Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

Nuclear density is a physical constant that is independent of the mass number $A$ of a nucleus. Because the volume of a nucleus increases in direct proportion to its mass number-a relationship established by the empirical formula $R = R_0 A^{1/3}$-the total mass of the nucleus increases at the same rate as its volume, resulting in a consistent density across all nuclides. This principle confirms that the reason provided is factually accurate, as it correctly identifies the dependency between nuclear radius and mass number. The assertion is false, however, because it erroneously claims that different nuclides possess varying nuclear densities, when in reality their densities are identical.

The energy absorbed by a hydrogen atom during a collision causes an excitation from the ground state to a higher principal quantum number, determined by the equation:
$E = 13.6 \left( 1 - \frac{1}{n^2} \right)$
Setting the excitation energy to 12.5 eV means $12.5 = 13.6(1 - 1/n^2)$, which simplifies to $\frac{1}{n^2} \approx 0.08$. Because $n^2$ is approximately 12.5, the atom reaches the $n = 3$ state, as it lacks sufficient energy to reach the $n = 4$ level.

As the excited hydrogen atoms transition back to lower energy levels, they emit radiation at various frequencies. The total number of distinct spectral lines produced is determined by the formula:
$\text{Number of lines} = \frac{n(n-1)}{2}$
Substituting $n = 3$ into this expression results in $\frac{3(3-1)}{2}$, which yields 3 possible spectral lines corresponding to transitions from level 3 to 2, 3 to 1, and 2 to 1.

In any nuclear disintegration where a stationary parent nucleus splits into two fragments, the conservation of linear momentum requires the two resultant parts to possess equal and opposite momenta. This implies that the product of mass and speed for both fragments must be balanced, written as $m_1 v_1 = m_2 v_2$.

Because nuclear density is essentially constant across different nuclei, the mass of each part is proportional to its volume, and consequently to the cube of its radius, or $m \propto r^3$. Substituting this proportionality into our momentum relation reveals that the ratio of the speeds is equivalent to the cube of the ratio of the radii, specifically $\frac{v_1}{v_2} = \left( \frac{r_2}{r_1} \right)^3$. Given that the ratio of the nuclear sizes is $1 : 2^{1/3}$, we evaluate $\left( \frac{2^{1/3}}{1} \right)^3$, which simplifies to 2. Since the speed ratio is defined as $n : 1$, we determine that $n = 2$.

Nuclear alpha decay is governed by the principle of mass-energy equivalence, where the energy $Q$ released during the reaction is directly proportional to the total mass lost by the system. This mass defect, $\Delta m$, is determined by subtracting the combined mass of the decay products-the ${}_{90}^{234}\mathrm{Th}$ nucleus at $234.04360 \mathrm{u}$ and the ${}_{2}^{4}\mathrm{He}$ alpha particle at $4.00260 \mathrm{u}$-from the mass of the parent ${}_{92}^{238}\mathrm{U}$ nucleus of $238.05060 \mathrm{u}$. The energy released is calculated by applying the conversion factor of $1 \mathrm{u} = 931.5 \frac{\mathrm{MeV}}{\mathrm{c}^2}$ to this mass defect:

$Q = \Delta m \times 931.5 \frac{\mathrm{MeV}}{\mathrm{u}}$

Using the provided mass values and this physical constant, the total energy released during the alpha decay of ${}^{238}\mathrm{U}$ is $4.25 \mathrm{MeV}$.

The mass of a nucleus with mass number $A$ is effectively $A u$, and its volume is defined by the sphere formula $V = \frac{4}{3} \pi R^3$. Substituting the empirical relationship for nuclear radius, $R = R_0 A^{1/3}$, into the volume expression gives:
$V = \frac{4}{3} \pi (R_0 A^{1/3})^3 = \frac{4}{3} \pi R_0^3 A$
When calculating the nuclear density $\rho$ by dividing mass by volume, the mass number $A$ appears in both the numerator and the denominator and cancels out completely:
$\rho = \frac{A u}{\frac{4}{3} \pi R_0^3 A} = \frac{3 u}{4 \pi R_0^3}$
Because this final expression depends only on physical constants and is independent of the mass number, the nuclear density is constant for all elements, resulting in a ratio of 1:1 for the oxygen and helium nuclei.

