Class 11 Chemistry (Chemistry Part-I) Chapter 2 Structure of Atom

Key Concept: Atomic Orbitals

b) Atomic orbitals

[Solution Description] The wave functions derived from solving the Schrödinger equation for hydrogen or hydrogen-like species are referred to as atomic orbitals. They describe the regions in space where there is a high probability of finding an electron.

Key Concept: Shielding Effect

b) Inner electrons shield outer electrons from the full nuclear charge.

[Solution Description] The inner-shell electrons shield the outer electrons from the full positive charge of the nucleus. This shielding effect reduces the effective nuclear charge experienced by the outer electrons, which in turn impacts their energy levels. They feel a reduced attractive force towards the nucleus, thus lowering the effective nuclear charge $(Z_{eff})$.

Key Concept: Transition Identification

b) Emission

[Solution Description] When an electron drops from a higher orbit $(n=6)$ to a lower orbit $(n=3)$, energy is emitted as light, resulting in an emission spectrum.

Key Concept: Speed of Light

a) $3.0 \times 10^8$ m/s

[Solution Description] The speed of light in a vacuum is universally recognized as $3.0 \times 10^8$ meters per second. This is a fundamental constant in physics, often denoted by the symbol $c$.

Key Concept: Simple Application

c) 3p$^3$

[Solution Description] Following Hund’s Rule, three electrons in 3p orbitals will occupy each orbital singly with parallel spins before any of them pairs up. Therefore, the electron configuration would be $3p^3$, where each of the three 3p orbitals has one electron.

Key Concept: Angular Momentum

c) $3h/2\pi$

[Solution Description] According to Bohr’s model, the quantized angular momentum $L$ is given by: $L = \frac{nh}{2\pi}$

For $n = 3$: $L = \frac{3h}{2\pi}$

Therefore, the angular momentum of an electron in the n=3 orbit is $\frac{3h}{2\pi}$.

Key Concept: Bohr’s Model

c) Energy levels

[Solution Description]

In Bohr’s model, the electron orbits are quantized, meaning that the energy levels of electrons in an atom are discrete. Electrons can only exist in certain allowed orbits with fixed energies.

Key Concept: Measurement Precision

a) Because their mass makes quantum effects negligible

[Solution Description] For macroscopic objects, the uncertainties involved in measurements become negligible. Consider the equation $\Delta x \Delta p \geq \frac{h}{4\pi}$. The Planck’s constant $h$ is extremely small, so when applied to large-scale objects where $\Delta x$ and $\Delta p$ are relatively large themselves, the product $\Delta x \Delta p$ greatly exceeds $\frac{h}{4\pi}$. Hence, the limitations imposed by the uncertainty principle do not significantly affect the precision of measurements for macroscopic objects, allowing them to behave as predicted classically.

Key Concept: Proton and Neutron Calculation

c) 20

[Solution Description]

The mass number $A$ is the sum of protons and neutrons, while the atomic number $Z$ represents only the protons. Therefore, to find the number of neutrons:

$\text{Number of neutrons} = A – Z = 40 – 20 = 20$

Key Concept: Angular Momentum

b) $m_evr = n\frac{h}{2\pi}$

[Solution Description] Bohr postulated that the angular momentum of an electron is an integral multiple of $\frac{h}{2\pi}$, hence $m_evr = n\frac{h}{2\pi}$.

Key Concept: Transition Understanding

d) $n_i > n_f$

[Solution Description] In an emission spectrum, energy is released when an electron falls to a lower energy level, so $n_i > n_f$.

Key Concept: Implications on Atomic Models, Rutherford’s Calculations

b) By demonstrating the existence of a massive, dense nucleus.

[Solution Description]

The plum pudding model suggested a uniform distribution of positive charge with electrons embedded within it. However, Rutherford observed that most alpha particles passed through undeflected, indicating vast empty space, and some were deflected sharply, suggesting a concentrated central charge rather than a uniform spread.

Key Concept: Black Body Characteristics

c) Emits and absorbs the same amount of energy

[Solution Description] When a black body is in thermal equilibrium with its surroundings, it maintains the condition that the amount of energy it absorbs is equal to the amount of energy it emits. Therefore, it neither gains nor loses energy over time; the absorption and emission are balanced.

