Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

The binding energy per nucleon remains approximately constant for mass numbers between 30 and 170 because nuclear forces are short-ranged, causing each nucleon to interact only with its nearest neighbors rather than the entire nucleus. This property, known as the saturation of nuclear forces, ensures that the total binding energy scales linearly with the number of nucleons, thereby making the binding energy per nucleon independent of the mass number in this range. Since the observed constancy of the binding energy per nucleon is a direct consequence of this saturation, which arises specifically because the nuclear force is short-ranged, the reason provided serves as the correct explanation for the assertion.

The half-life of a radioactive isotope relates to its decay constant $\lambda$ via the expression $T_{1/2} = \frac{\ln 2}{\lambda}$, whereas its average life is represented as $\tau = \frac{1}{\lambda}$. Given the condition that the half-life of element A matches the average life of element B, we equate these two expressions to get $\frac{\ln 2}{\lambda_A} = \frac{1}{\lambda_B}$. Rearranging this equality to isolate $\lambda_A$ yields the final relationship $\lambda_A = \lambda_B \ln 2$.

The stability of a nucleus is defined by its binding energy, which represents the mass lost when individual nucleons combine to form the nucleus, a phenomenon known as mass defect. For a helium nucleus, this mass defect is calculated by finding the difference between the combined mass of two free protons and two free neutrons and the actual mass of the helium nucleus.

$\Delta m = 2(1.0073) + 2(1.0087) - 4.0015 = 0.0305\,\mathrm{u}$

Applying the energy conversion factor to this mass defect yields the binding energy of 28.4 MeV.

The concept of half-life defines the duration required for half of the atoms in a radioactive sample to decay, meaning the quantity remaining is reduced by half after each interval. Consequently, the remaining fraction of the substance depends on the number of half-lives elapsed during the total time.

With a half-life of 30 min and a total elapsed time of 90 min, the number of half-lives is 3. The decay follows the relation $\frac{N}{N_0} = \left(\frac{1}{2}\right)^n$, where $n$ is the number of half-lives. Substituting 3 for $n$ leads to the calculation $\left(\frac{1}{2}\right)^3$, which results in a final remaining fraction of $\frac{1}{8}$.

The binding energy required to separate an electron from a hydrogen atom corresponds to the magnitude of the electron's total energy in its orbit, which is governed by the electrostatic potential energy relationship $E = \frac{ke^2}{2r}$, where $k$ is Coulomb's constant, $e$ is the elementary charge, and $r$ is the orbital radius. By rearranging this expression to solve for the radius, we obtain the formula $r = \frac{ke^2}{2E}$.

To determine the numerical value of the radius, we substitute the given constants $k = 9 \times 10^9\ \mathrm{Nm}^2/\mathrm{C}^2$ and $e = 1.6 \times 10^{-19}\ \mathrm{C}$, along with the total energy $E = 12.8 \times 1.6 \times 10^{-19}\ \mathrm{J}$. This substitution gives $r = \frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{2 \times (12.8 \times 1.6 \times 10^{-19})}$. We can simplify this expression by canceling out the common term $(1.6 \times 10^{-19})$ from both the numerator and the denominator, which results in $r = \frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{2 \times 12.8}$. Combining the powers of ten and simplifying the coefficients, we arrive at $r = \frac{9 \times 10^{-10}}{2 \times \frac{12.8}{1.6}}$. Since the ratio $\frac{12.8}{1.6}$ is equal to $8$, the denominator simplifies to $2 \times 8 = 16$. This leaves the radius as $r = \frac{9}{16} \times 10^{-10}\ \mathrm{m}$, which identifies $x$ as $16$.

The Rydberg formula for the wavelength of spectral lines in the Balmer series of hydrogen is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$, where $R$ is the Rydberg constant and $n$ is the principal quantum number of the higher energy level. For the $H_\alpha$ line, $n=3$, which yields $\frac{1}{\lambda_\alpha} = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36}$, resulting in $\lambda_\alpha = \frac{36}{5R}$. For the $H_\beta$ line, $n=4$, yielding $\frac{1}{\lambda_\beta} = R \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3R}{16}$, which results in $\lambda_\beta = \frac{16}{3R}$. The ratio of these wavelengths is $\frac{\lambda_\alpha}{\lambda_\beta} = \frac{36}{5R} \times \frac{3R}{16} = \frac{36 \times 3}{5 \times 16} = \frac{9 \times 3}{5 \times 4} = \frac{27}{20}$. Comparing this result to $\frac{x}{20}$ shows that $x=27$.

The release of energy in radioactive decay arises from Einstein's principle of mass-energy equivalence, where a reduction in the system's total mass during a reaction is converted into kinetic energy. In this alpha decay, the initial mass of the parent nucleus exceeds the combined mass of the resulting daughter nucleus and the emitted alpha particle, with this mass defect representing the energy liberated.