The radius of an electron's orbit in a hydrogen-like atom is determined by the principal quantum number $n$ and the atomic number $z$ according to the relationship $r_n = \frac{0.51 n^2}{z}$ Å. For the triply ionized lithium ion, $\mathrm{Li}^{++}$, the atomic number $z$ is 3, and we are calculating the radius for the fifth orbit, which corresponds to $n$ being 5.

Substituting these values into the expression gives $r_5 = \frac{0.51 \times 5^2}{3}$ Å. Simplifying the calculation, we have $r_5 = \frac{0.51 \times 25}{3}$ Å, which results in 4.25 Å. Because 1 Å is equivalent to $10^{-10}$ m, this distance is $4.25 \times 10^{-10}$ m. To express this value in the required unit of $10^{-12}$ m, we shift the decimal point two places to the right to obtain $425 \times 10^{-12}$ m.

Nuclear transformations strictly obey the law of conservation of charge, requiring the sum of the atomic numbers of the decay products to balance the atomic number of the parent nucleus. Each emission of an $\alpha$-particle reduces the atomic number by two, while the emission of an electron increases the atomic number by one and the emission of a positron decreases it by one. Considering the initial atomic number of 92, the final atomic number $P$ is determined by the net effect of these specific decay events:
$P = 92 - 2(2) + 1 - 2(1)$
This calculation confirms that $P = 87$, which is the atomic number of the resulting product nucleus.

Nuclear density is a fundamental constant that remains independent of the size of the nucleus because both the total mass and the volume of a nucleus scale linearly with the atomic mass number. Since density is defined as mass per unit volume, we determine the density of nuclear matter by dividing the mass of the nucleons by the spherical volume of the nucleus, and then compare this value to the density of water.

The mass of a nucleus is $A$ multiplied by the mass of a nucleon, $1.6 \times 10^{-27}$ kg, while its volume is calculated using the formula $V = \frac{4}{3} \pi R^3$. By substituting the given radius expression $R = 1.5 \times 10^{-15} A^{1/3}$ m, the density $\rho$ is determined by the equation:

$\rho = \frac{A \times 1.6 \times 10^{-27}}{\frac{4}{3} \pi (1.5 \times 10^{-15} A^{1/3})^3}$

Because the atomic mass number $A$ cancels out in the numerator and denominator, the expression confirms that nuclear density is constant. Computing the numerical values results in a nuclear density of approximately $0.113 \times 10^{18}$ kg/m³. Dividing this by the density of water, $10^3$ kg/m³, produces a ratio of $11.3 \times 10^{13}$, which leads to the value of $n$ being 11.

The radius of an orbit in a hydrogen-like species is proportional to the square of the principal quantum number divided by the atomic number, represented as $r \propto \frac{n^2}{Z}$. Determining the ratio $\frac{r_2}{r_1}$ for the fourth orbit of $\mathrm{Be}^{3+}$ ($n_2=4, Z_2=4$) against the second orbit of $\mathrm{He}^+$ ($n_1=2, Z_1=2$) requires the expression $\frac{r_2}{r_1} = \left( \frac{n_2}{n_1} \right)^2 \times \frac{Z_1}{Z_2}$. Substituting these values results in $\frac{r_2}{r_1} = \left( \frac{4}{2} \right)^2 \times \frac{2}{4}$, which simplifies to $2^2 \times 0.5$, yielding a final value of $2$. Given the ratio is defined as $x:1$, $x$ is 2.