Key Concept: Number of Orbitals in Subshell

b) 5

[Solution Description] The number of orbitals in a subshell is given by the formula $2l + 1$. For $l = 2$, the calculation is as follows: $2(2) + 1 = 5$

Therefore, there are 5 orbitals in the subshell when $l = 2$.

Key Concept: Energy Transition, Spectrum Calculation

c) $6.03 \times 10^{14} \text{ Hz}$

[Solution Description] The energy difference between two levels is given by $\Delta E = E_2 – E_1$. Here, $E_2 = 1.0 \times 10^{-19} \text{ J}$ and $E_1 = 5.0 \times 10^{-19} \text{ J}$. The energy of the emitted photon can be calculated using Planck’s equation: $\Delta E = h \nu$

where $h$ is Planck’s constant $(6.626 \times 10^{-34} \text{ J s})$. Plugging in the values:

$\Delta E = 5.0 \times 10^{-19} – 1.0 \times 10^{-19} = 4.0 \times 10^{-19} \text{ J}$

Now solving for frequency $\nu$: $\nu = \frac{\Delta E}{h} = \frac{4.0 \times 10^{-19}}{6.626 \times 10^{-34}} = 6.03 \times 10^{14} \text{ Hz}$

Hence, the frequency of the emitted photon is approximately $6.03 \times 10^{14} \text{ Hz}$.

Key Concept: Charge to Mass Ratio

b) The type of gas used in the tube

[Solution Description] The charge to mass ratio $(e/m)$ for positive ions in a cathode ray tube depends primarily on the type of gas used in the tube because different gases have atoms with different masses and numbers of protons, affecting their respective $e/m$ values.

Key Concept: Atomic Models

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description] Thomson’s model, also known as the “plum pudding model,” depicted the atom as a sphere of positive charge with negatively charged electrons scattered within it, much like plums in a pudding. This configuration explained the overall neutrality of the atom because the positive and negative charges would balance each other out. Thus, both the assertion that Thomson’s model could explain atomic neutrality and the reason describing the structure of the model are true. Moreover, the reason is the correct explanation for why the model accounted for atomic neutrality.

Key Concept: Basic Concept

d) $10^{-10}$ m

[Solution Description] According to the Plum Pudding Model, the atom is a sphere with a radius of approximately $10^{-10}$ m.

Key Concept: Bohr Model Limitations

d) Zeeman effect

[Solution Description] The Zeeman effect, which involves the splitting of spectral lines under a magnetic field, cannot be explained by Bohr’s model.

Key Concept: Spectrum Identification

c) Between microwaves and visible light

[Solution Description] The infrared region lies between microwaves and visible light in the electromagnetic spectrum, typically associated with frequencies around $10^{12}$ Hz to $10^{14}$ Hz.

Key Concept: Complex Analysis, Real-world Application

c) It leads to a misunderstanding of stellar atmosphere compositions.

[Solution Description] Bohr’s model assumes fixed and circular orbits for electrons around the nucleus, neglecting complex interactions such as those occurring in external fields. In a magnetic field, however, electron paths are altered due to the Lorentz force, causing different energy levels (Zeeman effect). This results in the splitting of spectral lines, which can be observed in astrophysical environments like star atmospheres. Thus, without accounting for these effects, Bohr’s model fails to predict the accurate spectra encountered in such settings.

Key Concept: Zeeman and Stark Effects, Quantum Numbers

c) It can’t explain line splitting in magnetic fields

[Solution Description] The Zeeman effect shows the splitting of spectral lines into different components in the presence of a magnetic field. In multi-electron atoms, this effect cannot be explained by Bohr’s model because it lacks quantum numbers that describe angular momentum coupling and magnetic effects, which are essential to understand such phenomena.

Key Concept: Emission vs Absorption

a) It shows bright lines on a dark background

[Solution Description] An emission spectrum is produced when electrons in atoms or molecules fall from a higher energy level to a lower one, emitting energy in the form of light. This typically results in bright lines on a dark background.

Key Concept: Model Characteristics

b) Electrons orbit the nucleus in fixed paths with defined energy levels.

[Solution Description] In Bohr’s model, electrons orbit the nucleus in fixed paths with quantized energy levels. This model explains the spectral lines of hydrogen because when an electron transitions between these orbits, it absorbs or emits a photon with energy equal to the difference between the energy levels.

