Take Class 12 Tuition from the Best Tutors
Affordable fees 1-1 or Group class Flexible Timings Verified Tutors Search in
Answered on 06 Apr Learn Unit 8-Atoms & Nuclei
Sadika
In the ground state of a hydrogen atom, the total energy is equal to the negative of the ground state energy, which is −13.6 eV−13.6eV. This total energy consists of the sum of the kinetic energy and the potential energy of the electron.
Given that the total energy is −13.6 eV−13.6eV, we can express this as the sum of the kinetic energy (KK) and the potential energy (VV) of the electron:
Total energy=K+VTotal energy=K+V
Since the electron is bound to the nucleus in the hydrogen atom, the total energy is negative, indicating a bound state.
Given that the total energy is −13.6 eV−13.6eV, and assuming the potential energy is negative (since it contributes to the binding of the electron), we can express the kinetic energy as:
K=Total energy−VK=Total energy−V
Substituting the given value of the total energy, we have:
K=−13.6 eV−VK=−13.6eV−V
Now, knowing that the potential energy is negative and contributes to the binding of the electron, and since the total energy is the sum of kinetic and potential energies, the potential energy (VV) is simply the negative of the total energy:
V=−13.6 eVV=−13.6eV
Substituting this value of potential energy into the expression for kinetic energy, we get:
K=−13.6 eV−(−13.6 eV)=0 eVK=−13.6eV−(−13.6eV)=0eV
So, in the ground state of the hydrogen atom, the kinetic energy of the electron is 0 eV0eV, and the potential energy is −13.6 eV−13.6eV.
Answered on 06 Apr Learn Unit 8-Atoms & Nuclei
Sadika
The Ha-line of the Balmer series in the emission spectrum of the hydrogen atom is obtained when an electron transitions from the n=3n=3 energy level to the n=2n=2 energy level.
In other words, when an electron that is initially in the third energy level (i.e., n=3n=3) of the hydrogen atom transitions to the second energy level (i.e., n=2n=2), it emits a photon of light corresponding to the Ha-line. This emitted photon has a specific wavelength characteristic of the Ha-line in the visible spectrum.
The transition from the n=3n=3 to n=2n=2 energy level is one of the spectral lines observed in the Balmer series of hydrogen, which consists of transitions of electrons to the second energy level (n=2n=2) from higher energy levels. The Ha-line is specifically associated with the transition from the third to the second energy level and corresponds to a visible spectral line with a specific wavelength in the red part of the spectrum.
Answered on 06 Apr Learn Unit 8-Atoms & Nuclei
Sadika
The classical Rutherford model, also known as the planetary model, proposed that electrons orbit the nucleus of an atom in well-defined circular orbits, much like planets orbiting around the sun. While this model successfully explained certain observations, such as the scattering of alpha particles by a gold foil and the existence of a dense, positively charged nucleus, it ultimately failed to explain several key aspects of atomic structure. Some reasons why the classical Rutherford model is insufficient to explain atomic structure include:
Electron Stability: According to classical electromagnetic theory, an accelerating charged particle, such as an electron orbiting around the nucleus, should continuously emit electromagnetic radiation. This emission of energy would cause the electron to lose energy and spiral into the nucleus, ultimately leading to the collapse of the atom. However, stable atoms exist, suggesting that electrons do not follow classical trajectories.
Quantization of Energy: Classical mechanics does not account for the quantization of energy levels observed in atomic spectra. In contrast to the continuous energy levels predicted by classical theory, experiments revealed that electrons in atoms can only occupy discrete energy levels. The classical model fails to explain why electrons are restricted to specific energy levels and why transitions between these levels result in discrete spectral lines.
Wave-Particle Duality: The classical Rutherford model treats electrons as classical particles with definite positions and momenta. However, experiments in quantum mechanics have shown that particles such as electrons exhibit wave-like behavior, described by wave functions. The classical model cannot explain phenomena such as electron diffraction, interference, and wave-particle duality observed in experiments.
Uncertainty Principle: The classical model does not account for the Heisenberg uncertainty principle, which states that it is impossible to simultaneously know both the exact position and momentum of a particle with certainty. In contrast to classical trajectories, quantum mechanics describes electron positions as probability distributions within certain regions of space.