When a nucleus of mass $M$ initially at rest splits, the law of conservation of linear momentum requires that the two resulting fragments, with masses $\frac{M'}{3}$ and $\frac{2M'}{3}$, must possess momenta equal in magnitude and opposite in direction to maintain a net system momentum of zero. This dictates that $|p_1| = |p_2|$, where $p_1$ and $p_2$ are the momenta of the two parts. Since the de Broglie wavelength is defined by the relationship $\lambda = \frac{h}{p}$, the wavelength of each fragment depends exclusively on its momentum magnitude. Because both fragments share the identical momentum magnitude regardless of their individual masses, their de Broglie wavelengths are necessarily equal, resulting in a ratio of $1:1$.

Radioactive decay follows a predictable pattern where the intensity of radiation is halved with every passing half-life period. To reach a safe working level from an initial intensity 64 times higher than the limit, the source must undergo enough half-lives such that its activity drops to $1/64$ of the original value. Because $2^6 = 64$, the activity must go through 6 complete half-life cycles to reach this threshold. With each half-life measured at 2.5 hours, the total duration required is $6 \times 2.5 = 15$ hours.

The energy emitted during an electronic transition in a hydrogen atom is directly proportional to the difference between the inverse squares of the principal quantum numbers of the initial and final states, defined by the relationship $\Delta E \propto \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$. For the transition from the third energy level to the second, the energy corresponds to $\left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \left( \frac{1}{4} - \frac{1}{9} \right)$, which simplifies to $\frac{5}{36}$. The transition from the highest permitted level, which is infinity, to the second level results in an energy proportional to $\left( \frac{1}{2^2} - \frac{1}{\infty ^2} \right) = \frac{1}{4}$. Calculating the ratio of these two energies gives $\frac{5/36}{1/4} = \frac{5}{9}$. Equating this ratio to the given expression $\frac{x}{x+4}$ leads to $\frac{x}{x+4} = \frac{5}{9}$, and solving this equation yields $9x = 5x + 20$, which simplifies to $x = 5$.

The Bohr atomic model defines the energy states of an electron as $K = \frac{Z^{2} m e^{4}}{8(n h \varepsilon_{0})^{2}}$, $P = \frac{-Z^{2} m e^{4}}{4(n h \varepsilon_{0})^{2}}$, and $E = \frac{-Z^{2} m e^{4}}{8(n h \varepsilon_{0})^{2}}$. When an electron transitions to a higher energy level, the principal quantum number $n$ increases, which reduces the magnitude of the $1/n^2$ term common to all three expressions. Because kinetic energy $K$ is a positive value, this change leads to a decrease in its magnitude, while the negative potential energy $P$ and total energy $E$ values become less negative, effectively increasing as they approach zero.

When an electron revolves around a nucleus, it functions as a microscopic current loop, creating a direct physical relationship between its orbital angular momentum and its magnetic dipole moment. Because the electron carries a negative elementary charge, the direction of the resulting magnetic moment vector is antiparallel to the orbital angular momentum vector, introducing a negative sign into the relationship. This fundamental interaction is expressed by the equation $\vec{\mu}_{L} = -\frac{e\vec{L}}{2m}$, where $e$ is the magnitude of the electron charge, $m$ represents its mass, and $\vec{L}$ denotes the orbital angular momentum.

The decay of radioactive substances follows an exponential pattern where the quantity remaining at time $t$ is calculated using the initial amount and the decay constant. For two materials $A$ and $B$ starting with the same initial number of nuclei $N_0$, the amounts remaining are $N_A = N_0 e^{-25 \lambda t}$ and $N_B = N_0 e^{-16 \lambda t}$. Determining the ratio of $B$ to $A$ results in:

$\frac{N_B}{N_A} = \frac{N_0 e^{-16 \lambda t}}{N_0 e^{-25 \lambda t}} = e^{(25 \lambda - 16 \lambda) t} = e^{9 \lambda t}$

Because the ratio is specified to be $e$, or $e^1$, we set the exponent equal to one, giving $9 \lambda t = 1$. Solving for $t$ provides $t = \frac{1}{9 \lambda}$. Comparing this expression to the given form $\frac{1}{a \lambda}$ identifies the value of $a$ as 9.

