Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

The size of an atomic nucleus depends directly on its mass number, where the radius $R$ scales with the cube root of the mass number $A$ according to the expression:

$R = R_0 A^{1/3}$

Because the proportionality constant $R_0$ remains consistent for all nuclei, we can relate the radii of two different atoms using the ratio $\frac{R_1}{R_2} = \left( \frac{A_1}{A_2} \right)^{1/3}$. By substituting the mass numbers of $27$ for aluminum and $125$ for tellurium into this equation, the ratio becomes $\left( \frac{27}{125} \right)^{1/3}$, which simplifies to $\frac{3}{5}$. Applying this ratio to the known radius of $3.6$ fermi, the calculation $R_2 = 3.6 \times \frac{5}{3}$ yields a final value of $6$ fermi.

Radioactive decay operates on the principle that the quantity of a radioactive isotope is reduced by half over every successive half-life period. When $\frac{7}{8}$ of a ${}^{66}\mathrm{Cu}$ sample has decayed, the remaining fraction is $1 - \frac{7}{8} = \frac{1}{8}$. Since this remaining amount is equal to $\left(\frac{1}{2}\right)^3$, the sample has completed exactly 3 half-life cycles. Dividing the total time of 15 minutes by these 3 cycles yields the duration of a single half-life:

$T = \frac{15}{3} = 5$

The resulting half-life period is 5 minutes.

The physical law governing this disintegration is the conservation of linear momentum, which requires that for an initially stationary nucleus, the product of mass and velocity for both fragments must be equal in magnitude, expressed as $m_1 v_1 = m_2 v_2$. This relationship dictates that the masses of the resulting fragments are inversely proportional to their velocities, so a velocity ratio of $2:1$ corresponds to a mass ratio of $\frac{1}{2}$. Because nuclear mass is directly proportional to volume, and volume scales with the cube of the radius, we can relate the mass ratio to the radii using $\frac{m_1}{m_2} = (\frac{r_1}{r_2})^3$. Substituting the mass ratio of $\frac{1}{2}$ into this equation results in $(\frac{r_1}{r_2})^3 = \frac{1}{2}$, identifying the ratio of their nuclear sizes as $1 : 2^{\frac{1}{3}}$.

The energy released in a nuclear fusion reaction is determined by the difference between the total binding energy of the resulting nucleus and the total binding energy of the initial reactants. Because binding energy represents the stability gained by nucleons within a nucleus, a higher total binding energy in the product indicates that energy has been released during the transition.

In this specific process, two deuteron nuclei combine to form a single helium nucleus. Each deuteron contains two nucleons, so the reactants comprise four nucleons in total, with each having a binding energy of $1.1 \text{ MeV}$. This results in an initial total binding energy of $4 \times 1.1 \text{ MeV} = 4.4 \text{ MeV}$. The product is a single helium nucleus also containing four nucleons, but with a binding energy of $7 \text{ MeV}$ per nucleon, leading to a final total binding energy of $4 \times 7 \text{ MeV} = 28 \text{ MeV}$. The energy released is the difference between these two states, calculated as $28 \text{ MeV} - 4.4 \text{ MeV} = 23.6 \text{ MeV}$.

When an alpha particle heads toward a uranium nucleus and scatters at 180 degrees, it momentarily stops at the distance of closest approach because its initial kinetic energy is entirely converted into electrostatic potential energy. By applying the principle of conservation of energy, we set the kinetic energy equal to the potential energy formula $E = \frac{Ze(2e)}{4\pi\varepsilon_0 r_0}$, where Z is 92, the Coulomb constant is 9 x 10^9 Nm^2/C^2, and e is 1.6 x 10^-19 C.

Using the kinetic energy of 5 MeV, which is 5 x 10^6 x 1.6 x 10^-19 J, we rearrange the equation to solve for the distance of closest approach:
$r_0 = \frac{9 \times 10^9 \times (92 \times 1.6 \times 10^{-19}) \times (2 \times 1.6 \times 10^{-19})}{5 \times 10^6 \times 1.6 \times 10^{-19}}$
This computation results in a distance of approximately 5.3 x 10^-12 cm, which is of the order of 10^-12 cm.

