Class 12 Physics Chapter 12 Atoms

Key Concept: Simple Application

a) -13.6 eV
[Solution Description] The energy levels of a hydrogen atom are quantized, and the energy of the electron at each level  n  is given by the formula:
$$E_n = -\frac{13.6 \text{ eV}}{n^2}$$
For the first energy level $$( n = 1 ),$$
$$E_1 = -\frac{13.6 \text{ eV}}{1^2} = -13.6 \text{ eV}.$$
Thus, the total energy of the electron in the ground state is $$-13.6 \text{ eV}$$.

Key Concept: Conceptual Understanding

b) Due to standing wave formation
[Solution Description]
De Broglie’s hypothesis states that electrons exhibit wave-like properties, forming standing waves around the nucleus. Only certain wavelengths (and thus orbits) are allowed because they form standing waves with integer multiples, leading to stable orbits.

Key Concept: Electron Interactions

b) It doesn’t account for electron-electron interactions.
[Solution Description]
Bohr’s model does not consider interactions between electrons in multi-electron atoms. Each electron experiences forces due to both the nucleus and other electrons, leading to electron-electron interactions. These interactions influence energy levels and spectral properties, which are not accounted for in Bohr’s model. This leads to inaccuracies when describing atoms with more than one electron.

Key Concept: Bohr’s Postulates

c) Angular momentum of electrons is quantized in integral multiples of $$\frac{h}{2\pi}$$.

[Solution Description] According to Bohr’s model, electrons move in fixed orbits around the nucleus. One of the key postulates states that the angular momentum of an electron is quantized and can be expressed by $$L = n\hbar$$, where n is the principal quantum number and $$\hbar$$ is the reduced Planck’s constant.

Key Concept: Nucleus Concentration

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

[Solution Description] In Rutherford’s model, the nucleus is dense with positive charge, leading to strong electrostatic interactions. Alpha particles, which are positively charged, experience significant deflection when they encounter this concentrated region due to repulsion. Hence, both the assertion and reason are true, and the reason correctly explains why alpha particles are deflected.

Key Concept: Electron Placement

b) Embedded within a positive charge cloud.
[Solution Description] In Thomson’s model, electrons are embedded within a spherical cloud of positive charge.

Key Concept: Bohr Model Basics

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

[Solution Description]

In Bohr’s model, one of the key postulates is that the energy levels of electrons are quantized, meaning electrons can only occupy certain allowed energy levels. This is true as it explains why atoms emit or absorb photons of discrete energies, corresponding to transitions between these levels.

The reason provided is also a fundamental aspect of Bohr’s model, which states that the angular momentum of an electron in its orbit is quantized and is expressed as $$L = n\hbar$$. This helps explain the stability of electrons in their orbits without radiating energy, aligning with the quantization concept.

Both assertion and reason statements are correct, and the reason provides a suitable explanation for the quantized nature of energy levels in the Bohr model.

Key Concept: Emission vs Absorption

d) Photons are absorbed as electrons transition to higher energy levels, creating dark lines.
[Solution Description] An absorption spectrum consists of dark lines on a continuous spectrum. These dark lines are produced when electrons in atoms absorb photons and transition to higher energy states, removing specific wavelengths from the spectrum.

Key Concept: Energy Transition Calculation

b) 10.2 eV
[Solution Description]
The energy difference between two levels in a hydrogen atom is given by the formula $$\Delta E = -13.6 \left(\frac{1}{{n_1}^2} – \frac{1}{{n_2}^2}\right)$$ eV, where $$n_1$$ and $$n_2$$ are the principal quantum numbers of the initial and final states, respectively.

Substituting $$n_1 = 1$$ and $$n_2 = 2$$, we have:

$$\Delta E = -13.6 \left(\frac{1}{1^2} – \frac{1}{2^2}\right) = -13.6 \left(1 – \frac{1}{4}\right)$$

$$\Delta E = -13.6 \times \frac{3}{4} = -10.2 \text{ eV}$$

Since the energy is absorbed for an excitation process, it is a positive value. Therefore, the energy required is 10.2 eV.

