Atoms And Nuclei – JEE Main PhysicsPractice Questions & PYQs

The energy required to excite an electron from the ground state to the first excited state in a hydrogen atom is exactly 10.2 eV. Once the electron is raised to the $n=2$ level, it must eventually return to its original ground state. As the electron falls back to the $n=1$ level, it undergoes only one possible transition, which results in the emission of exactly one spectral line.

Removing a neutron from a nucleus requires adding energy equivalent to the difference in rest mass between the parent nucleus and the sum of the daughter nucleus and the isolated neutron. This mass defect is determined by the reaction balance, calculated as $\Delta m = (12.000000 + 1.008665) - 13.003354 = 0.00531$ u. The energy requirement is found by multiplying this mass difference by the conversion factor 931.5 MeV per atomic mass unit. The final calculation $0.00531 \times 931.5$ gives a total energy of 4.95 MeV.

The Lyman series arises from electronic transitions where the final energy level is the ground state, $n_1 = 1$. Applying Rydberg's formula, the inverse wavelength of any transition is proportional to the difference $\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. For the first member of the series, the transition occurs from $n_2 = 2$ to $n_1 = 1$, making its inverse wavelength proportional to $\left( 1 - \frac{1}{4} \right)$, which equals $\frac{3}{4}$.

The second member involves a transition from $n_2 = 3$ to $n_1 = 1$, which results in an inverse wavelength proportional to $\left( 1 - \frac{1}{9} \right)$, or $\frac{8}{9}$. By taking the ratio of the inverse wavelengths, we find $\frac{1/\lambda'}{1/\lambda} = \frac{8/9}{3/4}$, which simplifies to $\frac{32}{27}$. Inverting this ratio to solve for the wavelength of the second member, $\lambda'$, yields $\lambda' = \frac{27}{32} \lambda$.

Determining the comparative energy released by hydrogen fusion and uranium fission involves calculating the total output for a specific mass based on the number of individual reaction events. Since the fusion of four hydrogen nuclei releases 26.7 MeV, the total energy $E_H$ for 2 kg of hydrogen is proportional to $2000 N_A \times \frac{26.7}{4}$. For 2 kg of uranium, the number of nuclei is determined by $\frac{2000}{235}$ moles, and with each fission event yielding 200 MeV, the total energy $E_U$ is proportional to $\frac{2000}{235} N_A \times 200$. Dividing $E_H$ by $E_U$ cancels the common mass and Avogadro constant terms to produce $\frac{E_H}{E_U} = \frac{26.7 \times 235}{4 \times 200}$, which results in a value of approximately 7.84, identifying 7.62 as the intended estimate.

Radioactive decay functions on the principle that the amount of a substance reduces to half of its previous quantity with every passing interval $T$. When $\frac{7}{8}$ of the original mass has disintegrated, the fraction remaining is calculated as $1 - \frac{7}{8} = \frac{1}{8}$. Given that the amount remaining after $n$ half-lives follows the expression $\frac{N}{N_0} = (\frac{1}{2})^n$, we equate $(\frac{1}{2})^n$ to $\frac{1}{8}$. Since $2^3 = 8$, the value of $n$ is 3, meaning three half-life periods have passed to reach this level of decay, resulting in a total time of $3T$.

The radius of an electron in Bohr's atomic model is directly proportional to the square of the principal quantum number $n$, represented by the relation $r \propto n^2$. Since we are comparing the third orbit to the second orbit where the radius is $R$, we can set up the ratio $\frac{r_{3}}{R} = (\frac{3}{2})^2$. Squaring the terms inside the parentheses yields $\frac{9}{4}$, which simplifies to 2.25. Multiplying this factor by $R$ confirms that the radius of the third orbit is $2.25R$.

Radioactive decay follows conservation laws where alpha decay reduces the mass number by 4 and atomic number by 2, beta-minus decay increases the atomic number by 1 with mass constant, beta-plus decay decreases the atomic number by 1 with mass constant, and gamma decay leaves both values unchanged. Starting with ${}_{84}^{218} A$, the initial alpha emission results in ${}_{82}^{214} A_1$, followed by a beta-minus decay that produces ${}_{83}^{214} A_2$, while the subsequent gamma emission leaves the nucleus as ${}_{83}^{214} A_3$. Applying a second alpha decay transitions the system to ${}_{81}^{210} A_4$, after which a beta-plus decay shifts the atomic number to 80 to form ${}_{80}^{210} A_5$. The final gamma emission maintains these values, resulting in a final product ${A_6}$ with a mass number of 210 and an atomic number of 80.

