Atoms And Nuclei – NEET UG PhysicsPractice Questions & PYQs

Nuclear reactions must strictly conserve both total mass number and total atomic number across the process. Equating the reactant and product atomic numbers gives the balance equation $92 + 0 = 36 + Z$, which yields $Z = 56$. Balancing the mass numbers requires $235 + 1 = 89 + 3 + A$, solving to $A = 144$. Combining these conserved values identifies the generated nuclide as $^{144}_{56}\mathrm{Ba}$.

Einstein's mass-energy equivalence principle dictates that mass converts directly into energy via the relation $E = mc^2$. Substituting the mass \in kilograms and the speed of light into the equation gives $E = (0.5 × 10^{-3})(3 × 10^8)^2$. Evaluating this product yields a total energy output of 4.5 × 10^13 J.

The Bohr model of atomic structure is exclusively valid for systems containing a single orbital electron, such as hydrogen or singly ionized helium. Singly ionized neon contains multiple electrons remaining in its shells, meaning inter-electron repulsions invalidate the fundamental assumptions of the Bohr model. Therefore, this model cannot accurately describe the behavior or energy levels of a singly ionized neon atom.

When an electron in a hydrogen atom is replaced by a muon, the system's dynamics shift due to the altered reduced mass, which is expressed as $\mu = \frac{mM}{M+m}$ and evaluates to $186 m_e$. Because the Bohr radius is inversely proportional to this reduced mass, the new first Bohr radius scales down to 2.56 × 10^-13 m. Conversely, the ground state energy is directly proportional to the reduced mass, multiplying the original energy to arrive at -2.8 eV.

In the Bohr model of a hydrogen atom, the total energy, kinetic energy, and potential energy of an orbiting electron maintain a proportional relationship where kinetic energy equals the negative of total energy, and potential energy equals twice total energy. Taking the negative of the given total energy of -3.4 eV yields a kinetic energy of 3.4 eV. Multiplying the total energy by two determines the potential energy to be -6.8 eV.

An alpha particle is fundamentally the nucleus of a helium atom stripped of its orbital electrons. Because it represents a helium-4 nucleus, it consists strictly of two protons and two neutrons bound tightly together by the strong nuclear force. Consequently, it contains no electrons in this bare nuclear state.

The rate of radioactive decay is linked to the number of active nuclei through the relation $R = N\lambda$, where $\lambda$ represents the decay constant defined by the half-life. Substituting the given half-life of $2.2 \times 10^9$ seconds yields the decay constant, which is then combined with the disintegration rate of $10^{10}$ per second. Rearranging the decay law equation gives $N = \frac{R}{\lambda}$, resulting in a total of $3.17 \times 10^{19}$ radioactive atoms present in the sample.

Radioactive decay follows an exponential reduction pattern where the fraction of remaining nuclei is expressed by the law $\frac{N}{N_0} = \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$. Since 450 out of the initial 600 nuclei have disintegrated, the remaining undecayed nuclei amount to 150. Substituting these values into the decay equation gives $\frac{150}{600} = \left(\frac{1}{2}\right)^{\frac{t}{10}}$, which simplifies to $\left(\frac{1}{2}\right)^2 = \left(\frac{1}{2}\right)^{\frac{t}{10}}$. Equating the exponents reveals a required time duration of 20 minutes.

In a Bohr hydrogen atom, the electrostatic force of attraction supplies the necessary centripetal force, establishing the relationship that kinetic energy is equal in magnitude to half of the potential energy and negative the total energy. Consequently, the total energy of مرs the orbiting electron is negative and matches the kinetic energy in magnitude but differs in sign. This direct inverse relationship dictates that the ratio of kinetic energy to total energy simplifies to $1 : -1$.

The wavelength of spectral lines in hydrogen-like atoms is determined using the Rydberg formula $\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. For the last line of the Balmer series, the transition occurs from infinity to the ground level $n_1 = 2$, yielding a wavelength of $\frac{4}{R}$. Similarly, the last line of the Lyman series corresponds to a transition from infinity to $n_1 = 1$, resulting in a wavelength of $\frac{1}{R}$. Taking the ratio of the Balmer wavelength to the Lyman wavelength gives a final value of 4.