Key Concept: Mathematical Analysis, Critical Thinking

c) 3.0 × 10–14 m
[Solution Description] Using conservation of energy, the initial kinetic energy of the alpha particle is converted into electrostatic potential energy at the point of closest approach:
$$\frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{r} = \text{Kinetic Energy},$$
where $$Z = 79$$ for gold. Solving for r:
$$r = \frac{1}{4\pi\epsilon_0} \frac{2(79)(1.6 \times 10^{-19})^2}{8 \times 10^6 \times 1.6 \times 10^{-13}} \approx 3.0 \times 10^{-14} \, \text{m}.$$

Key Concept: Rutherford’s Model Limitations

b) Instability of electron orbits causing radiation loss.
[Solution Description]
While Rutherford’s model could explain the existence of a nucleus and electron orbits, it failed to account for the stability of these orbits. According to classical physics, orbiting electrons would emit radiation and spiral into the nucleus, leading to atomic collapse. It also could not explain discrete spectral lines observed in atomic spectra. These limitations highlighted the need for quantum mechanics to describe atomic structure more accurately.

Key Concept: Standing Waves

a) $$2\pi r = n\lambda$$

[Solution Description] In Bohr’s model, electrons move in fixed orbits where their wave nature is described by standing waves. The condition for these standing waves is given by the relationship $$2\pi r = n\lambda$$. Here, $$\lambda$$ is the wavelength, r is the radius of the orbit, and n is an integer known as the principal quantum number.

Key Concept: Wave-Particle Duality

c) By creating standing waves
[Solution Description] De Broglie’s hypothesis posits that particles such as electrons have wave properties, and their wavelengths are given by:
$$\lambda = \frac{h}{mv}$$
In Bohr’s model, the stable orbits occur when the circumference of the orbit equals an integral multiple of the wavelength, i.e.,:
$$2\pi r = n\lambda$$
This results in standing waves around the nucleus, which explains the quantization condition.

Key Concept: Energy Levels

c) Energy levels are discrete
[Solution Description] Energy levels in an atom are discrete. Electrons can only exist at certain energy levels, not in between them. This discreteness leads to well-defined spectral lines.

Key Concept: Energy Transition

b) 0.66 eV
[Solution Description] To find the energy of the emitted photon, use Bohr’s third postulate: $$hn = E_i – E_f$$. The total energy of an electron in the nth stationary state is given by $$E_n = -13.6/n^2$$ eV.

First, calculate $$E_i$$ for $$n=4$$:
$$E_i = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \text{ eV}$$

Next, calculate $$E_f$$ for $$n=3$$:
$$E_f = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \text{ eV}$$

The energy of the emitted photon is:
$$hn = E_i – E_f = (-0.85) – (-1.51) = 0.66 \text{ eV}$$

Key Concept: Photon Emission

c) Photon
[Solution Description] When an electron transitions from a higher energy level to a lower energy level in an atom, it emits energy in the form of a photon. This process involves the release of light or electromagnetic radiation.

Key Concept: Energy Transition

b) A photon is emitted corresponding to the energy difference
[Solution Description] When an electron moves from a higher energy level to a lower one, the excess energy is emitted in the form of a photon. The energy of this photon corresponds directly to the difference between the two energy levels, resulting in emission lines at specific wavelengths.

Key Concept: Ionization Energy

b) 13.6 eV
[Solution Description]
Ionization energy is the energy needed to remove an electron from its orbit completely. In the ground state $$(n=1)$$, the total energy is:

$$E_1 = -\frac{13.6 \text{ eV}}{1^2} = -13.6 \text{ eV}$$

To ionize, the energy supplied must equal the magnitude of this negative energy to bring it to zero:

$$\text{Ionization Energy} = +13.6 \text{ eV}$$

Key Concept: Bohr’s Postulates

b) Second postulate
[Solution Description]
Bohr’s second postulate states that the electron revolves around the nucleus in certain stable orbits without radiating energy. The angular momentum of these stable orbits is quantized and given by $$L = nh/2\pi$$, where n is an integer. This leads to the quantization of angular momentum.

Key Concept: Impact Parameter

b) The perpendicular distance of the initial velocity vector from the nucleus center
[Solution Description]
In alpha-particle scattering experiments, the impact parameter is defined as the perpendicular distance from the center of the nucleus to the initial velocity vector of the incoming alpha particle. It gives information about how close the particle comes to the nucleus.

Thus, it doesn’t directly measure deflection angle nor kinetic energy but rather this perpendicular distance which has implications on scattering angles.

Key Concept: Bohr’s Postulates

b) The angular momentum of an electron orbiting the nucleus is quantized and given by $$L = n\hbar$$, where n is an integer.

[Solution Description] In Bohr’s model, one of the key postulates is that electrons revolve in specific stable orbits without emitting energy. This explains the stability of atoms despite being composed of charged particles.

