Bohr's Model of Hydrogen Atom Questions in English

(N/A) The velocity of an electron moving around a proton in a hydrogen atom in an orbit of radius $r = 5.3 \times 10^{-11} \, m$ is $v = 2.2 \times 10^{6} \, m/s$.
The frequency of the electron revolving around the proton is given by the formula $f = \frac{v}{2 \pi r}$.
Substituting the values:
$f = \frac{2.2 \times 10^{6} \, m/s}{2 \times 3.14 \times 5.3 \times 10^{-11} \, m}$
$f \approx 6.6 \times 10^{15} \, Hz$.
According to classical electromagnetic theory,the frequency of the electromagnetic waves emitted by a revolving electron is equal to the frequency of its revolution around the nucleus. Therefore,the initial frequency of the light emitted is $6.6 \times 10^{15} \, Hz$.

(A) The total energy $E$ of an electron in the ground state of a hydrogen atom is given as $E = -13.6 \; eV$.
The total energy is the sum of kinetic energy $(K)$ and potential energy $(U)$,i.e.,$E = K + U$.
For an electron in a hydrogen atom,the kinetic energy is related to the total energy by $K = -E$.
Therefore,$K = -(-13.6 \; eV) = 13.6 \; eV$.
The potential energy is related to the total energy by $U = 2E$.
Therefore,$U = 2 \times (-13.6 \; eV) = -27.2 \; eV$.
Thus,the kinetic energy is $13.6 \; eV$ and the potential energy is $-27.2 \; eV$.

(N/A) Let $v_{n}$ be the orbital speed of the electron in a hydrogen atom in the $n$-th level. The speed is given by $v_{n} = \frac{v_{1}}{n}$,where $v_{1} = \frac{e^{2}}{2 \epsilon_{0} h} \approx 2.18 \times 10^{6} \, m/s$.
For $n=1$: $v_{1} = 2.18 \times 10^{6} \, m/s$.
For $n=2$: $v_{2} = \frac{v_{1}}{2} = 1.09 \times 10^{6} \, m/s$.
For $n=3$: $v_{3} = \frac{v_{1}}{3} = 7.27 \times 10^{5} \, m/s$.
$(b)$ The orbital period $T_{n}$ is given by $T_{n} = \frac{2 \pi r_{n}}{v_{n}}$. Since $r_{n} = n^{2} r_{1}$ and $v_{n} = \frac{v_{1}}{n}$,we have $T_{n} = n^{3} T_{1}$,where $T_{1} = \frac{2 \pi r_{1}}{v_{1}} \approx 1.52 \times 10^{-16} \, s$.
For $n=1$: $T_{1} = 1.52 \times 10^{-16} \, s$.
For $n=2$: $T_{2} = 2^{3} T_{1} = 8 \times 1.52 \times 10^{-16} = 1.22 \times 10^{-15} \, s$.
For $n=3$: $T_{3} = 3^{3} T_{1} = 27 \times 1.52 \times 10^{-16} = 4.12 \times 10^{-15} \, s$.

(N/A) The radius of the innermost orbit of a hydrogen atom is given by $r_{1} = 5.3 \times 10^{-11} \;m$.
The radius of an orbit for a hydrogen atom is given by the formula $r_{n} = n^{2} r_{1}$,where $n$ is the principal quantum number.
For $n=2$:
$r_{2} = (2)^{2} \times r_{1} = 4 \times 5.3 \times 10^{-11} = 2.12 \times 10^{-10} \;m$.
For $n=3$:
$r_{3} = (3)^{2} \times r_{1} = 9 \times 5.3 \times 10^{-11} = 4.77 \times 10^{-10} \;m$.
Thus,the radii for $n=2$ and $n=3$ orbits are $2.12 \times 10^{-10} \;m$ and $4.77 \times 10^{-10} \;m$ respectively.

(D) Given:
Radius of the orbit,$r = 1.5 \times 10^{11} \; m$
Orbital speed,$v = 3 \times 10^{4} \; m/s$
Mass of the Earth,$m = 6.0 \times 10^{24} \; kg$
Planck's constant,$h = 6.626 \times 10^{-34} \; J \cdot s$
According to Bohr's quantization condition for angular momentum:
$L = mvr = \frac{nh}{2\pi}$
Rearranging for the quantum number $n$:
$n = \frac{2\pi mvr}{h}$
Substituting the values:
$n = \frac{2 \times 3.1416 \times (6.0 \times 10^{24}) \times (3 \times 10^{4}) \times (1.5 \times 10^{11})}{6.626 \times 10^{-34}}$
$n = \frac{169.646 \times 10^{39}}{6.626 \times 10^{-34}}$
$n \approx 25.603 \times 10^{73} \approx 2.56 \times 10^{74}$
Rounding to the nearest provided option,we get $n \approx 2.6 \times 10^{74}$.

(N/A) The radius of the first Bohr orbit is given by the relation:
$r_{1} = \frac{4 \pi \epsilon_{0} (\frac{h}{2 \pi})^{2}}{m_{e} e^{2}} \dots (i)$
Where:
$\epsilon_{0} = 8.854 \times 10^{-12} \, C^{2} N^{-1} m^{-2}$ (Permittivity of free space)
$h = 6.63 \times 10^{-34} \, J s$ (Planck's constant)
$m_{e} = 9.1 \times 10^{-31} \, kg$ (Mass of an electron)
$e = 1.6 \times 10^{-19} \, C$ (Charge of an electron)
$m_{p} = 1.67 \times 10^{-27} \, kg$ (Mass of a proton)
$G = 6.67 \times 10^{-11} \, N m^{2} kg^{-2}$ (Gravitational constant)
Coulomb force: $F_{c} = \frac{e^{2}}{4 \pi \epsilon_{0} r^{2}}$
Gravitational force: $F_{G} = \frac{G m_{p} m_{e}}{r^{2}}$
If the electron and proton were bound by gravitational attraction, we equate the centripetal force to the gravitational force:
$\frac{m_{e} v^{2}}{r} = \frac{G m_{p} m_{e}}{r^{2}}$
Using Bohr's quantization condition $m_{e} v r = \frac{h}{2 \pi}$, we substitute $v = \frac{h}{2 \pi m_{e} r}$ into the force equation:
$r_{1} = \frac{h^{2}}{4 \pi^{2} G m_{p} m_{e}^{2}}$
Substituting the values:
$r_{1} = \frac{(6.63 \times 10^{-34})^{2}}{4 \times (3.14)^{2} \times 6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times (9.1 \times 10^{-31})^{2}}$
$r_{1} \approx 1.21 \times 10^{29} \, m$
Since the observable universe is approximately $1.5 \times 10^{27} \, m$, the calculated radius is much larger than the size of the universe.