Calculating the mass difference involves subtracting the sum of the product masses from the mass of the parent nucleus: $238.05079 - (234.04363 + 4.00260) = 0.00456$. Multiplying this mass defect by the energy conversion factor provides the total Q-value: $Q = 0.00456 \times 931.5\,\text{MeV} = 4.25\,\text{MeV}$.

The wavelength of radiation emitted during a hydrogen atom transition is determined by the energy difference between orbital levels, expressed by the equation $\frac{hc}{\lambda} = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.

For the transition from the second excited state ($n=3$) to the first excited state ($n=2$), the wavelength $\lambda_0$ follows the expression $\frac{hc}{\lambda_0} = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$, while the transition from the third excited state ($n=4$) to the second orbit ($n=2$) follows $\frac{hc}{\lambda'} = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$. Dividing these expressions results in the ratio $\frac{\lambda_0}{\lambda'} = \frac{1/2^2 - 1/4^2}{1/2^2 - 1/3^2} = \frac{3/16}{5/36} = \frac{27}{20}$, which implies $\lambda' = \frac{20}{27} \lambda_0$. Comparing this result to the given expression $\lambda' = \frac{20}{x} \lambda_0$ confirms that $x = 27$.

The initial mole count for each substance is determined by dividing its mass by the atomic mass number. For substance A, this calculation is $320 / 16 = 20$ moles, and for substance B, it is $320 / 32 = 10$ moles.

The remaining quantity of radioactive material follows the exponential decay law:
$N = N_0 \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$
where $t$ is the elapsed time and $T_{1/2}$ is the half-life. Substance A decays for 2 days with a half-life of 1 day, leaving $20 \times (1/2)^{2/1} = 5$ moles. Substance B decays for 2 days with a half-life of 0.5 days, resulting in $10 \times (1/2)^{2/0.5} = 10 \times (1/2)^4 = 10 / 16 = 0.625$ moles.

The combined amount of both substances is 5.625 moles. Multiplying this total by Avogadro's number $6.023 \times 10^{23}$ yields $3.38 \times 10^{24}$ atoms.

The energy of an electron in a hydrogen-like ion is determined by the formula $E = -13.6 \frac{Z^2}{n^2}\,\text{eV}$, where $Z$ is the atomic number and $n$ is the principal quantum number. For the $\mathrm{He}^{+}$ ion, the atomic number is $Z = 2$, and the first excited state corresponds to $n = 2$. Substituting these values into the energy expression yields $E = -13.6 \left( \frac{2^2}{2^2} \right) = -13.6 \cdot 1 = -13.6\,\text{eV}$.

The wavelength of a photon emitted during an atomic transition is inversely proportional to the energy difference of that transition, as described by the equation $\lambda = \frac{hc}{E}$. To achieve the shortest possible wavelength, the system must undergo the transition associated with the maximum energy change. Among the available options, the transition labeled $D$ spans the largest energy gap, which therefore corresponds to the shortest emitted wavelength.

The Bohr model dictates that electrons in a hydrogen atom occupy discrete energy levels defined by the principal quantum number $n$. When an atom in its ground state, which has an energy of $-13.6\ \mathrm{eV}$, absorbs a photon of $12.75\ \mathrm{eV}$, it transitions to a higher energy state corresponding to the sum of its initial energy and the absorbed energy. This process results in a final energy of $-0.85\ \mathrm{eV}$.

By equating this result to the energy formula $E_n = -\frac{13.6}{n^2}\ \mathrm{eV}$, we find that $n^2$ equals 16, identifying the excited state as $n=4$. The angular momentum of the electron in this orbit is given by $L = \frac{nh}{2\pi}$. Substituting $n=4$ leads to $L = \frac{4h}{2\pi} = \frac{2h}{\pi}$. Inserting the provided value $h = 4.14 \times 10^{-15}\ \mathrm{eVs}$, we determine $L = \frac{2 \times 4.14 \times 10^{-15}}{\pi} = \frac{8.28 \times 10^{-15}}{\pi}\ \mathrm{eVs}$. Expressing this in the required format of $\frac{x}{\pi} \times 10^{-17}\ \mathrm{eVs}$ gives $\frac{828 \times 10^{-17}}{\pi}\ \mathrm{eVs}$, identifying the value of $x$ as 828.

Bohr's model states that the angular momentum of an electron in a hydrogen atom is quantized according to the formula $L = \frac{nh}{2\pi}$, where $n$ represents the orbit number. For the first orbit, where $n=1$, the angular momentum is defined as $L = \frac{h}{2\pi}$. When the electron occupies the second orbit, $n=2$, the angular momentum is calculated as $L_2 = \frac{2h}{2\pi}$, which simplifies to $2L$. Consequently, the change in angular momentum when moving between these orbits is determined by calculating the difference, resulting in $\Delta L = 2L - L = L$.