Key Concept: Quantization Concept

a) Discrete values at specific wavelengths

[Solution Description] A line spectrum consists of discrete lines, each representing specific wavelengths or frequencies unique to an element, whereas a continuous spectrum contains all possible wavelengths without interruptions.

Key Concept: Limitations of Bohr’s Model

b) Zeeman effect

[Solution Description] Bohr’s Model could not explain certain atomic phenomena such as the Zeeman effect, which involves the splitting of spectral lines in magnetic fields.

Key Concept: Real-world Application, Complex Charge Analysis

d) Hydrogen

[Solution Description] The charge-to-mass ratio $(\frac{q}{m})$ of positive rays depends on the nature of the gas used. Hydrogen, being the lightest element, has protons as its positive ions, which are the lightest and smallest possible positive ions, giving it the highest charge-to-mass ratio among all elements. Hence, using hydrogen will result in positive rays with the highest $\frac{q}{m}$.

Key Concept: Real-World Application

c) $^{14}C$

[Solution Description] Radiocarbon dating utilizes the isotope carbon-14 due to its radioactive nature and ability to incorporate into living organisms as part of their biological processes. Over time, $^{14}C$ decays and can provide a timeline for dating organic materials.

Key Concept: Real-World Application, Effective Nuclear Charge

c) Nitrogen

[Solution Description] The effective nuclear charge $(Z_{eff})$ is calculated using the equation $Z_{eff} = Z – S$, where $Z$ is the atomic number and $S$ is the shielding constant. For the 2p orbitals of these elements, we must consider the core electrons’ contribution to shielding. As we move from B to N, the number of protons increases, increasing $Z$. Meanwhile, the increase in electron repulsion due to additional electrons slightly increases $S$. Overall, $Z_{eff}$ increases across a period because the effect of added protons outweighs the additional shielding by extra electrons. Thus, nitrogen has the highest $Z_{eff}$ for its 2p orbital.

Key Concept: Subshell Identification

b) 4p

[Solution Description] The azimuthal quantum number $l = 1$ corresponds to the $p$ subshell. Thus, the subshell notation for $n = 4$ and $l = 1$ is $4p$.

Key Concept: Historical Context

c) Louis de Broglie

[Solution Description] Louis de Broglie proposed the theory of wave-particle duality, suggesting that particles such as electrons exhibit both wave and particle characteristics. This idea was fundamental in the development of quantum mechanics, where the Schrödinger equation incorporates these principles.

Key Concept: Planck’s Quantum Theory

c) Quanta

[Solution Description] Max Planck proposed that energy is emitted or absorbed in discrete amounts known as quanta or photons, rather than continuously.

Key Concept: Pauli Exclusion Principle

a) $1s$, $2s$, $2p$, $3s$, $3p$, $4s$, $3d$

[Solution Description] Calcium has 20 electrons. According to the Aufbau principle, they fill the orbitals in increasing energy: $1s^2$, $2s^2$, $2p^6$, $3s^2$, $3p^6$, $4s^2$. The progression shows that the $4s$ orbital fills before the $3d$ orbital, confirming the pattern.

Key Concept: Quantum Number Representation, Advanced Configuration Patterns

b) If both $2s$ electrons had identical spins.

[Solution Description] Nitrogen (N) with atomic number 7 has the electronic configuration $1s^2\, 2s^2\, 2p^3$. The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers. For nitrogen, in each orbital (like $1s$, $2s$, and three $2p$ orbitals), electrons must differ at least by their spin quantum number. Thus, the scenario where electrons could potentially have the same set of quantum numbers is if we incorrectly assume $2s$ electrons fill the same orbital ignoring different spins.

Key Concept: Order of Filling

b) 3d

[Solution Description] According to the aufbau principle, orbitals are filled in order of increasing energy levels. The sequence provided by the syllabus is: 1s, 2s, 2p, 3s, 3p, 4s, then 3d follows 4s.

Key Concept: Deflection Observation

c) They passed through undeflected

[Solution Description] In Rutherford’s gold foil experiment, it was observed that most $\alpha$-particles passed through the foil undeflected because most of the atom is empty space, allowing them to pass through without any interaction.