(N/A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{m e^4}{8 \epsilon_0^2 h^2 n^2}$.
When the atom de-excites from level $n$ to $(n-1)$,the energy of the emitted photon is $h \nu = E_n - E_{n-1}$.
$h \nu = \frac{m e^4}{8 \epsilon_0^2 h^2} \left( \frac{1}{(n-1)^2} - \frac{1}{n^2} \right)$.
$
u = \frac{m e^4}{8 \epsilon_0^2 h^3} \left( \frac{n^2 - (n-1)^2}{n^2(n-1)^2} \right) = \frac{m e^4}{8 \epsilon_0^2 h^3} \left( \frac{2n-1}{n^2(n-1)^2} \right)$.
For large $n$,$2n-1 \approx 2n$ and $(n-1) \approx n$,so $
u \approx \frac{m e^4}{8 \epsilon_0^2 h^3} \left( \frac{2n}{n^4} \right) = \frac{m e^4}{4 \epsilon_0^2 h^3 n^3}$.
The classical frequency of revolution $\nu_c$ is $\frac{v}{2 \pi r}$.
Using $v = \frac{e^2}{2 \epsilon_0 h n}$ and $r = \frac{\epsilon_0 h^2 n^2}{\pi m e^2}$,we get $\nu_c = \frac{v}{2 \pi r} = \frac{e^2 / (2 \epsilon_0 h n)}{2 \pi (\epsilon_0 h^2 n^2 / \pi m e^2)} = \frac{m e^4}{4 \epsilon_0^2 h^3 n^3}$.
Thus,for large $n$,the quantum frequency equals the classical frequency.

(N/A) The quantity having dimensions of length involving $e, m_{e},$ and $c$ is the classical electron radius,given by $r_{e} = \frac{1}{4 \pi \epsilon_{0}} \frac{e^{2}}{m_{e} c^{2}}$.
Using $e = 1.6 \times 10^{-19} \; C$,$m_{e} = 9.1 \times 10^{-31} \; kg$,$c = 3 \times 10^{8} \; m/s$,and $\frac{1}{4 \pi \epsilon_{0}} = 9 \times 10^{9} \; Nm^{2}C^{-2}$:
$r_{e} = (9 \times 10^{9}) \times \frac{(1.6 \times 10^{-19})^{2}}{9.1 \times 10^{-31} \times (3 \times 10^{8})^{2}} \approx 2.81 \times 10^{-15} \; m$.
This value is much smaller than the atomic size of $\sim 10^{-10} \; m$.
$(b)$ The quantity with dimensions of length involving $h, m_{e},$ and $e$ is the Bohr radius $a_{0} = \frac{4 \pi \epsilon_{0} \hbar^{2}}{m_{e} e^{2}}$,where $\hbar = \frac{h}{2 \pi}$.
Substituting $h = 6.63 \times 10^{-34} \; Js$:
$a_{0} = \frac{1}{9 \times 10^{9}} \times \frac{(6.63 \times 10^{-34} / (2 \times 3.14))^{2}}{9.1 \times 10^{-31} \times (1.6 \times 10^{-19})^{2}} \approx 0.53 \times 10^{-10} \; m$.
This value is of the order of $10^{-10} \; m$,which matches the known atomic size.

(N/A) The quantization of angular momentum is a fundamental law of nature, but its effects are only observable at the microscopic scale.
For planetary motion, the angular momentum $(L)$ is extremely large compared to Planck's constant $(h)$.
For example, the angular momentum of the Earth in its orbit is of the order of $10^{70} h$.
According to Bohr's postulate, $L = n(h / 2 \pi)$, which implies $n = 2 \pi L / h$.
Substituting the values, we find that the quantum number $(n)$ is of the order of $10^{70}$.
For such extremely large values of $(n)$, the difference between successive energy levels or angular momentum states is infinitesimally small.
Therefore, the discrete nature of the orbits becomes indistinguishable from a continuous distribution, and we treat planetary motion using classical mechanics.

(N/A) The mass of a negatively charged muon is $m_{\mu} = 207 m_{e}$.
According to Bohr's model,the radius of the $n^{th}$ orbit is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$. Since $r \propto \frac{1}{m}$,we have $r_{\mu} = r_{e} \times \frac{m_{e}}{m_{\mu}}$.
Given $r_{e} = 0.53 \times 10^{-10} \, m$,the radius of the muonic hydrogen atom is $r_{\mu} = \frac{0.53 \times 10^{-10}}{207} \approx 2.56 \times 10^{-13} \, m$.
The energy of the ground state is given by $E_n = -\frac{m Z^2 e^4}{8 \epsilon_0^2 n^2 h^2}$. Since $E \propto m$,we have $E_{\mu} = E_{e} \times \frac{m_{\mu}}{m_{e}}$.
Given $E_{e} = -13.6 \, eV$,the ground state energy of the muonic hydrogen atom is $E_{\mu} = -13.6 \times 207 \, eV = -2815.2 \, eV \approx -2.81 \, keV$.

(N/A) According to Bohr's theory of the hydrogen atom,the electrostatic force between the nucleus and the electron provides the necessary centripetal force for circular motion:
$F_e = F_c$
$\frac{1}{4\pi\epsilon_0} \cdot \frac{e^2}{r^2} = \frac{mv^2}{r}$
From this,we can derive the relationship between the orbital radius $r$ and the velocity $v$ of the electron:
$\frac{1}{4\pi\epsilon_0} \cdot \frac{e^2}{r} = mv^2$
Thus,the relationship is $r = \frac{e^2}{4\pi\epsilon_0 mv^2}$.

(N/A) Bohr's atomic model successfully explained the stability of the atom by postulating that electrons revolve in stationary orbits where they do not radiate energy.
It successfully calculated the energy levels and radii of hydrogen and hydrogen-like atoms (single-electron species).
It provided a theoretical basis for the Rydberg formula and the spectral series of hydrogen,given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
It successfully explained the spectra of hydrogen-like ions such as $He^+$,$Li^{2+}$,and $Be^{3+}$,where $Z$ is the atomic number.