Ionization potential increases as you move from left to right across a period and decreases as you descend a group, reflecting the increasing ease of removing a valence electron as the atomic radius grows. Because $\mathrm{^{133}_{55}Cs}$ is an alkali metal situated at the bottom of its group and to the far left of $\mathrm{^{14}_{7}N}$, $\mathrm{^{16}_{8}O}$, and $\mathrm{^{18}_{40}Ar}$, it possesses the largest atomic radius and the weakest electrostatic attraction between its nucleus and valence shell, resulting in the lowest energy requirement for ionization.

The wavelengths of spectral lines in hydrogenic atoms are determined by the Rydberg formula, where the Rydberg constant $R$ is influenced by the nuclear mass $M$ through the expression $R = \frac{1.097 \times 10^{7}}{1 + \frac{m}{M}}$. Since deuterium and ordinary hydrogen contain nuclei with different masses, this constant takes on a distinct value for each isotope, ultimately leading to the slight variations observed in their spectral wavelengths.

The electromagnetic spectrum is characterized by an inverse relationship between frequency and wavelength, meaning that the radiation with the highest frequency inherently possesses the shortest wavelength. Among the radiations listed, gamma-rays are situated at the extreme high-frequency end of this spectrum. Therefore, gamma-rays exhibit the least wavelength, which clearly distinguishes them from X-rays and the particulate nature of alpha-rays and beta-rays.

The principle of conservation of linear momentum applies here because the decay occurs without external influence. Since the initial momentum of the uranium nucleus at rest is zero, the combined momentum of the alpha particle and the residual nucleus after the decay must also equal zero. Given the mass of the alpha particle is $4$ units moving at velocity $u$ and the residual nucleus has a mass of $234$ units with velocity $v$, the balancing equation is $4u + 234v = 0$. Isolating $v$ results in $v = -\frac{4u}{234}$, which indicates the recoil velocity, making the magnitude of the recoil speed $\frac{4u}{234}$.

The decay process is governed by the first-order kinetics equation $\lambda = \frac{1}{t} \ln \frac{A_0}{A}$, which allows us to determine the decay constant from the change in activity over a specific time interval. By substituting the initial activity of 5000 disintegrations per minute and the final activity of 1250 disintegrations per minute into the equation, we get $\lambda = \frac{1}{5} \ln \frac{5000}{1250}$. Simplifying the ratio within the logarithm leads to $\frac{1}{5} \ln 4$, and since $\ln 4$ is equivalent to $2 \ln 2$, the calculation yields $0.4 \ln 2$.

For two nuclei to undergo fusion, their kinetic energy must be sufficient to overcome the significant repulsive potential energy barrier that exists due to their positive charges. This thermal energy is quantified by the relationship between temperature and kinetic energy, specifically $K.E. = \frac{3}{2} kT$, where $k$ is the Boltzmann constant.

Equating the average kinetic energy to the barrier energy of $7.7 \times 10^{-14} \mathrm{J}$ results in the equation $\frac{3}{2} kT = 7.7 \times 10^{-14} \mathrm{J}$. Solving for the temperature $T$ involves multiplying the potential energy by 2, dividing by 3, and then dividing by the Boltzmann constant $k = 1.38 \times 10^{-23} \mathrm{J/K}$. This process leads to $T = \frac{2 \times 7.7 \times 10^{-14}}{3 \times 1.38 \times 10^{-23}}$, which simplifies to a value of approximately $3.7 \times 10^{9} \mathrm{K}$, indicating that the necessary temperature is on the order of $10^{9} \mathrm{K}$.

Radioactive decay processes alter the atomic number $Z$ based on the specific particle emitted, where alpha decay reduces $Z$ by 2, beta-minus emission increases $Z$ by 1, and beta-plus emission decreases $Z$ by 1. By summing these individual contributions across the entire decay chain, we can precisely calculate the resulting atomic number from the initial value.