Key Concept: Energy Conservation, Mathematical Derivation

a) $$3.52 \times 10^{-14} \text{ m}$$
[Solution Description]
To find the distance of closest approach d, we use the given formula:

$$d = \frac{2Ze^2}{4\pi\epsilon_0 K}$$

The given values are:
– $$Z = 79$$
– $$K = 5$$ MeV

Converting the kinetic energy into joules as $$1 \text{ MeV} = 1.602 \times 10^{-13} \text{ J}$$:

$$K = 5 \times 1.602 \times 10^{-13} = 8.01 \times 10^{-13} \text{ J}$$

We know that:

$$e = 1.602 \times 10^{-19} \text{ C}, \quad \epsilon_0 = 8.854 \times 10^{-12} \text{ F/m}$$

Substitute these values into the formula:

$$d = \frac{2 \times 79 \times (1.602 \times 10^{-19})^2}{4\pi \times 8.854 \times 10^{-12} \times 8.01 \times 10^{-13}}$$

Solving for d:

$$d = 3.52 \times 10^{-14} \text{ m}$$

Hence, the distance of closest approach is $$3.52 \times 10^{-14} \text{ m}$$.

Key Concept: Energy Levels

b) $$-3.4 \, \text{eV}$$

[Solution Description] The energy levels in Bohr’s model are given by the formula $$E_n = -13.6 \, \text{eV}/n^2$$. For $$n = 2$$, we substitute into the equation to get $$E_2 = -13.6 \, \text{eV}/(2^2) = -13.6 \, \text{eV}/4 = -3.4 \, \text{eV}$$.

Key Concept: Energy Levels

b) -3.4 eV
[Solution Description]
The energy level of a hydrogen atom can be calculated using the formula:
$$E_n = -\frac{13.6}{n^2} \text{ eV}$$
For the first excited state, $$ n=2 $$, so:
$$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV}$$
Therefore, the energy of the first excited state is -3.4 eV.

Key Concept: Frequency of Emitted Photon

b) $$2.33 \times 10^{14}$$ Hz
[Solution Description]
First, find the energies for $$ n=5 $$ and $$ n=3 $$:

$$E_5 = -\frac{13.6}{5^2} = -\frac{13.6}{25} = -0.544\, \text{eV}$$

$$E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51\, \text{eV}$$

The energy difference $$( \Delta E )$$ is:
$$\Delta E = E_3 – E_5 = -1.51 – (-0.544) = -1.51 + 0.544 = -0.966\, \text{eV}$$

Convert this energy into joules:
$$\Delta E = 0.966 \times 1.6 \times 10^{-19}\, \text{J} = 1.5456 \times 10^{-19}\, \text{J}$$

Now, use Planck’s relation to find the frequency:
$$v_{if} = \frac{\Delta E}{h} = \frac{1.5456 \times 10^{-19}}{6.63 \times 10^{-34}} = 2.33 \times 10^{14}\, \text{Hz}$$

Thus, the frequency of the emitted photon is $$ 2.33 \times 10^{14}\, \text{Hz} $$.

Key Concept: Alpha Particle Trajectory

c) The gold nucleus remains stationary.
[Solution Description]
During the experiment, it was assumed that the gold nucleus remains stationary when alpha particles approach it. This assumption simplifies calculations since the massive nucleus does not move significantly upon impact with much lighter alpha particles. Therefore, the trajectory of alpha particles changes due to their interaction with the stationary nucleus.

Key Concept: Comparative Analysis, Advanced Quantum Concepts

a) Inclusion of wavefunctions over fixed orbits.
[Solution Description] Quantum mechanics introduces the concept of wave functions and probability densities, which explain electron behavior more accurately than fixed circular orbits. It also considers electron-electron interactions and spin, allowing for the description of complex spectra, unlike Bohr’s model.

Key Concept: Momentum and Wavelength

d) $$\lambda = h/p$$
[Solution Description] According to de Broglie’s hypothesis, the wavelength $$\lambda$$ associated with a particle such as an electron is given by:

$$\lambda = \frac{h}{mv}$$

Here, h is Planck’s constant and mv represents the momentum of the electron, denoted as p. Therefore, the de Broglie wavelength in terms of the momentum p is:

$$\lambda = \frac{h}{p}$$

Thus, option A is correct.

Key Concept: Trajectory Analysis

b) Small deflection
[Solution Description] A large impact parameter means that the alpha-particle approaches the gold nucleus at a greater distance. Since the strength of the repulsive Coulomb force is inversely proportional to the square of the distance, a larger impact parameter results in a weaker force and thus only a small deflection of the alpha-particle. With a large impact parameter, the trajectory will experience minimal deviation from its initial path.