(N/A) The Rydberg constant,denoted by $R$,is a physical constant related to the atomic spectra of hydrogen. Its value is approximately $R = 1.097 \times 10^7 \ m^{-1}$.

(N/A) In $1913$,Bohr concluded that despite the success of electromagnetic theory in explaining large-scale phenomena,it could not be applied to processes at the atomic scale.
Concepts different from classical mechanics and electromagnetism were needed to understand the structure of atoms and the relation of atomic structure to atomic spectra.
Bohr combined classical and early quantum concepts and proposed his theory in the form of three postulates:
$(1)$ An electron in an atom could revolve in certain stable orbits without the emission of radiant energy,contrary to the prediction of electromagnetic theory. Each atom has certain definite stable states in which it can exist,and each possible state has a definite total energy. These are called the stationary states of the atom.
$(2)$ The electron revolves around the nucleus only in those orbits for which the angular momentum is an integral multiple of $\frac{h}{2 \pi}$,where $h$ is Planck's constant $(6.626 \times 10^{-34} \ J \ s)$. Thus,the angular momentum $(L)$ of the orbiting electron is quantized.
That is,$L = \frac{nh}{2\pi} = mvr$,where $n = 1, 2, 3, ...$
$(3)$ Bohr's third postulate incorporated early quantum concepts developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so,a photon is emitted having energy equal to the energy difference between the initial and final states.
The frequency $(v)$ of the emitted photon is given by:
$hv = E_i - E_f$
$\therefore v = \frac{E_i - E_f}{h}$
Where $E_i$ is the energy of the initial state,$E_f$ is the energy of the final state,and $E_i > E_f$.

(N/A) The atomic model of Bohr is shown in the figure.
Let the mass of an electron be $m$,its charge be $e$,its linear speed in the $n^{th}$ orbit be $v_n$,the radius of the orbit be $r_n$,and the charge on the nucleus be $Ze$,where $Z$ is the atomic number.
The necessary centripetal force is provided by the Coulombic attractive force between the electron and the nucleus. Thus,
$\frac{m v_n^2}{r_n} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{(Ze)(e)}{r_n^2} = \frac{Z e^2}{4 \pi \epsilon_0 r_n^2}$ ... $(1)$
From Bohr's second postulate,the angular momentum $L_n$ of an electron in the $n^{th}$ orbit is given by:
$L_n = m v_n r_n = \frac{n h}{2 \pi}$ ... $(2)$
From equation $(1)$,we have $m v_n^2 = \frac{Z e^2}{4 \pi \epsilon_0 r_n}$.
Multiplying both sides by $r_n^2$,we get $m v_n^2 r_n^2 = \frac{Z e^2 r_n}{4 \pi \epsilon_0}$.
Taking the square root,$m v_n r_n = \sqrt{\frac{Z e^2 m r_n}{4 \pi \epsilon_0}}$.
Equating this to equation $(2)$:
$\frac{n h}{2 \pi} = \sqrt{\frac{Z e^2 m r_n}{4 \pi \epsilon_0}}$
Squaring both sides:
$\frac{n^2 h^2}{4 \pi^2} = \frac{Z e^2 m r_n}{4 \pi \epsilon_0}$
Solving for $r_n$:
$r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$

(N/A) The necessary centripetal force for an electron of mass $m$ and charge $e$ moving with speed $v_n$ in an orbit of radius $r_n$ around a nucleus of charge $Ze$ is provided by the electrostatic force of attraction:
$\frac{m v_n^2}{r_n} = \frac{1}{4 \pi \epsilon_0} \frac{(Ze)(e)}{r_n^2}$
Multiplying both sides by $\frac{r_n}{2}$,we get the kinetic energy $K$:
$K = \frac{1}{2} m v_n^2 = \frac{Ze^2}{8 \pi \epsilon_0 r_n} \quad \dots (1)$
The potential energy $U$ of the electron in the field of the nucleus is:
$U = -\frac{1}{4 \pi \epsilon_0} \frac{(Ze)(e)}{r_n} = -\frac{Ze^2}{4 \pi \epsilon_0 r_n} \quad \dots (2)$
The total energy $E_n$ is the sum of kinetic and potential energy:
$E_n = K + U = \frac{Ze^2}{8 \pi \epsilon_0 r_n} - \frac{Ze^2}{4 \pi \epsilon_0 r_n}$
$E_n = -\frac{Ze^2}{8 \pi \epsilon_0 r_n}$
Substituting the given expression for $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$ into the energy formula:
$E_n = -\frac{Ze^2}{8 \pi \epsilon_0} \left( \frac{\pi m Z e^2}{n^2 h^2 \epsilon_0} \right) = -\frac{m Z^2 e^4}{8 \epsilon_0^2 n^2 h^2}$

(A) The expression for the total energy of an electron $E_n = -\frac{Z^2me^4}{8\epsilon_0^2n^2h^2}$ is derived based on Bohr's hypothesis of circular orbits.
In this model,the electron is assumed to move in circular orbits around the nucleus under the influence of the electrostatic force.
Physicist Sommerfeld later extended this model to include elliptical orbits. He demonstrated that when the restriction of circular orbits is relaxed,the energy expression remains valid for elliptical orbits as well,provided the central force is an inverse-square force.

(N/A) According to the Bohr atomic model, the radius of the $n^{th}$ orbit of an electron in a hydrogen-like atom is given by the formula:
$r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$
Where:
$n$ is the principal quantum number (orbit number),
$h$ is Planck's constant,
$\epsilon_0$ is the permittivity of free space,
$m$ is the mass of the electron,
$Z$ is the atomic number,
$e$ is the elementary charge.

(N/A) The ground state of an atom is the lowest energy state of the system.
In the context of the Bohr model for a hydrogen atom,it corresponds to the state where the electron is in the innermost orbit,which is the orbit with the principal quantum number $n = 1$.
The energy of the electron in the ground state of a hydrogen atom is approximately $-13.6 \ eV$.