Key Concept: Nuclear Model Implications, Nucleus Size Comparison

a) 100 meters

[Solution Description] According to Rutherford’s nuclear model, the nucleus is very small compared to the entire atom. Typically, the radius of an atom is about 10,000 times that of its nucleus. Thus, if the nucleus is expanded to the size of a marble (1 cm), then the whole atom would be about 100 meters across, illustrating the vast empty space within atoms where electrons orbit.

Key Concept: Wavenumber Calculation

b) $33.33 \, \text{m}^{-1}$

[Solution Description]

Wavenumber $(k)$ is calculated using the formula: $k = \frac{1}{\lambda}$. First, convert the wavelength from centimeters to meters: $3 \, \text{cm} = 0.03 \, \text{m}$. Thus:

$k = \frac{1}{0.03 \, \text{m}} = 33.33 \, \text{m}^{-1}$

Key Concept: Electromagnetic Spectrum Analysis, Dual Nature of Light

d) X-rays

[Solution Description] X-rays are used in medical diagnostics because they can penetrate soft tissue while being absorbed by denser materials like bones. This property allows for detailed images, such as those obtained in radiography, showcasing wave-particle duality as X-rays behave both as waves and particles.

Key Concept: Cause and Effect

b) Ignores the impact of external magnetic fields

[Solution Description] The Zeeman effect involves the splitting of spectral lines when an external magnetic field is applied. Bohr’s model assumes that electrons move in fixed orbits and does not consider the external magnetic field effects on these orbits, hence failing to explain this phenomenon.

Key Concept: Rydberg Constant

d) 109,677 $cm^{-1}$

[Solution Description] The Rydberg constant for hydrogen is specified as 109,677 $cm^{-1}$ according to the syllabus.

Key Concept: Formula Application

b) $2.2087 \times 10^{-12}$ m

[Solution Description] Using de Broglie’s equation: $\lambda = \frac{h}{mv}$. Substituting the given values, $\lambda = \frac{6.626 \times 10^{-34}}{1 \times 10^{-30} \cdot 3 \times 10^8} = \frac{6.626 \times 10^{-34}}{3 \times 10^{-22}} = 2.2087 \times 10^{-12}$ m.

Key Concept: Angular Momentum Quantization, Energy Level Calculations

a) 6.17 $\times 10^{14}$ Hz

[Solution Description] Start by calculating the energy difference between the two levels using Bohr’s model formula for energy levels: $E_n = -R_H/n^2$. The Rydberg constant $R_H = 2.18 \times 10^{-18} \, \text{J}$.

For $n=4$, the energy is:  $E_4 = -\frac{2.18 \times 10^{-18}}{4^2} = -1.3625 \times 10^{-19} \, \text{J}$

For $n=2$, the energy is: $E_2 = -\frac{2.18 \times 10^{-18}}{2^2} = -5.45 \times 10^{-19} \, \text{J}$

The energy difference ($\Delta E$) is then:

$\Delta E = E_2 – E_4 = (-5.45 \times 10^{-19}) – (-1.3625 \times 10^{-19})$

$= -5.45 \times 10^{-19} + 1.3625 \times 10^{-19}$

$= -4.0875 \times 10^{-19} \, \text{J}$

The frequency can be calculated using Planck’s equation $\Delta E = h\nu$: $\nu = \frac{-4.0875 \times 10^{-19}}{6.626 \times 10^{-34}}$

$= 6.17 \times 10^{14} \, \text{Hz}$

Key Concept: Application of Concept

c) Electron Microscope

[Solution Description] Electron microscopes use the wave-like properties of electrons to achieve high magnification, which is not possible with traditional light microscopes.

Key Concept: Real-World Application, Quantum Mechanics Development

c) Zeeman Effect

[Solution Description] One significant limitation of Bohr’s model is its inability to address the Zeeman effect, which involves the splitting of spectral lines in the presence of a magnetic field. Schrödinger’s quantum mechanical model, using wave functions and quantum numbers, explains this phenomenon by considering the electron’s angular momentum and spin, allowing for the interaction with external fields.

Thus, while Bohr’s model fails to account for such complexities, Schrödinger’s approach successfully describes the observed spectral line splitting, showcasing quantum mechanics’ comprehensive applicability.

Key Concept: Basic Formula Recall

a) $r_n = n^2 a_0$

[Solution Description] The formula for the radius of the nth orbit in a hydrogen atom according to Bohr’s model is given by $r_n = n^2 a_0$.

Key Concept: Energy and Frequency Relationship

c) The energy triples.

