Atoms And Nuclei – NEET UG PhysicsPractice Questions & PYQs

The number of remaining nuclei in a radioactive sample decreases exponentially with time according to the decay law $N = N_0 e^{-\lambda t}$. Applying this exponential decay expression independently to both substances and taking their ratio eliminates the initial number of nuclei to yield $\frac{N_A}{N_B} = e^{-(\lambda_A - \lambda_B)t}$. Substituting the respective decay constants and the specified ratio leads to the equation $e^{-2} = e^{-4\lambda t}$, which easily simplifies to give the desired time interval.

Beta-minus decay is governed by the weak nuclear interaction where an uncharged neutron transforms into a positively charged proton, accompanied by the emission of an energetic electron and an antineutrino. Since electrons do not naturally reside within the atomic nucleus, this emitted beta particle is literally born at the moment of the neutron's transformation. The newly formed proton remains confined inside the nucleus while the electron is expelled at high speed to conserve charge and lepton number.

Radioactive decay obeys an exponential decay law where the activity $R$ at any given time $t$ is expressed as $R = R_0 e^{-\lambda t}$ in terms of the initial activity $R_0$ and the decay constant $\lambda$. Writing this relation for the two specific time instances gives $R_1 = R_0 e^{-\lambda t_1}$ and $R_2 = R_0 e^{-\lambda t_2}$. Dividing the first equation by the second eliminates the initial activity term and simplifies the exponents, resulting in the expression $R_1 = R_2 e^{-\lambda (t_1 - t_2)}$.

When a hydrogen atom absorbs a photon, the energy required to lift an electron from the ground state to an excited orbit $n$ corresponds to the energy difference between those specific levels, governed by the formula $E = E_n - E_1$. Substituting the given excitation energy of $12.1 \text{ eV}$ alongside the ground state energy of $-13.6 \text{ eV}$ allows us to solve for $n$, yielding $n^2 = \frac{13.6}{1.5} = 9$ and thus $n = 3$. As the excited electron transitions back down to the ground state, the total number of unique spectral lines emitted is calculated using the combinatorial formula $\frac{n(n-1)}{2}$, which evaluates to three lines.

Nuclear energy release in a fusion reaction equals the difference between the total binding energy of the product nucleus and the total binding energy of the reactant nuclei. The reaction combines two deuterons into one helium nucleus, so the total initial binding energy is twice that of a single deuteron, amounting to $2 \times 2.2 \text{ MeV} = 4.4 \text{ MeV}$. Subtracting this initial binding energy from the helium nucleus binding energy gives $28 \text{ MeV} - 4.4 \text{ MeV} = 23.6 \text{ MeV}$ of released energy.

The nuclear radius is directly proportional to the cube root of the mass number, expressed as $R = R_0 A^{1/3}$, where $A$ represents the total number of nucleons. Setting up the ratio of the radius of beryllium to germanium allows us to equate it to the cube root of their respective mass numbers. Since the radius of germanium is given as twice that of beryllium, substituting this relationship cancels the common radius terms and leaves the mass number relationship. Cube both sides of the resulting equation to solve for the mass number of germanium, yielding $A = 2^3 \times 9 = 72$.

The conservation of energy dictates that the energy difference for a direct transition between two levels equals the sum of energy differences across intermediate sequential transitions. Expressing each energy gap in terms of Planck's constant and the respective wavelength using the relation $E = \frac{hc}{\lambda}$, the equation transforms into $\frac{hc}{\lambda_{3}} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}}$. Canceling the common terms and rearranging algebraically yields the reciprocal sum relation, which directly simplifies to the product over sum formula for the third wavelength.

The energy released in a nuclear reaction is determined by the net increase in total binding energy when reactants transform into products. In this specific fusion process, the reactants are a deuterium and a tritium nucleus with binding energies $a$ and $b$, while the product helium-4 nucleus has a binding energy of $c$, with the free neutron contributing zero binding energy. Therefore, the total binding energy of the products is $c$, and the total binding energy of the reactants is $a + b$. Subtracting the reactant binding energy from the product binding energy gives the net energy output as $c - a - b$.

Isotones are nuclei that have the exact same number of neutrons, which is calculated by subtracting the atomic number from the mass number for each nuclide. Examining the first option, selenium-74 has an atomic number of 34, leaving it with $74 - 34 = 40$ neutrons. Gallium-71 has an atomic number of 31, which similarly gives it $71 - 31 = 40$ neutrons. Since both nuclei share an identical neutron count of 40, they satisfy the definition of isotones.