(N/A) Ionisation energy is defined as the minimum amount of energy required to remove an electron from an atom in its ground state to an infinite distance from the nucleus,effectively placing it in a state where it is no longer bound to the atom.
For a hydrogen atom,the energy of the ground state $(n = 1)$ is $E_1 = -13.6 \ eV$.
To ionise the atom,we must provide enough energy to bring the total energy to $0 \ eV$ (the energy of an electron at infinity).
Therefore,the ionisation energy is $E_{\infty} - E_1 = 0 - (-13.6 \ eV) = 13.6 \ eV$.

(N/A) According to Bohr's third postulate,when an electron makes a transition from a higher energy state with quantum number $n_{i}$ to a lower energy state with quantum number $n_{f}$ $(n_{i} > n_{f})$,a photon is emitted with energy equal to the difference in energy between the two states.
The energy of an electron in the $n_{i}$ state is given by:
$E_{n_{i}} = -\frac{m e^{4}}{8 \epsilon_{0}^{2} h^{2} n_{i}^{2}}$
The energy of an electron in the $n_{f}$ state is given by:
$E_{n_{f}} = -\frac{m e^{4}}{8 \epsilon_{0}^{2} h^{2} n_{f}^{2}}$
The energy of the emitted photon is $h \nu_{if} = E_{n_{i}} - E_{n_{f}}$.
Substituting the expressions:
$h \nu_{if} = -\frac{m e^{4}}{8 \epsilon_{0}^{2} h^{2} n_{i}^{2}} - \left( -\frac{m e^{4}}{8 \epsilon_{0}^{2} h^{2} n_{f}^{2}} \right)$
$h \nu_{if} = \frac{m e^{4}}{8 \epsilon_{0}^{2} h^{2}} \left[ \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right]$
Thus,the frequency of the emitted radiation is:
$\nu_{if} = \frac{m e^{4}}{8 \epsilon_{0}^{2} h^{3}} \left[ \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right]$
Since the wave number $\bar{\nu} = \frac{1}{\lambda} = \frac{\nu}{c}$,we divide the frequency by the speed of light $c$:
$\bar{\nu} = \frac{1}{\lambda_{if}} = \frac{m e^{4}}{8 \epsilon_{0}^{2} h^{3} c} \left[ \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right]$
Here,$R = \frac{m e^{4}}{8 \epsilon_{0}^{2} h^{3} c}$ is the Rydberg constant,with a theoretical value of approximately $1.097 \times 10^{7} \ m^{-1}$.

(N/A) The Rydberg constant $(R)$ is a fundamental physical constant related to the atomic spectra of hydrogen.
The theoretical value of the Rydberg constant is derived from the formula:
$R = \frac{m_e e^4}{8 \epsilon_0^2 h^3 c}$
where $m_e$ is the electron mass,$e$ is the elementary charge,$\epsilon_0$ is the vacuum permittivity,$h$ is Planck's constant,and $c$ is the speed of light.
Substituting the standard values:
$R \approx 1.0973731568 \times 10^7 \ m^{-1}$.
The experimental value is determined through high-precision spectroscopy of hydrogen atoms and is found to be consistent with the theoretical value:
$R \approx 1.0973731568 \times 10^7 \ m^{-1}$.

(N/A) In his second postulate,Bohr proposed that the angular momentum of an electron orbiting the nucleus is quantized.
Louis de Broglie provided a physical explanation for this in $1923$.
According to de Broglie's hypothesis,material particles exhibit wave-like properties. This wave nature was experimentally verified by Davisson and Germer.
Bohr suggested that an electron in a circular orbit behaves as a matter wave.
Similar to waves on a string,particle waves can form standing waves under resonant conditions.
When a string is plucked,many waves are excited,but only those that form standing waves (with nodes at the ends) persist. This occurs when the total distance traveled by the wave is an integral multiple of its wavelength.
Waves with other wavelengths interfere destructively upon reflection,and their amplitudes drop to zero; thus,an electron cannot exist in such an orbit.
For an electron in the $n^{\text{th}}$ circular orbit of radius $r_n$,the circumference of the orbit must be an integral multiple of the wavelength $\lambda$.
Therefore,$2 \pi r_n = n \lambda$ ... $(1)$,where $n = 1, 2, 3, \dots$
Using de Broglie's relation for wavelength and momentum,$\lambda = \frac{h}{p} = \frac{h}{m v_n}$ ... $(2)$.
Substituting $(2)$ into $(1)$:
$2 \pi r_n = n \left( \frac{h}{m v_n} \right)$
Rearranging the terms gives the quantization condition for angular momentum:
$m v_n r_n = \frac{n h}{2 \pi}$

(N/A) $(1)$ The Bohr model is applicable only to hydrogenic atoms. It cannot be extended even to simple two-electron atoms such as helium.
- The analysis of atoms with more than one electron was attempted using the lines of Bohr's model for hydrogenic atoms,but it did not meet with any success.
- The difficulty lies in the fact that each electron interacts not only with the positively charged nucleus but also with all other electrons.
- The formulation of the Bohr model involves the electrical force between the positively charged nucleus and the electron. It does not include the electrical force between electrons,which necessarily appears in multi-electron atoms.
$(2)$ While the Bohr model correctly predicts the frequencies of the light emitted by hydrogenic atoms,it is unable to explain the relative intensities of the frequencies in the spectrum.
- In the emission spectrum of hydrogen,some of the visible frequencies have weak intensity,while others are strong. Why?
- Experimental observations depict that some transitions are more favoured than others.
- The Bohr model is unable to account for these intensity variations.
- This model cannot be applied to complex atoms. For complex atoms,we must use a new theory based on quantum mechanics.

(N/A) According to Bohr's second postulate,the electron revolves around the nucleus only in those orbits for which its angular momentum is an integral multiple of $\frac{h}{2\pi}$.
Mathematically,this condition is expressed as:
$L = mvr = \frac{nh}{2\pi}$
Where:
$L$ is the angular momentum of the electron.
$m$ is the mass of the electron.
$v$ is the velocity of the electron in its orbit.
$r$ is the radius of the orbit.
$n$ is the principal quantum number $(n = 1, 2, 3, ...)$.
$h$ is Planck's constant.

(N/A) Both ${ }_{2} He^{3}$ and ${ }_{2} He^{4}$ are isotopes of helium.
When one electron is removed from each,they both become hydrogen-like ions $(He^+)$ with a single electron.
The energy levels of a hydrogen-like ion are given by the formula $E_n = -\frac{Z^2 R_y}{n^2}$,where $Z$ is the atomic number and $R_y$ is the Rydberg constant.
Since both isotopes have the same atomic number $Z = 2$,their energy levels are identical in the ideal Bohr model.
However,the slight difference in mass between the nuclei of $He^3$ and $He^4$ leads to a very small difference in the reduced mass of the electron-nucleus system,which is why their energy levels are very close but not perfectly identical.