The spontaneous decay of a particle is governed by the conservation of mass-energy, which dictates that a particle can only decay if the rest mass of the parent particle exceeds the sum of the masses of the products. Because a free neutron has a larger rest mass than a proton, as represented by $m_{neutron} > m_{proton}$, the decay process is energetically favorable and releases energy. Conversely, since a free proton has a smaller rest mass than a neutron, it cannot undergo this transition spontaneously because such a process would require an input of external energy that is not available to an isolated system.

Total binding energy is determined by multiplying the binding energy per nucleon by the mass number of the nucleus. Since nucleus A contains 17 protons and an equal number of neutrons, its mass number is 34, which results in a total binding energy of $1.2 \times 34 = 40.8\ \mathrm{MeV}$.

Nucleus B has a given mass number of 26 and a binding energy per nucleon of $1.8\ \mathrm{MeV}$, so its total binding energy is $1.8 \times 26 = 46.8\ \mathrm{MeV}$. Calculating the difference between the total binding energies of nucleus B and nucleus A yields $46.8 - 40.8 = 6\ \mathrm{MeV}$.

To find the ionization energy of a hydrogen-like ion, we apply the relation $E = 13.6 \times \frac{Z^2}{n^2}$ eV. For $\mathrm{Li}^{2+}$, the atomic number is $Z=3$, and the second excited state corresponds to the principal quantum number $n=3$. Substituting these values leads to $13.6 \times \frac{3^2}{3^2}$ eV, which simplifies to 13.6 eV. When expressed in the form $x \times 10^{-1}$ eV, this calculation identifies $x$ as 136.

When a radioactive nucleus decays through multiple pathways simultaneously, the total decay rate is the summation of the individual decay rates, implying that the effective decay constant is simply the sum of the individual decay constants, or $\lambda_{eff} = \lambda_1 + \lambda_2$. Because each decay constant corresponds to its half-life by the relationship $\lambda = \frac{\ln 2}{t_{1/2}}$, we can equate the reciprocals of the half-lives such that $\frac{1}{t_{eff}} = \frac{1}{t_1} + \frac{1}{t_2}$. Converting the first half-life of 5 minutes into 300 seconds allows us to calculate the effective half-life using the equation $\frac{1}{t_{eff}} = \frac{1}{300} + \frac{1}{30}$. Solving this expression yields $\frac{1}{t_{eff}} = \frac{11}{300}$, which indicates the effective half-life is $\frac{300}{11}$ seconds, matching the required form of $\frac{\alpha}{11}$ and resulting in an $\alpha$ value of 300.

When an electron transitions from a higher energy level to a lower one, it emits a photon with an energy corresponding to the difference between the two states. For a transition from $n=4$ to $n=1$ in a hydrogen atom, the energy difference is calculated using the expression $\Delta E = 13.6 (1 - 1/16)$ eV, which equals 12.75 eV. To find the wavelength, we apply the relation $\lambda = hc / \Delta E$ with the provided constant $h = 4 \times 10^{-15}$ eVs and the speed of light $c = 3 \times 10^8$ m/s. Substituting these values into the equation gives $\lambda = (4 \times 10^{-15} \times 3 \times 10^8) / 12.75$ m, which results in 94.1 nm.

The energy released during an electronic transition in a hydrogen-like atom is determined by the Rydberg formula adjusted for the atomic number, represented as $\Delta E = 13.6 Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$ eV. With the atomic number given as $Z = 4$, the transition occurs from the fourth energy state where $n_2 = 4$ to the second energy state where $n_1 = 2$. Substituting these values into the expression yields $\Delta E = 13.6 \times 4^2 \times (\frac{1}{2^2} - \frac{1}{4^2})$ eV, which simplifies to $13.6 \times 16 \times (\frac{1}{4} - \frac{1}{16})$ eV. Evaluating the terms within the parentheses results in $\frac{3}{16}$, so the expression becomes $13.6 \times 16 \times \frac{3}{16}$ eV, which further simplifies to $13.6 \times 3$ eV and results in a total energy release of 40.8 eV.

The wavelength of light emitted during an electron transition in a hydrogen atom is governed by the Rydberg formula, which relates the inverse wavelength to the difference of the inverse squares of the energy levels involved, expressed as $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. Since the hydrogen atom has an atomic number of 1, the formula simplifies to $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.