Key Concept: Scattering Angle

b) Small scattering angle
[Solution Description] A large impact parameter implies that the trajectory of the alpha-particle passes far from the nucleus, resulting in a small deflection or scattering angle. Thus, the scattering angle is small.

Key Concept: Transition Energy

c) 2.55 eV
[Solution Description]
To find the energy of the emitted photon, we use the formula for energy transition: $$h\nu = E_i – E_f$$. The energy of the electron in the nth orbit is given by $$E_n = -\frac{13.6 \, \text{eV}}{n^2}$$.

First, calculate $$E_i$$ for $$n=4$$:
$$E_i = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \, \text{eV}.$$

Next, calculate $$E_f$$ for $$n=2$$:
$$E_f = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, \text{eV}.$$

Now, find the energy difference:
$$h\nu = E_i – E_f = (-0.85) – (-3.4) = 2.55 \, \text{eV}.$$

Therefore, the energy of the emitted photon is 2.55 eV.

Key Concept: Bohr Radius Calculation

b) $$4.76 \times 10^{-10}$$ m
[Solution Description] The Bohr radius formula is given by $$r_n = \frac{n^2h^2}{4\pi^2m_ee^2}$$. We know that the Bohr radius for $$n=1$$ is approximately $$5.29 \times 10^{-11}$$ m. For $$n=3$$, the radius is:
$$r_3 = 3^2 \cdot r_1 = 9 \cdot 5.29 \times 10^{-11} = 47.61 \times 10^{-11}\text{ m}.$$
Therefore, $$r_3 \approx 4.76 \times 10^{-10}$$ m.

Key Concept: Photon Emission

b) Emits a photon
[Solution Description]
When an electron transitions from a higher energy state $$(n_i)$$ to a lower energy state $$(n_f)$$, it emits a photon with energy equal to the difference in energies of these states. This process is characterized by the emission of light or other electromagnetic radiation.

Key Concept: Intensity Variations

c) Assertion is true, but Reason is false.

[Solution Description] The assertion is correct as per Bohr’s postulate, which states that angular momentum $$(mvr)$$ is quantized and given by $$mvr = n(h/2\pi)$$, where n is an integer. However, this postulate is related to the discrete energy levels and does not explain intensity variations in spectral lines. Therefore, while the quantization of angular momentum is true, it does not address why some spectral lines are more intense than others. The reason provided is incorrect because it attributes the explanation of intensity variations to angular momentum quantization, which is not accurate.

Key Concept:  Model Limitations

d) It cannot explain the stability of atoms

[Solution Description]

Rutherford’s nuclear model successfully demonstrated that most of the mass and positive charge of an atom is concentrated in a small nucleus, which explains many observed phenomena like scattering patterns in experiments. However, one major limitation was its inability to account for the emission of light at discrete wavelengths by atoms. According to classical physics, electrons spiraling around the nucleus should lose energy continuously and eventually fall into the nucleus, but this does not occur in stable atoms. The model could not explain why atoms are stable or how they emit light at specific wavelengths rather than a continuous spectrum.

Key Concept: Alpha-Particle Scattering

c) That atoms have a dense central nucleus
[Solution Description] The large deflection of alpha particles during Rutherford’s gold foil experiment indicated that a small fraction of these particles encountered something dense and highly positively charged, leading to significant deflections. This led to the conclusion that most of the mass of an atom and its positive charge is concentrated in a small central region called the nucleus, while the rest of the atom is mostly empty space.

Key Concept: Energy Transition

b) 2.55 eV
[Solution Description]
The energy levels in a hydrogen atom are given by the formula:
$$E_n = -13.6 \, \text{eV} / n^2$$
For $$n=4$$, the energy is:
$$E_4 = -13.6 \, \text{eV} / 4^2 = -13.6 \, \text{eV} / 16 = -0.85 \, \text{eV}$$
For $$n=2$$, the energy is:
$$E_2 = -13.6 \, \text{eV} / 2^2 = -13.6 \, \text{eV} / 4 = -3.4 \, \text{eV}$$
The energy difference $$\Delta E$$ is:
$$\Delta E = E_2 – E_4 = (-3.4 \, \text{eV}) – (-0.85 \, \text{eV}) = -2.55 \, \text{eV}$$

Key Concept: Historical Comparison

c) The Nuclear Model describes a central positive nucleus
[Solution Description] Rutherford’s Nuclear Model introduced the concept of a central nucleus containing all the positive charge and most of the mass of the atom, whereas Thomson’s Plum Pudding Model depicted electrons embedded in a diffuse sphere of positive charge without a central nucleus. This key difference allowed the Nuclear Model to explain the scattering of alpha particles observed in experiments.

