Atoms And Nuclei – NEET UG PhysicsPractice Questions & PYQs

Einstein's mass-energy relation $E = mc^2$ connects the energy produced in a nuclear reactor to the corresponding mass defect. The mass decay rate per second is expressed through the formula $\frac{\Delta m}{\Delta t} = \frac{P}{c^2}$, where $P$ represents power and $c$ represents the speed of light. Substituting the given reactor power and speed of light yields the decay rate per second, which when multiplied by the 3600 seconds in an hour gives the total mass loss. This calculation results in a mass decay of $4 \times 10^{-8}$ kg per hour, equivalent to $40\text{ }\mu\text{g}$.

Given that the ratio of the remaining isotope to the stable product is one to fifteen, the fraction of the original radioactive substance left in the rock is $\frac{N}{N_0} = \frac{1}{1 + 15}$, which simplifies to $\frac{1}{16}$. Equating this fraction to the decay expression $\left(\frac{1}{2}\right)^n$ determines that exactly four half-lives have elapsed. Multiplying this count of half-lives by the given duration of fifty years per half-life calculates the total age of the rock as two hundred years.

Binding energy per nucleon is determined by converting the mass defect into energy using the relation $E = \Delta m c^2$ and dividing the result by the total mass number of the nucleus. Multiplying the given mass defect of 0.042 u by the atomic mass unit conversion factor, the square of the speed of light, and dividing by the mass number 7 yields the total binding energy per nucleon. Converting this value from joules to electron-volts by dividing by the elementary charge results in approximately 5.6 MeV.

The energy levels of single-electron atomic systems are determined by the formula $E_n = -13.6 Z^2 / n^2$, where $Z$ is the atomic number and $n$ is the principal quantum number. For a helium ion, the atomic number is two, and its first excited state corresponds to the principal quantum number two. Substituting these values into the energy expression causes the squared atomic number in the numerator and the squared quantum number in the denominator to cancel each other out completely. Consequently, the resulting energy matches the ground state energy of a hydrogen atom, yielding a value of -13.6 eV.

The distance of closest approach occurs when a moving charged particle converts all of its initial kinetic energy into electrostatic potential energy due to repulsion from the target nucleus, expressed as $\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \frac{(2e)(Ze)}{d}$. Rearranging this conservation equation to solve for the distance $d$ places the kinetic energy term in the denominator. Since the kinetic energy contains the mass of the alpha particle, the distance of closest approach is inversely proportional to this mass. Therefore, the scaling relationship shows direct dependence on the reciprocal of the mass.

Nuclear fusion energy is determined by calculating the net increase in total binding energy when lighter nuclei combine to form a heavier product. Each deuterium nucleus possesses two nucleons with a binding energy of 1.1 MeV per nucleon, yielding a total initial binding energy of 4.4 MeV for the two reacting nuclei. The resulting helium-4 nucleus contains four nucleons with a binding energy of 7.0 MeV each, establishing a final total binding energy of 28.0 MeV. Subtracting the initial combined binding energy from the final binding energy reveals an energy release of 23.6 MeV.

Radioactive decay follows the exponential law described by the formula $R = R_0 e^{-\lambda t}$, which relates the activity at any time to the initial activity and the decay constant. Substituting the condition that activity reduces to $N_0/e$ at five minutes allows the decay constant to be evaluated as one-fifth per minute. The half-life of the sample is reached when the activity drops to half its initial value, leading to the relation where the product of the decay constant and half-life equals the natural logarithm of two. Rearranging this relation using the calculated decay constant yields a half-life of $5 \log_e 2$ minutes.

Radioactive activity is defined by the relation $A = \lambda N$, where $A$ represents activity, $\lambda$ is the decay constant, and $N$ is the number of active nuclei present. The population of undecayed nuclei at times $t_1$ and $t_2$ can therefore be written as $N_1 = \frac{A_1}{\lambda}$ and $N_2 = \frac{A_2}{\lambda}$ respectively. The total number of nuclei that undergo decay during the specified interval is found by taking the difference between these two population values, which directly results in $\frac{A_1 - A_2}{\lambda}$.

The total number of spectral lines emitted when an electron drops from an excited state $n$ is determined by the formula $\frac{n(n-1)}{2}$, which for six wavelengths points to an initial excitation level of $n = 4$. Because wavelength is inversely proportional to the energy released during a transition, the longest wavelength corresponds to the smallest energy step available. Within the cascades from $n = 4$, the transition from $n = 4$ to $n = 3$ yields the minimum energy difference, producing the maximum wavelength.

At the distance of closest approach in Rutherford scattering, the incoming projectile's kinetic energy is completely converted into electrostatic potential energy given by $U = \frac{1}{4 \pi \varepsilon_0} \frac{z_1 z_2 e^2}{r_0}$. Since total mechanical energy is conserved, the initial energy of the projectile must equal this maximum potential energy value at the turning point. Analyzing this relationship shows that the projectile's energy depends directly on the product of the charges of the projectile and the target nucleus. Thus, the energy is directly proportional to $z_1 z_2$.

Radioactive alpha decay decreases a nucleus's mass number by four and atomic number by two, while beta-minus decay leaves the mass number unchanged and increases the atomic number by one. When a parent nucleus undergoes emission of a certain number of alpha particles alongside twice as many beta particles, the net change in atomic number evaluates to zero because the increases from beta decays perfectly balance the decreases from alpha decays. Therefore, the resulting daughter nucleus shares the identical atomic number as the parent while differing in mass, classifying it as an isotope.