The decay chain consists of 8 alpha particles, 4 beta-minus particles, and 2 beta-plus particles. The 8 alpha emissions result in a total reduction of $8 \times 2 = 16$ in the atomic number, whereas the 4 beta-minus emissions increase it by $4 \times 1 = 4$, and the 2 beta-plus emissions provide a further decrease of $2 \times 1 = 2$. Accounting for these changes starting from the initial value of $Z=92$, the final atomic number is:

$Z_{final} = 92 - 16 + 4 - 2 = 78$

Radioactive decay occurs when an unstable nucleus transitions to a more stable state by releasing energy and matter in the form of specific particles. Known emission types include $\alpha$-particles, which are identified as a Helium nucleus, $\beta$-particles, which are identified as electrons, and neutrinoes. Since protons are not among the standard particles released during these radioactive transformations, they are the one item from the list that cannot be emitted.

The energy levels of a hydrogen-like ion are determined by its atomic number and the principal quantum number. An electron's binding energy in a particular state is equivalent to the amount of energy required to remove it from that state, which is equal to the magnitude of the total energy at that level.

For a lithium ion, the atomic number is $Z = 3$. The first excited state corresponds to the quantum number $n = 2$. We apply the energy formula, given by $E_n = - \frac{13.6 Z^2}{n^2}$ eV. Substituting the values for $Z$ and $n$, we find $E_2 = - \frac{13.6 \times 3^2}{2^2}$ eV, which simplifies to -30.6 eV. Since the binding energy is the absolute value of the total energy of the electron in this state, it requires 30.6 eV to remove the electron completely.

When a particle moves through a magnetic field, it only experiences a magnetic Lorentz force if it carries an electric charge, which is defined by the relationship $F = q(\vec{v} \times \vec{B})$. Since electrons, protons, and $He^{2+}$ ions all possess a non-zero electric charge, they interact with the magnetic field and are consequently deflected from their path. Conversely, because neutrons are neutral and have no charge, they travel through the magnetic field without experiencing any force or change in their trajectory.

The energy states of a hydrogen atom are quantized such that the energy of an electron in any orbit $n$ is defined by the expression $E_n = -\frac{13.6}{n^2}\ \mathrm{eV}$. This negative value signifies that the electron is held within the atomic potential well, and removing it requires adding sufficient energy to bring its total energy to zero, which represents a completely free, unbound electron. By substituting $n=2$ into the energy formula, we find that the electron's energy in that specific state is $E_2 = -\frac{13.6}{4} = -3.4\ \mathrm{eV}$. Consequently, to overcome this binding and completely strip the electron from the atom, an input of $3.4\ \mathrm{eV}$ is necessary to bridge the energy gap from the current bound state to the zero-energy level.

Radioactive decay dictates that a substance reduces to half its current mass over every duration of its half-life. Since the half-life is 5 years, the total 15-year duration covers exactly three such intervals. After the first 5 years the substance reduces to $\frac{N_0}{2}$, after 10 years it becomes $\frac{N_0}{4}$, and after the final 5-year period it reaches $\frac{N_0}{8}$.

The impact parameter in Rutherford scattering is proportional to the cotangent of half the scattering angle, governed by the relation $b \propto \cot \frac{\theta}{2}$. Given that the impact parameters for the scattering angles of $60^{\circ}$ and $90^{\circ}$ are represented by $\sqrt{d_1}$ and $\sqrt{d_2}$, their ratio is determined by the expression $\frac{\sqrt{d_1}}{\sqrt{d_2}} = \frac{\cot(60^{\circ}/2)}{\cot(90^{\circ}/2)}$. Substituting the specific trigonometric values, where $\cot(30^{\circ}) = \sqrt{3}$ and $\cot(45^{\circ}) = 1$, leads to the equation $\sqrt{\frac{d_1}{d_2}} = \sqrt{3}$. Squaring both sides of this equation results in $\frac{d_1}{d_2} = 3$, which confirms that $d_1 = 3d_2$ and consequently identifies $x$ as 3.