Key Concept: de Broglie’s Hypothesis

c) By introducing wave-particle duality, explaining quantised orbits through standing waves.
[Solution Description] De Broglie hypothesized that particles such as electrons exhibit wave-particle duality, meaning they have both particle and wave characteristics. For an electron in an atom, its wave nature leads to standing waves forming in certain orbits. These standing waves correspond to specific, quantised values of angular momentum, $$L_n = n\hbar$$, as postulated by Bohr. Thus, de Broglie’s wave concept provides a rationale for Bohr’s quantisation condition.

Key Concept: Quantum Number Calculation, Real World Application

a) $$2.86 \times 10^{74}$$
[Solution Description]
To solve this problem, we will use Bohr’s quantization condition: the angular momentum is quantized and given by $$mvr = n\hbar$$, where $$m$$ is the mass of the object (here considered to be equivalent to Earth’s mass for simplicity), v is the velocity, and r is the radius of the orbit. The reduced Planck’s constant $$\hbar$$ is approximately $$1.05 \times 10^{-34}\, \text{Js}$$.

Let’s assume the mass of the planet $$m = 6 \times 10^{24}$$ kg, similar to Earth’s mass.

$$mvr = n\hbar \implies n = \frac{mvr}{\hbar}$$
Substituting the known values:
$$n = \frac{(6 \times 10^{24}\, \text{kg}) \cdot (5 \times 10^{4}\, \text{m/s}) \cdot (1 \times 10^{12}\, \text{m})}{1.05 \times 10^{-34}\, \text{Js}}$$

Calculating this gives:
$$n = \frac{3 \times 10^{41}\, \text{kg m}^2/\text{s}}{1.05 \times 10^{-34}\, \text{Js}} = 2.86 \times 10^{74}$$

Thus, the quantum number characterizing the planet’s motion is approximately $$2.86 \times 10^{74}$$.

Key Concept: Wave-Particle Duality

d) By introducing the concept of wave-particle duality

[Solution Description] According to Bohr’s second postulate of quantisation, the angular momentum $$(L)$$ of an electron in an orbit is quantised and given by $$L = n\hbar$$, where $$\hbar = \frac{h}{2\pi}$$. De Broglie’s hypothesis introduces the wave nature of electrons with wavelength $$\lambda = \frac{h}{mv}$$. For standing waves, the circumference of the electron orbit must be an integral multiple of the electron’s wavelength: $$2\pi r_n = n\lambda$$. Substituting $$\lambda = \frac{h}{mv}$$ into this condition gives $$2\pi r_n = n\frac{h}{mv}$$, which simplifies to $$mvr_n = n\hbar$$, matching Bohr’s second postulate.

Key Concept:  Trajectory Calculation, Real-World Application

a) $$2.84 \times 10^{-14}$$ m

[Solution Description]

The distance of closest approach d can be found using the formula:

$$d = \frac{2Ze^2}{4\pi\epsilon_0 K}.$$

Substituting the given values:

– $$Z = 79$$

– $$e = 1.6 \times 10^{-19} \, C$$

– $$K = 8 \, MeV = 8 \times 1.6 \times 10^{-13} \, J = 12.8 \times 10^{-13} \, J$$

– $$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, Nm^2/C^2$$

Calculating d:

$$d = \frac{2 \cdot 79 \cdot (1.6 \times 10^{-19})^2 \cdot 9 \times 10^9}{12.8 \times 10^{-13}}.$$

$$d = \frac{2 \cdot 79 \cdot 2.56 \times 10^{-38} \cdot 9 \times 10^9}{12.8 \times 10^{-13}}.$$

$$d = \frac{364.32 \times 10^{-29}}{12.8 \times 10^{-13}}.$$

$$d = \frac{364.32}{12.8} \times 10^{-16} = 28.44 \times 10^{-16} = 2.844 \times 10^{-14} m.$$

Hence, the distance of closest approach is approximately $$2.84 \times 10^{-14}$$ m.

Key Concept: Emission Lines

c) Emission lines
[Solution Description] When electrons transition from a higher energy state to a lower energy state, they emit photons, resulting in emission lines. These lines are part of the atomic spectrum observed in spectroscopy.