Radioactive decay obeys the exponential law $N = N_0 e^{-\lambda t}$, allowing us to express the population of each sample at any given time $t$. Writing the expressions for both materials gives $N_A = N_0 e^{-8\lambda t}$ and $N_B = N_0 e^{-\lambda t}$. Setting up the required ratio between the remaining nuclei of material B and material A results in the equation $\frac{e^{-\lambda t}}{e^{-8\lambda t}} = \frac{1}{e}$, which simplifies to $e^{7\lambda t} = e^{-1}$. Solving this expression for time yields the required duration of $\frac{1}{7\lambda}$.

The distance of closest approach is found by equating the initial kinetic energy of the alpha particle to the electrostatic potential energy it possesses at the turning point, expressed through the conservation relation $\frac{1}{2} m v^2 = \frac{1}{4\pi\varepsilon_0} \frac{(2e)(Ze)}{r}$. Rearranging this equation to solve for the distance $r$ shows that it is inversely proportional to the mass of the projectile when velocity is held constant. Consequently, the dependence of the closest approach distance on mass follows a reciprocal relationship where $r$ scales as one over $m$.

Radioactive decay follows an exponential reduction pattern where the remaining substance halves every half-life period. When forty percent of the substance has decayed, sixty percent of the original nuclei remain un-decayed as the starting quantity for this interval. Similarly, when eighty-five percent has decayed, only fifteen percent of榴 the original nuclei are left remaining. Comparing these two remaining amounts shows that the quantity has reduced by a factor of four, which corresponds exactly to two successive half-lives passing and results in a total elapsed time of sixty minutes.

The wave number of a spectral line is determined using the Rydberg formula $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. Since the Balmer series terminates at a lower energy level of two and the last line comes from an infinite upper level, the terms simplify to one over four multiplied by the Rydberg constant. Inserting the specified value of $10^7$ for the constant yields a wave number of $0.25 \times 10^7$.

The wavelength of an emitted photon is determined by evaluating the energy difference between orbits using the Rydberg relation $\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$. For the transition from the third to the second orbit, the inverse wavelength evaluates to $\frac{5}{36}R$. When the electron falls from the fourth to the third orbit, the corresponding inverse expression becomes $\frac{7}{144}R$, and comparing these two equations yields the new wavelength as $\frac{20}{7}\lambda$.

When a stationary body decays into two fragments, conservation of momentum dictates that both fragments possess equal and opposite linear momenta. The kinetic energy of a particle is related to its momentum by the expression $KE = \frac{p^2}{2m}$, showing that kinetic energy is inversely proportional to mass when momentum is constant. Because the helium nucleus has a significantly smaller mass than the thorium nucleus, it must possess a higher kinetic energy to maintain that equal momentum.

The nuclear radius scales directly with the cube root of the mass number according to the scaling law $R = R_0 A^{1/3}$. Setting up the ratio of the radius of aluminum to that of tellurium using their respective mass numbers twenty-seven and one hundred twenty-five allows the constant factor to cancel out. Evaluating the cube root of this mass number ratio yields three-fifths, which rearranges directly to give the tellurium radius as five-thirds times the aluminum radius.

The wavelength of emitted radiation in hydrogen spectral series is determined by the Rydberg formula, where the longest wavelength in any series corresponds to a transition from the immediately adjacent higher energy level. For the Lyman series, this transition occurs from the second energy level to the ground state, producing a reciprocal wavelength equal to three-fourths of the Rydberg constant. For the Balmer series, the transition occurs from the third to the second level, resulting in a reciprocal wavelength equal to five thirty-sixths of the Rydberg constant, which leads to a wavelength ratio of 5/27.

The velocity of an electron in the $n$-th orbit of a hydrogen-like ion is governed by the relation $v = \frac{Z}{n} v_0$. Substituting the atomic number for helium and the specified orbit number into this scaling expression scales the baseline velocity constant. Performing this arithmetic yields the final speed of 1.46 $\times$ 10^6 m/s.

The fraction of a radioactive substance remaining after a certain time is related to the number of elapsed half-lives through the decay equation $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$. Since the ratio of isotope X to isotope Y is $1:7$, the remaining amount of X relative to the total initial amount is $\frac{1}{1 + 7}$, which equals $\frac{1}{8}$. Equating this fraction to $\left(\frac{1}{2}\right)^n$ shows that $n = 3$ half-lives have passed. Multiplying this number of half-lives by the given half-life duration of $1.4 \times 10^9$ years gives the total age of the rock as $4.20 \times 10^9$ years.