In a hydrogen-like atom, the total energy of an electron is equal in magnitude to its kinetic energy but opposite in sign, while the potential energy is negative and twice the magnitude of the kinetic energy. Because the total energy is the sum of the kinetic and potential energies, it simplifies to the negative of the kinetic energy value. Given that the total energy in the first excited state is -3.4 eV, setting this equal to the negative kinetic energy directly reveals that the kinetic energy must be positive 3.4 eV.

Nuclear mass defect dictates that energy released during fission is accompanied by a conversion of a fraction of mass into pure energy. Because the parent nucleus loses mass in the form of binding energy during the splitting process, the total mass of the resulting fission products remains lower than the original mass. Consequently, dividing the mass of the fission products by the mass of the parent nucleus yields a fraction less than unity.

Nuclear stability is directly determined by the binding energy per nucleon, where higher values indicate a more tightly bound and stable configuration. For heavy nuclei with large mass numbers, electrostatic repulsion among protons increasingly counteracts the short-range strong nuclear force, causing the binding energy per nucleon to decrease. When these heavy nuclei split into smaller fragments, the resulting products possess a higher binding energy per nucleon, releasing the excess energy that drives the fission process.

Radioactive decay follows an exponential decrease where the remaining amount of a substance depends on the number of half-lives that have elapsed. The fraction of the sample remaining is given by the relation $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$, where $n$ represents the number of half-lives. Substituting the given masses into this relation gives $\frac{25}{100} = \left(\frac{1}{2}\right)^n$, which simplifies to $n = 2$. Multiplying this number of half-lives by the given half-life of 1600 years yields a total time of 3200 years.

Nuclear fusion releases an immense amount of energy by combining lighter nuclei into a heavier product, which requires a conversion of mass into energy. Because a portion of the initial mass is permanently transformed and radiated away during the reaction, the combined mass of the initial fusing nuclei is always greater than the mass of the resultant nucleus, expressed mathematically as $m_{3} < (m_{1} + m_{2})$.

Niels Bohr introduced a modification to the Rutherford atomic model by proposing that electrons can only occupy specific orbits where their orbital angular momentum is an integral multiple of $\frac{h}{2\pi}$. This quantization condition restricts electrons to stable, non-radiating stationary states rather than radiating energy continuously. Consequently, transitions between these discrete energy levels result in unique line spectra characteristic of each element.

The atomic number designates the exact count of protons inside an atomic nucleus, which uniquely identifies the chemical element and its position on the periodic table. Meanwhile, the mass number represents the total nucleon count, combining both protons and neutrons together. Subtracting the atomic number from the mass number therefore gives the exact number of neutrons present in that specific nucleus.

The binding energy of a nucleus is determined by calculating its mass defect, which represents the difference between the total mass of its individual constituent nucleons and the actual mass of the assembled nucleus. For a helium nucleus containing two protons and two neutrons, the total mass of the separate nucleons is calculated via the equation $2(1.0073\,\mathrm{u}) + 2(1.0087\,\mathrm{u}) = 4.0320\,\mathrm{u}$. Subtracting the given helium nucleus mass of 4.0015 u yields a mass defect of 0.0305 u. Multiplying this mass defect by the energy equivalent of 931 MeV per atomic mass unit gives the final binding energy of 28.4 MeV.

Coulomb's law describes the electrostatic force between two point charges, acting along the line joining them and directed inward for opposite charges. For a hydrogen atom, the nucleus has a charge of $+e$ and the orbiting electron has a charge of $-e$. When writing this vector force using the position vector $\vec{r}$ instead of the unit vector $\hat{r}$, the radius term in the denominator becomes cubed because the unit vector is expressed as $\frac{\vec{r}}{r}$. Including the negative sign to account for the attractive nature of the force between the positive proton and negative electron yields $-\frac{1}{4\pi\varepsilon_0} \frac{e^{2}}{r^{3}} \vec{r}$.

The radius of an electron orbit in a hydrogen-like atom is inversely proportional to the atomic number of the element according to the relation $r \propto \frac{1}{Z}$. Among the given systems, doubly ionized lithium has the highest atomic number with $Z = 3$, whereas hydrogen, deuterium, and singly ionized helium have lower atomic numbers of 1, 1, and 2 respectively. A higher nuclear charge exerts a stronger electrostatic attraction on the orbiting electron, pulling it into a tighter orbit and resulting in the minimum radius.

Stellar energy generation is driven by thermonuclear fusion reactions occurring under extreme temperatures and pressures deep within a star's core. Inside the Sun, immense gravitational forces compress hydrogen nuclei, forcing individual protons to overcome electrostatic repulsion and fuse together in a multi-step sequence known as the proton-proton chain. This continuous synthesis of lighter hydrogen into heavier helium nuclei converts a fraction of mass into an immense amount of energy according to mass-energy equivalence.