[Solution Description] According to Planck’s theory, the energy of a quantum of electromagnetic radiation is given by $E = h\nu$. Tripling the frequency $\nu$ results in three times the original energy since energy is directly proportional to frequency.

Key Concept: Series Identification

c) Balmer series

[Solution Description] The Balmer series occurs when electrons transition from higher energy levels to the second energy level (n=2). It is responsible for the lines in the visible spectrum of hydrogen.

Key Concept: Dual Nature of Light

c) Interference

[Solution Description] The wave nature of light can be demonstrated through phenomena such as interference and diffraction. Interference occurs when two or more light waves superpose to form a resultant wave of greater, lower, or the same amplitude. It is one of the key pieces of evidence that supports the wave nature of light.

Key Concept: Basic Concept

a) Electrons are emitted instantly

[Solution Description] When light of sufficient frequency strikes a metal surface, electrons are immediately ejected from the surface due to the photoelectric effect. This is one of the fundamental observations described in the study of the photoelectric effect.

Key Concept: Charge Distribution

b) In the nucleus

[Solution Description] According to Rutherford’s model, most of the positive charge and mass of the atom is concentrated in a small central region called the nucleus. This was concluded from the observation that alpha particles were deflected at large angles, implying a dense positively charged center.

Key Concept: Energy Levels

b) $E_n = -\frac{R_H}{n^2}$

[Solution Description] The energy of a stationary state in a hydrogen atom is given by the expression $E_n = -\frac{R_H}{n^2}$, where $R_H$ is the Rydberg constant and $n$ is the principal quantum number.

Key Concept: Space in Atom

b) It is mostly empty

[Solution Description]

Rutherford concluded that atoms are mostly empty space since most alpha particles passed through the gold foil without any deflection.

Key Concept: Advanced Isobar Calculation, Isobaric Decay Processes

c) $^{82}_{36}Kr$

[Solution Description] Beta-minus decay increases the atomic number by one unit while keeping the mass number the same, as a neutron converts into a proton. Starting with bromine ($^{82}_{35}Br$):

Post-decay, the atomic number becomes:

$Z_{new} = Z_{original} + 1 = 35 + 1 = 36$

Hence, the new element with atomic number 36 is krypton. The isotopic notation remains unchanged for the mass number, so the new element is $^{82}_{36}Kr$.

Key Concept: Bohr’s Postulates

b) Energy is emitted

[Solution Description]

Bohr’s model postulates that electrons move in fixed orbits around the nucleus without radiating energy. When an electron transitions from a higher energy level (higher orbit) to a lower energy level (lower orbit), it emits energy in the form of electromagnetic radiation.

This emission of energy corresponds to the difference in energy between the two levels involved in the transition. The frequency of the emitted radiation is given by Bohr’s frequency rule:

$\Delta E = h\nu$

where $h$ is Planck’s constant and $\nu$ is the frequency of the emitted radiation.

Therefore, according to Bohr’s model, energy is emitted when an electron moves from a higher to a lower stationary state.

Key Concept: Quantum Number Set

b) $n = 2, l = 1, m_l = -2, m_s = +\frac{1}{2}$

[Solution Description]

To determine which set of quantum numbers is not possible, we need to understand that each value of $l$ depends on the value of $n$, and $m_l$ ranges from $-l$ to $+l$. Additionally, $m_s$ must be either $+\frac{1}{2}$ or $-\frac{1}{2}$. Evaluate each set:

– For option a: $n = 3, l = 1, m_l = +1, m_s = -\frac{1}{2}$, all values are valid.

– For option b: $n = 2, l = 1, m_l = -2, m_s = +\frac{1}{2}$, $m_l = -2$ is invalid as it exceeds the range $-l$ to $+l$ for $l = 1$.

– For option c: $n = 4, l = 2, m_l = 0, m_s = +\frac{1}{2}$, all values are valid.

– For option d: $n = 5, l = 0, m_l = 0, m_s = -\frac{1}{2}$, all values are valid.

Therefore, the set of quantum numbers in option b is not possible.

Key Concept: Basic Concept

c) Eugen Goldstein

[Solution Description] Canal rays were discovered by Eugen Goldstein. He identified these positively charged particles in a cathode ray tube, leading to the discovery of ions.