Spectral Line Fine Structure: The classical model cannot explain the fine structure observed in atomic spectra, including the splitting of spectral lines in the presence of external magnetic or electric fields, known as the Zeeman and Stark effects, respectively. These effects arise from the interaction of electrons' intrinsic magnetic moments and electric dipoles with external fields, which are not accounted for in the classical model.
In summary, while the classical Rutherford model provided valuable insights into atomic structure, its inability to explain phenomena such as electron stability, quantization of energy levels, wave-particle duality, uncertainty principle, and spectral line fine structure led to its replacement by quantum mechanical models, such as the Bohr model and the modern quantum mechanical model of the atom. These models successfully reconcile experimental observations with the principles of quantum mechanics, providing a more accurate description of atomic structure and behavior.
Take Class 12 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 06 Apr Learn Unit 8-Atoms & Nuclei
Sadika
Ionization energy is the energy required to remove an electron from an atom or molecule in its ground state to infinity, resulting in the formation of a positively charged ion. It is commonly expressed in units of electron volts (eV) or kilojoules per mole (kJ/mol).
For a hydrogen atom, the ionization energy corresponds to the energy required to completely remove its single electron from the ground state to infinity. The ionization energy of a hydrogen atom can be calculated using the Rydberg formula or obtained from experimental measurements. The exact value of the ionization energy of a hydrogen atom is known to be approximately 13.6 electron volts (eV) or 1312 kilojoules per mole (kJ/mol). This value represents the minimum energy required to completely remove the electron from the hydrogen atom, resulting in the formation of a hydrogen cation (H+H+).
Answered on 06 Apr Learn Unit 8-Atoms & Nuclei
Sadika
In the Bohr model of the hydrogen atom, the radii of the electron orbits are given by the formula:
rn=n2⋅h24π2⋅m⋅k⋅e2rn=4π2⋅m⋅k⋅e2n2⋅h2
where:
The ratio of the radii of two orbits corresponding to different energy levels n1n1 and n2n2 can be expressed as:
rn2rn1=n12n22rn1rn2=n22n12
For the first excited state (n2=2n2=2) and the ground state (n1=1n1=1) of the hydrogen atom, the ratio of radii is:
r2r1=1222=14r1r2=2212=41
Therefore, the ratio of the radii of the orbits corresponding to the first excited state and the ground state in a hydrogen atom is 1441.
Answered on 06 Apr Learn Unit 8-Atoms & Nuclei
Sadika
In the Bohr model of the hydrogen atom, the radius of the electron orbit corresponding to the nn-th energy level (rnrn) is given by the formula:
rn=n2⋅h24π2⋅m⋅k⋅e2rn=4π2⋅m⋅k⋅e2n2⋅h2
where:
Given that the radius of the innermost electron orbit (corresponding to the ground state, n=1n=1) is 5.3×10−115.3×10−11 m, we can use this information to find the radius of the orbit in the second excited state (n=3n=3).
We can rewrite the formula for rnrn as follows:
rn=n2r1⋅r1rn=r1n2⋅r1
where:
Substituting the given values, we have:
r3=3212⋅(5.3×10−11 m)r3=1232⋅(5.3×10−11m)
r3=9×(5.3×10−11 m)r3=9×(5.3×10−11m)
r3=47.7×10−11 mr3=47.7×10−11m
r3=4.77×10−10 mr3=4.77×10−10m
Therefore, the radius of the orbit in the second excited state of the hydrogen atom is 4.77×10−104.77×10−10 m.
Take Class 12 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 06 Apr Learn Unit 8-Atoms & Nuclei
Sadika
The expression for Bohr's radius (r) in a hydrogen atom is given by the Bohr radius formula:
r=4πε0ℏ2mee2r=mee24πε0ℏ2
Where:
Note: In LaTeX, the symbols are represented as ε0ε0, ℏℏ, meme, and ee.
Answered on 06 Apr Learn Unit 8-Atoms & Nuclei
Sadika
Bohr's quantization condition for defining stationary orbits in the hydrogen atom states that the angular momentum (LL) of the electron orbiting the nucleus is quantized and can only take on discrete values which are integer multiples of Planck's constant divided by 2π2π:
L=nh2πL=n2πh
Where:
This condition implies that electrons in the hydrogen atom can only occupy certain allowed orbits, each characterized by a specific value of nn. These allowed orbits are termed stationary orbits.