(N/A) According to classical electrodynamics,an accelerated charged particle emits energy in the form of electromagnetic waves.
When an electron transitions from a higher energy orbit to a lower energy orbit,it moves closer to the nucleus.
Due to the change in the electrostatic potential energy and the requirement to maintain angular momentum,the electron undergoes a change in its velocity,effectively performing an accelerated motion.
Since the electron is a charged particle,this accelerated motion results in the emission of energy in the form of periodically oscillating electric and magnetic fields,which we observe as electromagnetic radiation.
Other forms of energy,such as mechanical or thermal energy,do not satisfy the fundamental conservation laws and the specific interaction dynamics between the electron's charge and the electromagnetic field during such transitions.

(A) Yes,the Bohr formula for the $H-$ atom would remain unchanged in this case because the electrostatic force between the proton and the electron remains the same.
In the original case,the electrostatic force is given by $F = \frac{1}{4 \pi \epsilon_{0}} \frac{(e)(e)}{r^{2}} = \frac{1}{4 \pi \epsilon_{0}} \frac{e^{2}}{r^{2}}$.
In the new case,the magnitude of the charges are $q_{1} = \frac{4e}{3}$ and $q_{2} = \frac{3e}{4}$. The new electrostatic force $F^{\prime}$ is given by:
$F^{\prime} = \frac{1}{4 \pi \epsilon_{0}} \frac{(\frac{4e}{3})(\frac{3e}{4})}{r^{2}} = \frac{1}{4 \pi \epsilon_{0}} \frac{e^{2}}{r^{2}}$.
Since $F^{\prime} = F$,the centripetal force required for the orbit remains identical,and therefore,the Bohr model energy levels and radii formulas remain unchanged.

(B) According to the Bohr model,the total energy of an electron in the $n^{\text{th}}$ energy level of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
If the energies $E_n$ are different,the principal quantum numbers $n$ must be different.
The orbital angular momentum of an electron in the $n^{\text{th}}$ orbit is given by $L_n = \frac{nh}{2\pi}$.
Since the angular momentum $L_n$ is directly proportional to $n$,if the values of $n$ are different,the values of the angular momentum $L_n$ must also be different.
Therefore,it is not possible for the electrons to have different energies but the same orbital angular momentum.

(B) Positronium is a binary system consisting of an electron and a positron. Both particles revolve around their common center of mass.
To treat this as a single-particle system,we use the reduced mass $\mu$ of the binary system:
$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m_e \times m_e}{m_e + m_e} = \frac{m_e}{2} = \frac{m}{2}$
The ground state energy of an electron in a hydrogen atom is given by:
$E_1 = -\frac{m e^4}{8 \epsilon_0^2 h^2} = -13.6 \text{ eV}$
Since the energy of the ground state is directly proportional to the mass of the orbiting particle,the ground state energy of positronium $E_1'$ is:
$E_1' = \frac{\mu}{m} E_1 = \frac{m/2}{m} E_1 = \frac{1}{2} E_1$
Substituting the value of $E_1$:
$E_1' = \frac{1}{2} (-13.6 \text{ eV}) = -6.8 \text{ eV}$

(N/A) As per the assumption made in the statement,since the $He$ atom behaves like a hydrogen-like atom (where electron-electron repulsion is ignored),we can apply Bohr's atomic model to the $He$ atom.
The energy of an electron in an orbit is given by:
$E_{n} = -\frac{m Z^{2} e^{4}}{8 \epsilon_{0}^{2} n^{2} h^{2}}$
This can be simplified as:
$E_{n} = -13.6 \frac{Z^{2}}{n^{2}} \text{ eV}$
For a $He$ atom,the atomic number $Z = 2$. For the ground state,the principal quantum number $n = 1$.
Substituting these values:
$E_{1} = -13.6 \times \frac{(2)^{2}}{(1)^{2}} \text{ eV}$
$E_{1} = -13.6 \times 4 \text{ eV}$
$E_{1} = -54.4 \text{ eV}$
Since there are two electrons in the $He$ atom and we assume no repulsion between them,the total ground state energy of the $He$ atom is the sum of the energies of both electrons:
$E_{\text{total}} = 2 \times E_{1} = 2 \times (-54.4 \text{ eV}) = -108.8 \text{ eV}$.

(A) The orbital radius of an electron in the $n^{\text{th}}$ orbit of an $H$-atom is given by $r_{n} = \frac{\epsilon_{0} n^{2} h^{2}}{\pi m Z e^{2}}$.
For the ground state,$n = 1$ and $Z = 1$,so $r_{1} = a_{0} = 0.53 \times 10^{-10} \text{ m}$.
The velocity of the electron in the $n^{\text{th}}$ orbit is $v_{n} = \frac{Z e^{2}}{2 \epsilon_{0} n h}$. For $n = 1$,$v_{1} = 2.187 \times 10^{6} \text{ m/s}$.
The frequency of revolution $f_{1}$ is given by $f_{1} = \frac{v_{1}}{2 \pi r_{1}}$.
The electric current $I$ is $I = f_{1} e = \frac{v_{1} e}{2 \pi r_{1}}$.
Substituting the values: $I = \frac{(2.187 \times 10^{6} \text{ m/s}) \times (1.6 \times 10^{-19} \text{ C})}{2 \times 3.1416 \times (0.53 \times 10^{-10} \text{ m})}$.
$I \approx 1.05 \times 10^{-3} \text{ A} = 1.05 \text{ mA}$.