Key Concept: Basic Concept

c) The minimum frequency required to eject an electron

[Solution Description]

The threshold frequency, denoted as $\nu_0$, is the minimum frequency of incident light required to eject electrons from a metal surface in the photoelectric effect phenomenon.

Key Concept: Basic Concept

c) Interference

[Solution Description] Interference is a classic demonstration of the wave nature of light. It occurs when two waves superimpose and either amplify or cancel each other, a behavior typical of waves.

Key Concept: Quantum Mechanics

a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.

[Solution Description]

Quantum mechanics is indeed essential for understanding the behavior of subatomic particles such as electrons. Its principles, including wave-particle duality and quantization, provide insights into atomic structure that classical mechanics cannot account for. While classical mechanics is successful at explaining macroscopic object motions, it cannot accurately describe phenomena at the quantum level due to its deterministic nature. Both statements are true, and the reasoning accurately explains why quantum mechanics is necessary for electron behavior analysis.

Key Concept: Stability Reasoning

d) Larger coulombic repulsion

[Solution Description] The stability of half-filled subshells is attributed to symmetrical distribution, maximized exchange energy, and decreased coulombic repulsion. Larger coulombic repulsion, on the other hand, contributes negatively to stability.

Key Concept: Model Features

b) The mass is spread uniformly over the atom.

[Solution Description] In Thomson’s model, the atom is considered to have a uniform distribution of positive charge with electrons embedded within it. The mass of the atom is also assumed to be uniformly distributed throughout the entire volume of the atom.

Key Concept: Advanced Symbol Interpretation, Isotope Stability

a) $^{204}_{82}Pb$

[Solution Description] The stability of isotopes often depends on their neutron-to-proton (n/p) ratio. Calculate n/p for both isotopes:

For $^{204}_{82}Pb$:

$n = A – Z = 204 – 82 = 122$

Neutron-to-proton ratio $n/p = \frac{122}{82} \approx 1.49$

For $^{207}_{82}Pb$:

$n = 207 – 82 = 125$

Neutron-to-proton ratio $n/p = \frac{125}{82} \approx 1.52$

Between similar isotopes, those with an n/p ratio closer to 1 are generally more stable. Thus, $^{204}_{82}Pb$ is slightly more stable due to its lower n/p ratio.

Key Concept: Concept Identification

c) $^{40}_{18}Ar$ and $^{42}_{20}Ca$

[Solution Description] Calculate the number of neutrons in each atom:

For $^{40}_{18}Ar$: $n_1 = 40 – 18 = 22$

For $^{42}_{20}Ca$: $n_2 = 42 – 20 = 22$

Since both atoms have 22 neutrons, they are isotones.

Key Concept: Boundary Surface

d) 2pz

[Solution Description] p orbitals have two lobes which are oriented along the x, y, or z axes. Hence, any p orbital like 2px, 2py, or 2pz will have such a diagram.

Key Concept: Comparative Analysis

d) Photoelectric effect

[Solution Description] The particle nature of light is primarily observed in phenomena where individual photons interact with matter, such as the photoelectric effect.

Key Concept: Wavenumber Calculation

a) $1.03 \times 10^{7} \text{ m}^{-1}$

[Solution Description]

To find the wavenumber, we first calculate the energy change $(\Delta E)$:

$\Delta E = -2.18 \times 10^{-18} \left( \frac{1}{1^2} – \frac{1}{4^2} \right)$

$= -2.18 \times 10^{-18} \left( 1 – \frac{1}{16} \right)$

$= -2.18 \times 10^{-18} \times \frac{15}{16}$

$= -2.04375 \times 10^{-18} \text{ J}.$

Using $\Delta E = hc\tilde{\nu}$, solve for $\tilde{\nu}$:

$\tilde{\nu} = \frac{-2.04375 \times 10^{-18}}{6.626 \times 10^{-34} \times 3.00 \times 10^8}$

$\approx 1.03 \times 10^7 \text{ m}^{-1}.$

Key Concept: Synthesis and Evaluation

c) Schrödinger wave equation

[Solution Description] Bohr’s model is limited to single-electron systems due to its reliance on classical orbits and quantized angular momentum without considering electron-electron interactions. The Schrödinger equation provides a more robust framework for multi-electron systems. It treats electrons as wavefunctions, allowing for the consideration of electron cloud distributions and interactions between electrons, leading to accurate spectral predictions.

