specific element is Tin ($Z=50$), which completes its outer electron arrangement after the preceding noble gas core to arrive at $[\mathrm{Kr}] 4d^{10} 5s^{2} 5p^{2}$.

Nuclear stability depends on the ratio of neutrons to protons within a nucleus. Among the isotopes of hydrogen, Tritium is unique because its nucleus contains two neutrons and one proton, resulting in an $n/p$ ratio of 2. This high ratio renders the nucleus unstable and radioactive, prompting it to seek a more stable state. To achieve this, Tritium undergoes decay by emitting a low-energy $\beta^{-}$ particle, which is represented by the nuclear reaction ${}_{1}^{3}\mathrm{H} \longrightarrow {}_{2}^{3}\mathrm{He} + {}_{-1}^{0}\mathrm{e}$.

When electromagnetic radiation is just sufficient to ionize a metal, the energy of a single photon is equivalent to the ionization energy of the metal atom. This energy is expressed by $E = \frac{hc}{\lambda}$, and calculating the total ionization energy for a mole of atoms requires scaling this value by Avogadro's constant, $N_A$, according to the relation $IE = \frac{hcN_A}{\lambda}$.

Placing the given values into this formula leads to the following calculation:
$IE = \frac{(6.63 \times 10^{-34}\ \mathrm{J\cdot s}) \times (3.00 \times 10^{8}\ \mathrm{m/s}) \times (6.02 \times 10^{23}\ \mathrm{mol}^{-1})}{663 \times 10^{-9}\ \mathrm{m}}$
The multiplication in the numerator produces $119.7378 \times 10^{-3}\ \mathrm{J\cdot m\cdot mol^{-1}}$, and dividing this product by the wavelength of $663 \times 10^{-9}\ \mathrm{m}$ results in $180,600\ \mathrm{J\cdot mol^{-1}}$. Converting this value into kilojoules by dividing by $1000$ yields $180.6\ \mathrm{kJ\cdot mol^{-1}}$, which rounds to $180\ \mathrm{kJ\cdot mol^{-1}}$.

In Bohr's model, the velocity of an electron in a specific orbit is determined by the relationship $v \propto \frac{Z}{n}$, where $Z$ represents the atomic number corresponding to the nuclear charge and $n$ denotes the principal quantum number. Since velocity is directly proportional to the nuclear charge, a decrease in the magnitude of the positive charge would cause the velocity to decrease rather than increase, making the first statement false. Conversely, because velocity is inversely proportional to the principal quantum number, a decrease in $n$ results in a higher velocity for the electron, which confirms the second statement is true.