(N/A) In an atom of an element with atomic number $Z$,when an electron makes a transition from the $(n+x)^{\text{th}}$ orbit (where $x=1, 2, 3, \dots$) to the $n^{\text{th}}$ orbit,the frequency $f$ of the emitted radiation is given by the Rydberg formula:
$\frac{1}{\lambda} = R Z^{2} \left( \frac{1}{n^{2}} - \frac{1}{(n+x)^{2}} \right)$
Since $f = \frac{c}{\lambda}$,we have:
$f = c R Z^{2} \left( \frac{(n+x)^{2} - n^{2}}{n^{2}(n+x)^{2}} \right)$
$f = c R Z^{2} \left( \frac{n^{2} + 2nx + x^{2} - n^{2}}{n^{2}(n+x)^{2}} \right)$
$f = c R Z^{2} \left( \frac{2nx + x^{2}}{n^{2}(n+x)^{2}} \right)$
Given $n >> 1$ and $x$ is small $(x=1, 2, 3, \dots)$,we can approximate $(n+x) \approx n$:
$f \approx c R Z^{2} \left( \frac{2nx}{n^{2} \cdot n^{2}} \right)$
$f \approx \left( \frac{2 R c Z^{2}}{n^{3}} \right) x$
Since $\frac{2 R c Z^{2}}{n^{3}}$ is constant for a fixed $n$,we have $f \propto x$.
Therefore,the ratio of frequencies for $x=1, 2, 3$ is $f_{1} : f_{2} : f_{3} = 1 : 2 : 3$.

(D) For the Balmer series of the atomic spectrum of hydrogen,the transition ends at $n=2$. The lines are defined as $H_{\alpha} (n=3 \to 2)$,$H_{\beta} (n=4 \to 2)$,$H_{\gamma} (n=5 \to 2)$,and so on.
To emit an $H_{\gamma}$ line,the electron must be excited from the ground state $(n=1)$ to the $n=5$ energy level.
The energy required is $\Delta E = E_{5} - E_{1}$.
Using $E_{n} = -\frac{13.6 \text{ eV}}{n^{2}}$,we get:
$\Delta E = -\frac{13.6}{5^{2}} - (-13.6) = 13.6 \left(1 - \frac{1}{25}\right) = 13.6 \times \frac{24}{25} = 13.056 \text{ eV}$.
The angular momentum of the emitted photon is equal to the change in the orbital angular momentum of the electron during the transition from $n=5$ to $n=2$.
Using Bohr's quantization condition $L = \frac{nh}{2\pi}$:
$L_{\text{photon}} = L_{5} - L_{2} = \frac{5h}{2\pi} - \frac{2h}{2\pi} = \frac{3h}{2\pi}$.
Substituting $h = 6.626 \times 10^{-34} \text{ Js}$:
$L_{\text{photon}} = \frac{3 \times 6.626 \times 10^{-34}}{2 \times 3.14159} \approx 3.165 \times 10^{-34} \text{ Js}$.

(A) The Rydberg constant is given by $R = \frac{\mu e^4}{8 \epsilon_0^2 h^3 c}$,where $\mu$ is the reduced mass of the electron-nucleus system,$\mu = \frac{M m_e}{M + m_e}$.
For Hydrogen $(H)$,$M_H \approx 1836 m_e$,so $\mu_H = \frac{1836 m_e^2}{1837 m_e} \approx 0.99945 m_e$.
For Deuterium $(D)$,$M_D \approx 3670 m_e$,so $\mu_D = \frac{3670 m_e^2}{3671 m_e} \approx 0.99973 m_e$.
Since $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,we have $\lambda \propto \frac{1}{R} \propto \frac{1}{\mu}$.
Thus,$\frac{\lambda_D}{\lambda_H} = \frac{\mu_H}{\mu_D} \approx \frac{0.99945}{0.99973} \approx 0.99972$.
The shift is $\Delta \lambda = \lambda_H - \lambda_D = \lambda_H (1 - 0.99972) = 0.00028 \lambda_H$.

(A) The orbital radius of an electron in the ground state of a $H$-atom is the Bohr radius $a_0 = 0.53 \mathring{A} = \frac{\epsilon_0 h^2}{\pi m e^2}$.
For case $(i)$,$R = 0.1 \mathring{A}$. Since $R < a_0$,the proton can be treated as a point charge. The ground state energy is $E_1 = -13.6 \text{ eV}$.
For case $(ii)$,$R = 10 \mathring{A}$. Since $R > a_0$,the electron moves inside the proton. The effective charge $e'$ at radius $b_0$ is $e' = e \left( \frac{b_0^3}{R^3} \right)$.
The Bohr radius condition is $b_0 = \frac{\epsilon_0 h^2}{\pi m e e'} = \frac{\epsilon_0 h^2}{\pi m e^2} \left( \frac{R^3}{b_0^3} \right) = a_0 \frac{R^3}{b_0^3}$.
Thus,$b_0^4 = a_0 R^3 \Rightarrow b_0 = (a_0 R^3)^{1/4} = (0.53 \times 10^3)^{1/4} \approx 4.8 \mathring{A}$.
The potential energy inside a uniform charge sphere is $U = \frac{e}{4\pi\epsilon_0 R^3} \left( \frac{3R^2 - b_0^2}{2} \right) (-e)$.
The total energy $E = K + U = \frac{e^2}{8\pi\epsilon_0 R^3} (3R^2 - b_0^2) - \frac{e^2}{4\pi\epsilon_0 R^3} (3R^2 - b_0^2) = -\frac{e^2}{8\pi\epsilon_0 R^3} (3R^2 - b_0^2)$.

(D) The energy of an electron in a hydrogen-like atom is given by $E_{n} = -13.6 \frac{Z^{2}}{n^{2}} \ eV$.
For Chromium $(Z = 24)$:
Energy of $n = 2$ state: $E_{2} = -13.6 \times \frac{24^{2}}{2^{2}} = -13.6 \times 144 = -1958.4 \ eV$.
Energy of $n = 1$ state: $E_{1} = -13.6 \times \frac{24^{2}}{1^{2}} = -13.6 \times 576 = -7833.6 \ eV$.
The energy released during the $n = 2$ to $n = 1$ transition is $\Delta E = E_{2} - E_{1} = -1958.4 - (-7833.6) = 5875.2 \ eV$.
This energy is transferred to an electron in the $n = 4$ state.
The energy of the $n = 4$ state is $E_{4} = -13.6 \times \frac{24^{2}}{4^{2}} = -13.6 \times 36 = -489.6 \ eV$.
The kinetic energy $K$ of the ejected Auger electron is $K = \Delta E - |E_{4}| = 5875.2 - 489.6 = 5385.6 \ eV$.