Answered on 06 Apr Learn Unit 8-Atoms & Nuclei
Sadika
In the Rutherford scattering experiment, the distance of closest approach (d0d0) is determined by the balance between the electrostatic repulsion between the incident particle (alpha particle or proton) and the positively charged nucleus, and the kinetic energy of the incident particle.
Let's denote the kinetic energy required to achieve the distance of closest approach d0d0 for an alpha particle as KαKα, and the kinetic energy required for a proton to achieve the same distance of closest approach as KpKp.
The kinetic energy of a particle can be related to its charge and mass by the equation:
K=12mv2K=21mv2
Where:
In both cases, for the alpha particle and the proton, the electrostatic repulsion with the nucleus is the same, since both have the same charge (one charge unit). So, the kinetic energy required to achieve the same distance of closest approach is only dependent on the mass of the particle.
Given that the mass of an alpha particle (mαmα) is approximately 4 times the mass of a proton (mpmp), we can use the fact that kinetic energy is directly proportional to mass. Therefore, to achieve the same distance of closest approach d0d0, the kinetic energy required for the proton (KpKp) would be 1441 times the kinetic energy required for the alpha particle (KαKα).
In mathematical terms: Kp=14KαKp=41Kα
So, to achieve the same distance of closest approach, the kinetic energy required for the proton will be one-fourth of the kinetic energy required for the alpha particle.
Take Class 12 Tuition from the Best Tutors
Affordable fees Flexible Timings Choose between 1-1 and Group class Verified Tutors Answered on 06 Apr Learn Unit 8-Atoms & Nuclei
Sadika
The energy levels of a hydrogen atom are given by the formula for the energy levels of the hydrogen atom:
En=−RHn2En=−n2RH
Where:
The energy of a photon emitted or absorbed during a transition between two energy levels is given by the difference in energy between the initial and final energy levels:
ΔE=Efinal−EinitialΔE=Efinal−Einitial
For a transition from the second permitted energy level (n=2n=2) to the first permitted energy level (n=1n=1), the energy of the photon emitted (ΔE2→1ΔE2→1) is:
ΔE2→1=E1−E2ΔE2→1=E1−E2
For a transition from the highest permitted energy level (n=∞n=∞) to the first permitted energy level (n=1n=1), the energy of the photon emitted (ΔE∞→1ΔE∞→1) is:
ΔE∞→1=E1−E∞ΔE∞→1=E1−E∞
Substituting the expressions for the energy levels (EnEn) into these equations, we get:
ΔE2→1=−(−RH12)−(−RH22)ΔE2→1=−(−12RH)−(−22RH) ΔE2→1=−(−13.6 eV)+(−13.64 eV)ΔE2→1=−(−13.6eV)+(−413.6eV) ΔE2→1=13.6 eV−3.4 eVΔE2→1=13.6eV−3.4eV ΔE2→1=10.2 eVΔE2→1=10.2eV
ΔE∞→1=−(−RH12)−(−RH∞2)ΔE∞→1=−(−12RH)−(−∞2RH) ΔE∞→1=−(−13.6 eV)−(0 eV)ΔE∞→1=−(−13.6eV)−(0eV) ΔE∞→1=13.6 eVΔE∞→1=13.6eV
Now, to find the ratio of energies of photons:
ΔE2→1ΔE∞→1=10.2 eV13.6 eVΔE∞→1ΔE2→1=13.6eV10.2eV
ΔE2→1ΔE∞→1=10.213.6ΔE∞→1ΔE2→1=13.610.2
ΔE2→1ΔE∞→1≈0.75ΔE∞→1ΔE2→1≈0.75
So, the ratio of energies of photons produced due to the transition of an electron from the second permitted energy level to the first permitted level to the transition from the highest permitted energy level to the first permitted level is approximately 0.75.
UrbanPro.com helps you to connect with the best Class 12 Tuition in India. Post Your Requirement today and get connected.
Ask a Question
The best tutors for Class 12 Tuition Classes are on UrbanPro
The best Tutors for Class 12 Tuition Classes are on UrbanPro
1 
