Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

Bohr's atomic model postulates that an electron's angular momentum, or moment of momentum, is quantized and restricted to integral multiples of $\frac{h}{2\pi}$ based on its orbit number $n$. This relationship is defined by the formula $L = \frac{nh}{2\pi}$.

For an electron situated in the fourth orbit, substituting $n = 4$ into the formula yields $L = \frac{4h}{2\pi}$. Simplifying this fraction results \in a value of $\frac{2h}{\pi}$.

The principle of mass-energy equivalence dictates that the energy equivalent to the mass defect in a nuclear fission reaction is fully converted into the kinetic energy of the resulting daughter nuclei. In this specific scenario, a parent nucleus of mass $M$ disintegrates into three fragments of equal mass, so each fragment has a mass of $M/3$. The total energy released, defined by $\Delta M c^2$, is shared among these three daughter nuclei as kinetic energy. Setting the released energy equal to the combined kinetic energy of the products, we have the relation $\Delta M c^2 = 3 \times (\frac{1}{2} \times \frac{M}{3} \times v^2)$. Simplifying the right side, the three in the numerator and denominator cancels out, which reduces the equation to $\Delta M c^2 = \frac{1}{2} M v^2$. To find the speed $v$, we multiply both sides by two and divide by the mass $M$, resulting in $v^2 = \frac{2 \Delta M c^2}{M}$. Extracting the square root provides the final expression $v = c \sqrt{\frac{2 \Delta M}{M}}$.

Nuclear binding energy represents the energy equivalent of the mass defect, which is the difference between the total mass of the constituent nucleons and the actual mass of the nucleus. The isotope ${}_{5}^{12}\mathrm{B}$ consists of 5 protons and 7 neutrons, so the sum of the masses of these individual particles is $(5 M_p + 7 M_n)$. By subtracting the nuclear mass $M_0$ from this sum to determine the mass defect and multiplying the result by $C^2$ according to Einstein's mass-energy equivalence, we find the binding energy is $(5 M_p + 7 M_n - M_0) C^2$.

According to the Bohr model, the angular momentum of an electron in a stable orbit is quantized and restricted to integral multiples of $\frac{h}{2\pi}$, providing the fundamental condition for electronic stability. In the realm of nuclear physics, nuclear forces are distinct from electromagnetic or gravitational interactions; they are short-range forces that do not adhere to the inverse square law. These interactions are inherently spin-dependent, meaning the magnitude of the force varies based on whether the spins of the interacting nucleons are parallel or antiparallel. While nuclear forces are charge-independent, they contain non-central components, which invalidates the claim that they are strictly central. Furthermore, the stability of a nucleus is inversely proportional to its packing fraction, as a lower packing fraction corresponds to a greater mass defect and thus a higher binding energy per nucleon.

The volume of an atomic nucleus is directly proportional to its mass number because the nuclear radius scales with the cube root of the total number of nucleons. Starting with the standard volume formula for a sphere, $V = \frac{4}{3} \pi R^3$, and substituting the empirical relationship for the radius $R = R_0 A^{1/3}$, we obtain the simplified expression
$V = \frac{4}{3} \pi R_0^3 A$
This derivation confirms that the volume depends linearly on the mass number $A$. Consequently, the ratio of the volumes of two nuclei is simply the ratio of their mass numbers, so substituting the given relationship $A_2 = 4 A_1$ into the expression $\frac{V_2}{V_1} = \frac{A_2}{A_1}$ leads directly to a value of 4.

The energy released during this nuclear chain reaction is fundamentally governed by the mass defect of the net transformation. By combining the two provided reactions, we observe that the neutron acts as an intermediate-consumed in the first step and produced in the second-which allows us to define the net reaction as the fusion of a Lithium-6 nucleus and a Deuterium nucleus into two Helium-4 nuclei. This simplification demonstrates that the total energy output depends solely on the difference in mass between the initial reactants and the final products.

To determine the energy, we calculate the mass defect by subtracting the total mass of the two resulting Helium-4 nuclei from the sum of the masses of the Lithium-6 and Deuterium isotopes. The energy released, $Q$, is determined by the formula:

$Q = \$[M(\mathrm{Li}) + M({}_1^2\mathrm{H}) - 2 \times M({}_2^4\mathrm{He})]\$ \times 931.5$

Substituting the provided mass values of 6.01690, 2.01471, and 4.00388 into this expression gives the total mass defect, which is then multiplied by 931.5 to obtain the energy in mega-electron volts. Completing this calculation yields a total energy release of approximately 22.22 MeV.

Einstein's mass-energy equivalence principle relates mass and energy through the formula $E = mc^2$. To find the energy equivalent of 1 g, we convert the mass to 10^-3 kg and calculate the energy as 9 × 10^13 J using a speed of light of 3 × 10^8 m/s. Converting this value to electron volts with the factor 6.241 × 10^18 eV per Joule gives approximately 56.169 × 10^31 eV. Finally, dividing by 10^6 to express the result in mega-electron volts yields 5.6 × 10^26 MeV.