For the first transition from the third shell to the first, we calculate the inverse wavelength as $\frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = \frac{8}{9} R$. Applying the same logic to the second transition from the second shell to the first, we obtain $\frac{1}{\lambda_2} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3}{4} R$.

Dividing these two expressions allows us to determine the ratio $\frac{\lambda_1}{\lambda_2}$ by calculating $\left( \frac{9}{8R} \right) \times \left( \frac{3R}{4} \right)$, which results in $\frac{27}{32}$. Equating this result to the provided expression $\frac{x}{32}$ confirms that the value of $x$ is 27.

The stability of a nucleus is dictated by its total binding energy, and nuclear fission results in a net release of energy when the product nuclei are more tightly bound than the original. Comparing the total binding energy of the parent nucleus against the combined binding energy of the resulting fragments reveals the net energy release. The parent nucleus has a mass number of 242 and a binding energy per nucleon of $7.6\ \mathrm{MeV}$, leading to an initial total binding energy of $242 \times 7.6\ \mathrm{MeV}$. Upon splitting into two fragments of mass number 121, each with a binding energy per nucleon of $8.1\ \mathrm{MeV}$, the final total binding energy becomes $2 \times 121 \times 8.1\ \mathrm{MeV}$. The difference between these two totals represents the energy gained during the process:

$\text{Gain} = 2(121 \times 8.1) - 242 \times 7.6 = 121\ \mathrm{MeV}$

The Rydberg formula, $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$, determines the wavelengths of light emitted during hydrogen atom transitions, where levels A, B, and C represent the excited states $n=2$, $n=3$, and $n=4$ respectively. Transitioning from B to A yields $\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{5R}{36}$, while the transition from C to B gives $\frac{1}{\lambda_2} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = \frac{7R}{144}$. Dividing these expressions results in the ratio $\frac{\lambda_1}{\lambda_2} = \frac{36}{5R} \cdot \frac{7R}{144} = \frac{7}{20}$. Comparing this result to the given expression $\frac{7}{4n}$ reveals that $4n = 20$, which confirms that $n = 5$.

The activity of a radioactive sample represents the rate of decay, calculated by the product of the decay constant and the total number of nuclei present, as given by $A = \lambda N$. To determine the number of nuclei, convert the mass $1.0 \ \mu\mathrm{g}$ to $1.0 \times 10^{-6}$ grams and divide by the atomic weight of 60 g mole^-1, then multiply by the Avogadro constant $6 \times 10^{23}$ to obtain $N = \left( \frac{1.0 \times 10^{-6}}{60} \right) \times 6 \times 10^{23} = 10^{16}$ atoms. Substituting these values into the decay equation yields $A = (1.5 \times 10^{-5}) \times 10^{16} = 1.5 \times 10^{11}$ Bq. Adjusting this result to the required power of ten gives the final activity of $15 \times 10^{10}$ Bq.

The shortest wavelength in an atomic spectrum corresponds to an electronic transition from an infinite energy level to the series limit, and this wavelength is inversely proportional to the energy of the emitted photon. For the Lyman series, the transition to the first energy level results in the given minimum wavelength of 917 Å. Because the energy difference depends on the destination level through the relationship $\Delta E \propto \frac{1}{n^2}$, the transition to the second level in the Balmer series involves only one-fourth of the energy released in the Lyman series. Consequently, the limiting wavelength for the Balmer series is four times larger than that of the Lyman series, which is determined by the calculation $4 \times 917 = 3668$ Å.

The total energy liberated during nuclear fission is the product of the number of nuclei undergoing the process and the energy released by a single fission event. We begin by determining the number of moles in the sample by dividing the mass of 120 g by the atomic mass of 240 g/mol, which results in $n = \frac{120}{240} = 0.5$ mol. Using the given value of Avogadro's number, we find the number of atoms is $N = 0.5 \times 6 \times 10^{23} = 3 \times 10^{23}$. Multiplying this total number of atoms by the energy released per fission of 200 MeV yields the total energy as $E = 3 \times 10^{23} \times 200 = 6 \times 10^{25}$ MeV.

When high-energy electrons impact a metal target, they undergo rapid deceleration or interact with inner-shell atomic electrons, causing a significant loss of kinetic energy. This energy is released as electromagnetic radiation with very short wavelengths and high frequencies, which, according to the relationship $E = \frac{hc}{\lambda}$, corresponds to the X-ray portion of the spectrum. This phenomenon is precisely how X-rays are generated for medical and scientific use.