(A) Given $\lambda = \frac{m_p c}{\hbar} = \frac{2 \pi m_p c}{h}$. With $m_p = 10^{-6} m_e$,$\lambda = \frac{2 \pi (10^{-6} \times 9.1 \times 10^{-31} \text{ kg}) (3 \times 10^8 \text{ m/s})}{6.63 \times 10^{-34} \text{ J s}} \approx 2.58 \times 10^6 \text{ m}^{-1}$.
Since the Bohr radius $r_B \approx 5.3 \times 10^{-11} \text{ m}$,we have $\lambda r_B \approx 1.37 \times 10^{-4} \ll 1$.
Using the modified potential $U(r) = -\frac{e^2}{4\pi \epsilon_0} \frac{e^{-\lambda r}}{r}$,we expand the exponential for small $\lambda r$: $U(r) \approx -\frac{e^2}{4\pi \epsilon_0} \frac{1-\lambda r}{r} = -\frac{e^2}{4\pi \epsilon_0} (\frac{1}{r} - \lambda)$.
The change in potential energy is $\Delta U = U_{modified} - U_{coulomb} = \frac{e^2}{4\pi \epsilon_0} \lambda$.
Substituting $\frac{e^2}{4\pi \epsilon_0} = k e^2 \approx 1.44 \text{ eV nm} = 1.44 \times 10^{-9} \text{ eV m}$,the energy shift $\Delta E \approx \Delta U = (1.44 \times 10^{-9} \text{ eV m}) (2.58 \times 10^6 \text{ m}^{-1}) \approx 3.7 \times 10^{-3} \text{ eV}$.

For $r < R_0$,the modified force is $F = \frac{k e^2}{R_0^{2-\epsilon} r^{\epsilon}}$.
Equating this to the centripetal force: $\frac{m v^2}{r} = \frac{k e^2}{R_0^{2-\epsilon} r^{\epsilon}}$.
Thus,$v^2 = \frac{k e^2}{m R_0^{2-\epsilon} r^{\epsilon-1}}$.
The Bohr quantization condition is $mvr = n\hbar$. For the ground state,$n=1$,so $v = \frac{\hbar}{mr}$.
Substituting $v^2 = \frac{\hbar^2}{m^2 r^2}$ into the force equation: $\frac{\hbar^2}{m r^3} = \frac{k e^2}{R_0^{2-\epsilon} r^{\epsilon}}$.
$r^{3-\epsilon} = \frac{\hbar^2 R_0^{2-\epsilon}}{m k e^2} = a_0 R_0^{2-\epsilon}$,where $a_0$ is the Bohr radius.
$r = (a_0 R_0^{2-\epsilon})^{1/(3-\epsilon)}$.
The total energy $E = K + U = \frac{1}{2} m v^2 + \int_{\infty}^{r} F dr$.
Calculating the potential energy $U = -\int_{\infty}^{R_0} \frac{k e^2}{r^2} dr - \int_{R_0}^{r} \frac{k e^2}{R_0^{2-\epsilon} r^{\epsilon}} dr$.
Evaluating these integrals gives the ground state energy $E$.

(D) The kinetic energy gained by a charged particle accelerated through a potential difference $V$ is given by $K = qV = \frac{1}{2}mv^2$.
From this,the speed $v$ is given by $v = \sqrt{\frac{2qV}{m}}$.
For a hydrogen ion $(H^+)$,charge $q_H = e$ and mass $m_H \approx m$.
For a singly ionized helium atom $(He^+)$,charge $q_{He} = e$ and mass $m_{He} \approx 4m$.
The ratio of speeds is $\frac{v_H}{v_{He}} = \frac{\sqrt{2q_H V / m_H}}{\sqrt{2q_{He} V / m_{He}}} = \sqrt{\frac{q_H}{q_{He}} \cdot \frac{m_{He}}{m_H}}$.
Substituting the values: $\frac{v_H}{v_{He}} = \sqrt{\frac{e}{e} \cdot \frac{4m}{m}} = \sqrt{4} = 2$.
Thus,the ratio is $2:1$.

(D) Let $m = 200 \, MeV/c^2$ be the mass of the particle and $M = 1000 \, MeV/c^2$ be the mass of the hydrogen atom.
Let $v$ be the initial velocity of the particle and $V$ be the recoil velocity of the atom.
By conservation of linear momentum: $mv = MV = p$,where $p$ is the momentum.
The energy required to excite the hydrogen atom to its first excited state is $\Delta E = 13.6 \, eV \times (1 - 1/4) = 10.2 \, eV$.
By conservation of energy: $K_{initial} = K_{final} + \Delta E$.
$K_{initial} = \frac{p^2}{2m}$ and $K_{final} = \frac{p^2}{2M}$.
Thus,$\frac{p^2}{2m} - \frac{p^2}{2M} = 10.2 \, eV$.
$\frac{p^2}{2m} (1 - \frac{m}{M}) = 10.2 \, eV$.
Given $m/M = 200/1000 = 0.2$,we have $K_{initial} (1 - 0.2) = 10.2 \, eV$.
$K_{initial} (0.8) = 10.2 \, eV$.
$K_{initial} = \frac{10.2}{0.8} = 12.75 \, eV$.
We are given $K_{initial} = \frac{N}{4}$,so $\frac{N}{4} = 12.75$.
$N = 12.75 \times 4 = 51$.

(A) The Bohr model is strictly applicable only to hydrogen-like species,which are atoms or ions containing exactly one electron.
$1$. Hydrogen atom $(H)$ has $1$ electron.
$2$. Singly ionized helium atom $(He^+)$ has $2 - 1 = 1$ electron.
$3$. Deuteron atom (an isotope of hydrogen) has $1$ electron.
$4$. Singly ionized neon atom $(Ne^+)$ has $10 - 1 = 9$ electrons.
Since $Ne^+$ has more than one electron,the Bohr model is not valid for it.

(D) The magnetic field at the center of the orbit due to a revolving electron is given by $B = \frac{\mu_0 i}{2r}$.
Here,the current $i = \frac{e}{T} = \frac{ev}{2\pi r}$.
Substituting $i$,we get $B = \frac{\mu_0 ev}{4\pi r^2}$.
For the $n$-th orbit of a hydrogen atom,the velocity is $v_n = \frac{v_1}{n} = \frac{2.18 \times 10^6}{3} \approx 7.27 \times 10^5 \ m/s$.
The radius is $r_n = r_1 n^2 = 0.529 \times 10^{-10} \times 3^2 = 4.761 \times 10^{-10} \ m$.
Substituting these values into the formula:
$B = \frac{(10^{-7}) \times (1.6 \times 10^{-19}) \times (7.27 \times 10^5)}{(4.761 \times 10^{-10})^2}$.
$B = \frac{1.1632 \times 10^{-20}}{22.667 \times 10^{-20}} \approx 0.0513 \ T$.
Rounding to the nearest option,the value is $0.05 \ T$.