The energy released by the triple-alpha process originates from the mass difference between the reactants and the product, which is given as $7.266\,\mathrm{MeV}$ per reaction. To relate this to the star's power output, we divide the total power of $5.808 \times 10^{30}\,\mathrm{W}$ by the energy released per reaction, after converting the latter into joules by multiplying by $1.6 \times 10^{-19}\,\mathrm{J/eV}$. This calculation yields a reaction rate of $5 \times 10^{42}$ occurrences per second. Because every individual reaction consumes three ${}^{4}\mathrm{He}$ nuclei to form one ${}^{12}\mathrm{C}$ nucleus, the total conversion rate of helium is $3 \times 5 \times 10^{42}$, which confirms $n = 15$.

The energy of an electron in a hydrogen atom is governed by the principal quantum number $n$ through the relation $E_n = -\frac{13.6}{n^2}$ eV. By setting the energy of the excited state equal to -0.85 eV and solving the equation, we determine that the electron is in the $n = 4$ orbit.

The total number of spectral lines emitted during transitions to lower energy levels is calculated using the formula $\frac{n(n-1)}{2}$. Substituting the value $n = 4$ into this expression gives $\frac{4(4-1)}{2}$, which results in a maximum of 6 allowed transitions.

At the distance of closest approach, the alpha particle momentarily stops as its initial kinetic energy is entirely converted into electrostatic potential energy. This relationship is defined by equating the kinetic energy $\frac{1}{2}mv^2$ to the potential energy $\frac{k(2e)(Ze)}{r_{\min}}$, which simplifies to the expression $v = \sqrt{\frac{4kZe^2}{mr_{\min}}}$.

By substituting the known values into this formula, where $k = 9 \times 10^9$ SI units, $Z = 80$, $r_{\min} = 4.5 \times 10^{-14}$ m, $m = 6.72 \times 10^{-27}$ kg, and $e = 1.6 \times 10^{-19}$ C, we can compute the velocity. Evaluating the expression $\sqrt{\frac{4 \times 9 \times 10^9 \times 80 \times (1.6 \times 10^{-19})^2}{6.72 \times 10^{-27} \times 4.5 \times 10^{-14}}}$ yields a final velocity of 156 x 10^5 m/s.

For an electron revolving in a stable circular orbit around a nucleus of charge $+Ze$, the potential energy $U$ is defined by the electrostatic attraction as $U = -\frac{k(Ze)(e)}{r}$. The kinetic energy is determined by the requirement that the electrostatic force provides the necessary centripetal acceleration, satisfying the balance $\frac{mv^2}{r} = \frac{k(Ze)(e)}{r^2}$, which yields a kinetic energy of $\frac{1}{2}mv^2 = \frac{k(Ze)(e)}{2r}$. By combining these to calculate the total energy $E$, we perform the addition $E = \frac{k(Ze)(e)}{2r} - \frac{k(Ze)(e)}{r}$, which simplifies to $E = -\frac{k(Ze)(e)}{2r}$. Comparing this result to the potential energy shows that the total energy is exactly half of the potential energy, establishing the relationship $2E = U$.

Nuclear fission is governed by the conservation of total atomic number and total mass number, ensuring that the sum of protons and the sum of nucleons remain constant before and after the interaction. For the reaction involving uranium and a neutron, the initial totals are 92 atomic units and 236 mass units. The combination of products involving barium-144, krypton-89, and three neutrons maintains this balance, as the resulting total atomic number is $56 + 36 = 92$ and the total mass number is $144 + 89 + 3 = 236$.

Electron transitions between energy levels in a hydrogen atom emit photons with energy inversely proportional to the wavelength, a relationship governed by the Rydberg principle. The shortest wavelength of the Lyman series occurs when an electron falls from an infinite energy level down to $n=1$, effectively capturing the total ionization energy of 13.6 eV, which is described by the relation $\frac{hc}{\lambda_{Lyman}} = 13.6$.

The Balmer series exhibits its longest wavelength during the smallest energy drop possible for that series, which is the transition from $n=3$ down to $n=2$. This transition follows the relation $\frac{hc}{\lambda_{Balmer}} = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right)$. Simplifying the energy term results \in $\frac{hc}{\lambda_{Balmer}} = 13.6 \times \frac{5}{36}$. By substituting the energy constant from the Lyman limit into this expression, the calculation for the wavelength becomes $\lambda_{Balmer} = \frac{915 \times 36}{5}$, which results in 6588 Å.