Key Concept: Limitations

b) It fails to explain the Zeeman effect.

[Solution Description] Bohr’s model has certain limitations, including its inability to account for the Stark effect and the Zeeman effect, and it only accurately describes the spectrum of hydrogen-like atoms.

Key Concept: Discrete Energy Levels

b) $2hν$

[Solution Description] In Planck’s theory, the energy levels are quantized and can be given by multiples of $hν$: $E = 0, hν, 2hν, 3hν, \ldots, nhν$. Therefore, an energy level that matches one of these multiples is correct.

Key Concept: Basic Configuration

b) $1s^2$

[Solution Description] Helium has 2 electrons. According to the sa pbdc notation, these 2 electrons will fill the 1s subshell, which can hold a maximum of 2 electrons. Therefore, the electronic configuration for helium is $1s^2$.

Key Concept: Quantization Concept

b) Discrete energy levels for electrons

[Solution Description]

Quantization in atomic physics refers to the concept that electrons can only occupy certain discrete energy levels within an atom. These energy levels are quantized rather than continuous, meaning electrons cannot exist between these defined levels.

Key Concept: Energy Calculation

c) $-3.872 \times 10^{-18} , \text{J}$

[Solution Description]

The energy level of a hydrogen-like ion is given by the formula: $E = -\frac{Z^2}{n^2} \times 2.18 \times 10^{-18} , \text{J}$

Substituting $Z=4$, $n=3$: $E = -\frac{4^2}{3^2} \times 2.18 \times 10^{-18}$

Simplifying gives: $E = -\frac{16}{9} \times 2.18 \times 10^{-18}$

Calculating further: $E = -3.872 \times 10^{-18} , \text{J}$

Key Concept: Discovery Challenges, Neutron Applications

d) Detecting uncharged particles; enabling nuclear fission reactions in reactors

[Solution Description] The key challenge for Chadwick was detecting a particle that is electrically neutral because it does not leave a direct trace like charged particles do. This required careful observation of the effects of neutrons on other materials, such as causing protons to eject from paraffin wax. One significant real-world application of neutrons is their use in nuclear reactors, where they initiate fission reactions crucial for energy production.

Key Concept: Charge to Mass Ratio

b) Both Assertion and Reason are true, but Reason is NOT the correct explanation of Assertion.

[Solution Description] The assertion states that cathode rays are made up of electrons possessing a constant charge to mass $(\frac{e}{m})$ ratio, which is true because the properties of electrons, including this ratio, do not change. The reason given is that cathode rays can be deflected by electric and magnetic fields, which is also true. However, the ability of cathode rays to be deflected is due to their charge and velocity, not directly related to them having a constant $\frac{e}{m}$ ratio. Thus, both statements are true, but the reason does not explain the assertion.

Key Concept: Advanced Calculation, Conceptual Understanding

b) $3.41 \times 10^{-5} \, \text{m}$

[Solution Description] To find the radius of the path, we use the formula for radius in a magnetic field: $r = \frac{mv}{qB},$ where $m$ is the mass of the electron, $v$ is the velocity, $q$ is the charge, and $B$ is the magnetic field strength. Substituting the given values: $r = \frac{(9.1094 \times 10^{-31} \, \text{kg})(3 \times 10^6 \, \text{m/s})}{(1.602176 \times 10^{-19} \, \text{C})(0.5 \, \text{T})}.$ Calculating gives $r = \frac{2.73282 \times 10^{-24}}{8.01088 \times 10^{-20}} \approx 3.41 \times 10^{-5} \, \text{m}.$

Key Concept: Photon Interaction

c) The electron’s velocity changes unpredictably.

[Solution Description] When high-momentum photons are used to observe an electron, they impart significant energy and momentum to the electron due to their own high momentum. This interaction can cause a noticeable change in the electron’s velocity, making accurate measurement difficult. Therefore, using high-momentum photons increases the uncertainty in the electron’s velocity.

Key Concept: Mass Number

b) 31

[Solution Description] The mass number $(A)$ is calculated by adding the number of protons and neutrons in the nucleus. For this atom, it is $15 + 16 = 31$.

Key Concept: Energy Transition

b) Energy is emitted

[Solution Description] According to Bohr’s model, when an electron transitions from a higher energy level to a lower one, energy is emitted in the form of radiation.