The semi-empirical mass formula models the binding energy of a nucleus by summing independent contributions, where the volume energy term $a_3 A$ accounts for the short-range, attractive nuclear forces that scale linearly with the total number of nucleons $A$. The surface energy term represents a reduction in binding energy because nucleons at the periphery have fewer neighbors than those in the interior, making this correction proportional to the surface area, which scales as $A^{2/3}$. In contrast, the other proposed expressions, such as the surface energy per nucleon or the Coulomb contribution, fail to align with the specific dependencies on $A^{1/3}$ found in the standard nuclear model, confirming that only the statements regarding volume and surface energy area proportionality are accurate.

For a particle moving in a circular orbit within a potential $U = \frac{1}{2} m \omega^2 r^2$, the centripetal force is provided by the radial gradient of the potential, $F = -\frac{dU}{dr} = -m \omega^2 r$, resulting in the balance $\frac{mv^2}{r} = m \omega^2 r$ which simplifies to $v = \omega r$. By substituting this into Bohr's angular momentum quantization condition, $mvr = \frac{nh}{2\pi}$, we obtain $m(\omega r)r = \frac{nh}{2\pi}$, leading to the expression $m \omega r^2 = \frac{nh}{2\pi}$. Solving for the radius $r$ reveals that $r^2 \propto n$, which implies that the radius of the $n^{\text{th}}$ orbit is proportional to $\sqrt{n}$.

Radioactive decay follows an exponential pattern where the remaining quantity of a substance decreases by a specific fraction over given time intervals. Since the material reduces to $\frac{1}{8}$ of its original amount in 3 days, and knowing that $\frac{1}{8} = (\frac{1}{2})^3$, we can deduce that the material completes three half-lives in 3 days, meaning the half-life is 1 day.

With a half-life of 1 day, the remaining amount after any time $t$ is given by $m = m_0 \left( \frac{1}{2} \right)^t$, where $m_0$ represents the initial amount. The problem states that 8 g of the material remains after 5 days. Applying the half-life formula leads to the equation $8 = m_0 \left( \frac{1}{2} \right)^5$ Since the value of $\left( \frac{1}{2} \right)^5$ is $\frac{1}{32}$, the calculation becomes $8 = \frac{m_0}{32}$, resulting in an initial mass of 256 g.

Bohr's theory for the hydrogen atom establishes that the speed of an electron is inversely proportional to its orbital number, represented as $v \propto \frac{1}{n}$. Because the speed is 3.6 × 10⁶ m/s in the seventh orbit, we establish the relationship between the third and seventh orbits using the equation $\frac{v_3}{v_7} = \frac{7}{3}$. Applying this ratio to the known speed, the velocity in the third orbit calculates to 8.4 × 10⁶ m/s.

The disintegration of a stationary nucleus relies on the conservation of linear momentum, requiring the two resulting fragments to have equal momentum magnitudes. This relationship, expressed as $m_1 v_1 = m_2 v_2$, dictates that the mass ratio must be the inverse of the velocity ratio. Given the velocity ratio is 3 to 2, the mass ratio is $m_1 / m_2 = 2/3$.

Nuclear density is a constant property of matter, meaning mass is directly proportional to the cube of the radius. Consequently, the mass ratio of the two fragments corresponds to the cube of their radius ratio, as shown in the equation:
$\frac{m_1}{m_2} = \left(\frac{r_1}{r_2}\right)^3$
By substituting the mass ratio, we arrive at $(r_1 / r_2)^3 = 2/3$, which implies $r_1 / r_2 = (2/3)^{1/3}$. Aligning this with the given form $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ reveals that $x = 2$.

The energy released during an atomic transition manifests as a photon, with the relationship between the energy difference of the levels, $\Delta E$, and the photon's wavelength, $\lambda$, defined by $\Delta E = \frac{hc}{\lambda}$. When working with atomic transitions in electron-volts and nanometers, this calculation is efficiently simplified to $\lambda = \frac{1241 \text{ eV} \cdot \text{nm}}{\Delta E}$ based on the physical constants provided.

To determine which transition produces a photon of $124.1 \text{ nm}$, we rearrange the expression to solve for the energy gap: $\Delta E = \frac{1241 \text{ eV} \cdot \text{nm}}{124.1 \text{ nm}}$, which calculates to exactly $10 \text{ eV}$. Comparing this result to the transitions shown in the figure, transition D corresponds to an energy difference of $10 \text{ eV}$, confirming that this specific path results in the required photon emission.

The concept of half-life dictates that a radioactive substance reduces to one-half of its initial quantity after each specific interval of time. Given a half-life of 5 years and a total elapsed time of 15 years, the sample undergoes 3 full half-life cycles.

We determine the remaining substance using the relation $N = N_0 \left(\frac{1}{2}\right)^n$, where $n$ counts the number of half-lives passed. With 3 cycles complete, the fraction remaining is $N = N_0 \left(\frac{1}{2}\right)^3$, which simplifies to $\frac{N_0}{8}$. Since the question asks for the decayed portion, we subtract this remainder from the original total to get $1 - \frac{1}{8} = \frac{7}{8}$.