Classical electromagnetic theory dictates that a charged particle moving in a circular orbit undergoes continuous acceleration, which necessarily results in the emission of electromagnetic radiation. As the electron radiates this energy, it experiences a steady loss of orbital energy, causing its trajectory to spiral inward toward the nucleus. Because this energy dissipation would happen rapidly, a classical atom designed according to the Rutherford model cannot maintain stability and is inherently doomed to collapse.

The relationship between the size of a nucleus and its mass number $A$ is governed by the expression $r = r_0 A^{1/3}$, where $r_0$ is a constant. By applying the formula for the volume of a sphere, $v = \frac{4}{3} \pi r^3$, and substituting the given radius equation, we find $v = \frac{4}{3} \pi r_0^3 A$. This derived result shows that the volume of a nucleus increases in direct proportion to its mass number.

To determine nuclear density, one must divide the total nuclear mass, which scales linearly with the number of nucleons $A$, by the volume of the nucleus. Because the volume itself is proportional to $A$, the mass number appears in both the numerator and the denominator of the density calculation, causing it to cancel out entirely. Consequently, nuclear density remains a constant value regardless of the mass number, establishing that it is independent of $A$.

Radioactive decay follows a consistent pattern where a sample loses half of its remaining mass with each passing half-life period, governed by the relationship $N = N_0(1/2)^n$, where $n$ represents the number of half-lives elapsed. To determine the time, we track how many halving steps are required to reduce the initial amount to 6.25%. Since 6.25% is equivalent to 1/16, and this fraction equals 1/2 raised to the power of 4, the sample has undergone four full half-life cycles. Given that each half-life lasts 5 years, the total time $x$ is calculated as:

$x = 4 \times 5 = 20$

This calculation shows that the substance reaches the target value after 20 years.

Exciting an electron in a hydrogen-like atom requires an incident photon with energy exactly equal to the difference in energy between the initial and final states. For a lithium ion $\mathrm{Li}^{++}$, the atomic number $Z$ is $3$, and the allowed energy levels follow the relationship
$E_n = -13.6 \frac{Z^2}{n^2} \, \text{eV}$
The energy required to move the electron from the first orbit, $n_1 = 1$, to the third orbit, $n_2 = 3$, is calculated using the transition energy formula
$\Delta E = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 13.6 \times 3^2 \times \left( 1 - \frac{1}{9} \right) = 13.6 \times 8 = 108.8 \, \text{eV}$
Since the energy of the photon is given by $\Delta E = \frac{hc}{\lambda}$, we can determine the wavelength by substituting the provided constant $hc = 1242 \, \text{eV nm}$ into the expression
$\lambda = \frac{1242}{108.8} \, \text{nm} \approx 11.415 \, \text{nm}$
Converting this wavelength into units of meters, we obtain $11.415 \times 10^{-9} \, \text{m}$, which is equivalent to $114.15 \times 10^{-10} \, \text{m}$. Rounding this to the nearest integer as required by the format $x \times 10^{-10} \, \text{m}$ gives $x = 114$.

The Rydberg formula establishes the relationship between the wavelength of an emitted photon and the quantum states of a hydrogen atom. When an electron transitions from an excited state $n$ to the ground state $1$, the wavelength $\lambda$ is governed by the equation:

$\frac{1}{\lambda} = R \left(1 - \frac{1}{n^2}\right)$

To solve for $n$, divide both sides by the Rydberg constant $R$ to obtain $\frac{1}{\lambda R} = 1 - \frac{1}{n^2}$. Rearranging this expression to isolate $\frac{1}{n^2}$ results in $\frac{1}{n^2} = 1 - \frac{1}{\lambda R}$, which simplifies to $\frac{1}{n^2} = \frac{\lambda R - 1}{\lambda R}$. Inverting this final fraction provides $n^2 = \frac{\lambda R}{\lambda R - 1}$, which confirms the principal quantum number of the excited state is $n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$.