Radioactive decay processes alter the atomic and mass numbers of a nucleus in distinct ways, where beta emission increases the atomic number by one, alpha emission reduces the mass number by four and the atomic number by two, and gamma emission releases excess energy without changing either value. Observing the first transformation from $_Z^A X$ to $_{Z+1}^A Y$, the mass number stays constant while the atomic number increases by one, confirming a beta particle emission. The subsequent change to $_{Z-1}^{A-4} B^*$ drops the mass number by four, which signifies an alpha particle, and the final transition from the excited state $_{Z-1}^{A-4} B^*$ to the ground state $_{Z-1}^{A-4} B$ is achieved through gamma ray emission.

The energy of an electron in any orbit of a hydrogen atom is determined by the relation $E_n = -13.6/n^2$, where $n$ represents the principal quantum number. In the ground state, the electron resides at $n=1$ with an energy of $-13.6 eV$, while in the first excited state, it moves to $n=2$ with an energy of $-3.4 eV$. The excitation energy is the precise difference between these two energy levels, found by subtracting the ground state energy from the first excited state energy. This yields an energy requirement of $-3.4 eV - (-13.6 eV)$, resulting in $10.2 eV$.

Radioactive decay is governed by the exponential law $N(t) = N_0 e^{-\lambda t}$, which describes how the number of remaining nuclei decreases over time based on a given decay constant. Applying this law to both materials with initial nuclei $N_0$ gives the individual remaining quantities as $N_0 e^{-5\lambda t}$ for the first substance and $N_0 e^{-\lambda t}$ for the second substance. Taking the ratio of the first quantity to the second quantity simplifies the expression to $e^{-4\lambda t}$. Equating this resulting ratio to $1/e$ allows the exponents to be matched, since $1/e$ is equivalent to $e^{-1}$, and solving the resulting linear equation for time produces a duration of $\frac{1}{4\lambda}$.

A nucleus always weighs slightly less than the total mass of its constituent nucleons because a fraction of their mass is converted into binding energy. The combined mass of the individual protons and neutrons is given by $ZM_p + (A-Z)M_n$, and the difference between this sum and the actual nuclear mass represents the mass defect. Applying Einstein's mass-energy equivalence yields the relation $BE = \left[ZM_p + (A-Z)M_n - M(A,Z)\right]c^2$. Rearranging this formula to express the nuclear mass isolates $M(A,Z)$ on one side, resulting \in $M(A,Z) = ZM_p + (A-Z)M_n - \frac{BE}{c^2}$.

Nuclear density is expressed as $\rho = \frac{m A}{\frac{4}{3}\pi R^3}$, where $m$ is the average nucleon mass, $A$ is the mass number, and $R$ is the nuclear radius given by $R = R_0 A^{1/3}$. Substituting this radius relation into the volume denominator yields $\rho = \frac{3m}{4\pi R_0^3}$, demonstrating that nuclear density is entirely independent of the mass number $A$. Because the mass number cancels out completely, any two nuclei will possess identical densities regardless of their individual mass numbers. Therefore, the ratio of their nuclear densities remains unaffected by their mass number proportion of $1:3$.

Nuclear radius scales uniformly with the cube root of the atomic mass number according to the empirical relation $R = R_0 A^{1/3}$, implying that the ratio of radii between any two nuclei depends purely on the cube root of the ratio of their mass numbers. For aluminum with a mass number of $27$ and tellurium with a mass number of $125$, the radius ratio simplifies to the cube root of $125$ divided by $27$, which evaluates to $\frac{5}{3}$. Multiplying this scaling factor by the given aluminum radius of $3.6 \text{ fm}$ yields $3.6 \times \frac{5}{3}$, resulting in a final nuclear radius of $6.0 \text{ fm}$ for tellurium.

The binding energy of a nucleus is fundamentally tied to its mass defect, which is the energy equivalent of the mass difference between constituent nucleons and the assembled nucleus as dictated by Einstein's mass-energy equivalence principle. A nucleus with mass number $A$ and atomic number $Z$ contains $Z$ protons and $(A-Z)$ neutrons, meaning the combined mass of these separate nucleons is given by $Z M_p + (A-Z) M_n$. Subtracting the actual nuclear mass $M(A, Z)$ yields the total mass defect of the system. Multiplying this resulting mass defect by $c^2$ scales it into energy units, producing the expression $\text{B.E.} = [Z M_p + (A - Z) M_n - M(A, Z)] c^2$ for the total binding energy.

The total energy of an electron in a hydrogen atom is quantized according to the principal quantum number $n$ via the relation $E_n = \frac{-13.6}{n^2} \text{ eV}$, where the kinetic energy of the bound electron is always equal in magnitude and opposite in sign to its total energy. For the first excited state, the quantum number is $n = 2$, which gives a total energy of $E_2 = \frac{-13.6}{2^2} = -3.4 \text{ eV}$. Taking the negative of this total energy yields a kinetic energy of $3.4 \text{ eV}$ for the electron in that state.

The charge-to-mass ratio of an ion in a mass spectrometer is derived by equating the electrical work done during acceleration to its kinetic energy, and then balancing the magnetic Lorentz force with the centripetal force in the curved region. Equating the potential energy to the kinetic energy and combining it with the circular motion radius formula yields the expression $\frac{q}{m} = \frac{2V}{B^2 R^2}$. Since the accelerating potential and magnetic field strength are held constant, the charge-to-mass ratio varies inversely with the square of the orbital radius.