(D) According to the Bohr quantization condition,the circumference of the $n^{th}$ orbit is an integer multiple of the de-Broglie wavelength: $2 \pi r_n = n \lambda$.
Therefore,the wavelength is given by $\lambda = \frac{2 \pi r_n}{n}$.
The radius of the $n^{th}$ orbit for a hydrogen-like ion is $r_n = a_0 \frac{n^2}{Z}$,where $a_0 \approx 0.529 \ \mathring{A}$.
For $He^{+1}$ ion,the atomic number $Z = 2$. For the $3^{rd}$ orbit,$n = 3$.
Substituting these values into the radius formula: $r_3 = 0.529 \times \frac{3^2}{2} = 0.529 \times 4.5 = 2.3805 \ \mathring{A}$.
Now,substitute $r_3$ and $n$ into the wavelength formula: $\lambda = \frac{2 \pi (2.3805)}{3}$.
$\lambda = \frac{2 \times 3.14159 \times 2.3805}{3} \approx \frac{14.958}{3} \approx 4.986 \ \mathring{A}$.
Rounding to the nearest integer,we get $\lambda \approx 5 \ \mathring{A}$.

(B) The centripetal acceleration $a$ of an electron moving in a circular orbit is given by the formula:
$a = \frac{v^{2}}{r}$
Given values:
Velocity $v = 2.18 \times 10^{6} \; m/s$
Radius $r = 0.528 \; \mathring{A} = 0.528 \times 10^{-10} \; m$
Substituting these values into the formula:
$a = \frac{(2.18 \times 10^{6})^{2}}{0.528 \times 10^{-10}}$
$a = \frac{4.7524 \times 10^{12}}{0.528 \times 10^{-10}}$
$a \approx 9 \times 10^{22} \; m/s^{2}$

(C) The magnetic field $B$ at the center of the orbit due to an electron revolving in a circular path is given by $B = \frac{\mu_{0} I}{2r}$.
The current $I$ is given by $I = \frac{ev}{2\pi r}$,where $e$ is the charge of the electron,$v$ is the velocity,and $r$ is the radius.
Substituting $I$ into the magnetic field formula: $B = \frac{\mu_{0} ev}{4\pi r^{2}}$.
Given values: $\mu_{0} = 4\pi \times 10^{-7} \, T \cdot m/A$,$e = 1.6 \times 10^{-19} \, C$,$v = 2.18 \times 10^{6} \, m/s$,and $r = 5 \times 10^{-11} \, m$.
$B = \frac{(4\pi \times 10^{-7}) \times (1.6 \times 10^{-19}) \times (2.18 \times 10^{6})}{4\pi \times (5 \times 10^{-11})^{2}}$
$B = \frac{1.6 \times 10^{-19} \times 2.18 \times 10^{6} \times 10^{-7}}{25 \times 10^{-22}}$
$B = \frac{3.488 \times 10^{-20}}{25 \times 10^{-22}} = \frac{348.8}{25} = 13.952 \, T$.
Thus,the magnetic field is approximately $13.95 \, T$.

(B) The velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $v_n = v_0 \frac{Z}{n}$,where $v_0$ is the velocity of the electron in the ground state of the hydrogen atom.
The value of $v_0$ is $2.18 \times 10^{6} \text{ m/s}$.
For a hydrogen atom ($H$-atom),the atomic number $Z = 1$. We need to find the speed in the fourth orbit,so $n = 4$.
Substituting these values into the formula:
$v_4 = (2.18 \times 10^{6} \text{ m/s}) \times \frac{1}{4}$
$v_4 = 0.545 \times 10^{6} \text{ m/s}$
$v_4 = 5.45 \times 10^{5} \text{ m/s}$.
Thus,the speed of the electron in the fourth orbit is $5.45 \times 10^{5} \text{ m/s}$.

(A) The radius of an electron in a hydrogen-like ion is given by the formula:
$r = r_{0} \frac{n^{2}}{Z}$
where $r_{0}$ is the Bohr radius,$n$ is the principal quantum number,and $Z$ is the atomic number.
For $Be^{3+}$,the atomic number $Z = 4$.
For the ground state of a Hydrogen atom,the radius is $r = r_{0}$ (where $n = 1$ and $Z = 1$).
Setting the radii equal:
$r_{0} = r_{0} \frac{n^{2}}{4}$
$n^{2} = 4$
$n = 2$
The principal quantum number $n = 2$ corresponds to the first excited state.
Therefore,at the first excited state of $Be^{3+}$,the radius of the electron is the same as that of a Hydrogen atom in the ground state.

(B) According to Bohr's theory of the hydrogen atom,the velocity $(v_n)$ of an electron in the $n^{\text{th}}$ orbit is given by the formula:
$v_n = \frac{2 \pi k Z e^2}{n h}$
Where:
$k$ is Coulomb's constant,
$Z$ is the atomic number (for hydrogen,$Z=1$),
$e$ is the charge of the electron,
$h$ is Planck's constant,
$n$ is the principal quantum number (orbit number).
Since $2, \pi, k, Z, e^2,$ and $h$ are constants,the velocity is inversely proportional to the orbit number $n$.
Therefore,$v_n \propto \frac{1}{n}$.

(B) The energy levels of a hydrogen-like atom are given by $E_n = -\frac{m e^4}{8 \epsilon_0^2 n^2 h^2}$,where $m$ is the mass of the orbiting particle.
Since $E \propto m$,the ionization potential $IP$ is directly proportional to the mass of the particle.
For a standard hydrogen atom,$IP_e = 13.6 \ eV$.
When the electron is replaced by a muon with mass $m_{\mu} = 207 m_e$,the new ionization potential $IP_{\mu}$ is given by:
$IP_{\mu} = IP_e \times \frac{m_{\mu}}{m_e}$
$IP_{\mu} = 13.6 \ eV \times 207$
$IP_{\mu} = 2815.2 \ eV$.