The energy released during nuclear fission arises from the principle of mass-energy equivalence, where a portion of the system's total mass is converted directly into energy. Specifically, the disintegration energy $Q$ corresponds to the difference between the combined mass of the initial reactants and the combined mass of the final products. When the mass of the starting nucleus is greater than the sum of the masses of the resulting particles, this discrepancy, known as the mass defect, is released as energy based on the relationship
$Q = (m_{initial} - m_{final}) c^2$

In this specific fission process, the reactant is a ${}^{235}\mathrm{U}$ nucleus with a mass of 235.0439 u. On the product side, the total mass is the sum of the ${}^{140}\mathrm{Ce}$ nucleus at 139.9054 u, the ${}^{94}\mathrm{Zr}$ nucleus at 93.9063 u, and the neutron at 1.0086 u, which equals 234.8203 u. Subtracting the product mass from the reactant mass yields a mass defect of 0.2236 u. Multiplying this mass difference by the energy equivalent 931 MeV/u gives the total disintegration energy, resulting in a value of 208.1716 MeV. Rounding this value provides the final energy release of approximately 208 MeV.

Rutherford's atomic model established the foundation of modern physics by proposing that an atom's total mass and positive charge are concentrated within a tiny, dense central nucleus, while electrons orbit around it, confirming that the first statement is accurate. The description of an atom as a spherical cloud of positive charge with electrons embedded inside corresponds to J.J. Thomson's atomic model, which is fundamentally different from Rutherford's, rendering the second statement false.

The Rydberg formula $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ dictates the wavelength of spectral lines in a hydrogen atom, where $Z=1$. The shortest wavelength in any series corresponds to the maximum energy transition, occurring when an electron drops from an infinite energy level ($n_2 = \infty$) to the destination orbit $n_1$.

For the Lyman series, the electron transitions to $n_1 = 1$, resulting in a wave number of $\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} \right) = R$, which implies $\lambda_L = \frac{1}{R}$. For the Balmer series, the electron transitions to $n_1 = 2$, yielding a wave number of $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} \right) = \frac{R}{4}$, which implies $\lambda_B = \frac{4}{R}$. Calculating the ratio of the Balmer wavelength to the Lyman wavelength gives $\frac{4/R}{1/R}$, resulting in $4:1$.

In the Bohr model of the atom, the radius of an electron's orbit is directly proportional to the square of the principal quantum number $n$ and inversely proportional to the atomic number $Z$, represented as $r \propto \frac{n^2}{Z}$. Since the atomic number remains constant for a single atom, the relationship between the radii of the fourth and third orbits is determined by the ratio of their squared quantum numbers, $\frac{r_4}{r_3} = \frac{4^2}{3^2}$. By substituting the given radius of the third orbit $R$ into this ratio, the radius of the fourth orbit is calculated as $\frac{16}{9} R$.

In hydrogen-like ions, the frequency of emitted radiation during an electronic transition is proportional to the difference between the inverse squares of the principal quantum numbers, defined by $\nu \propto \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$, where the proportionality constant is the same for a specific ion. This relationship allows for the direct comparison of different transitions by analyzing the geometric factors associated with each energy level jump.

For the transition from $n=2$ to $n=1$, the frequency depends on the factor $\left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \left( 1 - \frac{1}{4} \right) = \frac{3}{4}$, and this value corresponds to a frequency of $3 \times 10^{15} \text{Hz}$. The transition from $n=3$ to $n=1$ involves a frequency proportional to $\left( \frac{1}{1^2} - \frac{1}{3^2} \right) = \left( 1 - \frac{1}{9} \right) = \frac{8}{9}$. Taking the ratio of these proportionality factors gives $\frac{8/9}{3/4} = \frac{8}{9} \times \frac{4}{3} = \frac{32}{27}$. Multiplying this ratio by the initial frequency yields $\frac{32}{27} \times 3 \times 10^{15} \text{Hz} = \frac{32}{9} \times 10^{15} \text{Hz}$, which indicates that $x=32$.

The radius of an atomic nucleus is related to its mass number by the expression $R = R_0 A^{1/3}$, where $R_0$ is a constant. Because this value is consistent for all nuclei, the ratio of the radii for two nuclei corresponds to the cube root of the ratio of their mass numbers, which is represented by:
$\frac{R_1}{R_2} = \left( \frac{A_1}{A_2} \right)^{1/3}$
Given that the radius of the first nucleus is half that of the second, which has a mass number of 192, we substitute $1/2 = (A_1 / 192)^{1/3}$ into the equation. Cubing both sides produces $1/8 = A_1 / 192$, which identifies the mass number as $A_1 = 24$.

The radius of an electron's orbit in a hydrogen atom is determined by the relationship $r = a_0 n^2$, where $n$ is the principal quantum number and $a_0$ is the Bohr radius. Substituting the given radius and the Bohr constant into this formula gives $n^2 = \frac{8.48}{0.529}$, which simplifies to $n^2 = 16$. The energy of an electron in any orbit is related to the ground state energy $E$ by the inverse square relationship $E_n = \frac{E}{n^2}$. By substituting the value of $n^2$ into this expression, we obtain $E_n = \frac{E}{16}$. Since the problem defines this energy as $\frac{E}{x}$, the value of $x = 16$.