Bohr's Model of Hydrogen Atom Questions in English

(C) The electric potential $V$ due to a point charge $Q$ at a distance $r$ is given by the formula $V = \frac{kQ}{r}$,where $k = 9 \times 10^9 \, N \cdot m^2/C^2$.
For a hydrogen nucleus,the charge $Q = +e = 1.6 \times 10^{-19} \, C$.
The radius of the orbit is $r = 0.53 \times 10^{-10} \, m$.
Substituting these values into the formula:
$V = \frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{0.53 \times 10^{-10}}$
$V = \frac{14.4 \times 10^{-10}}{0.53 \times 10^{-10}}$
$V \approx 27.17 \, V \approx 27.2 \, V$.

(B) The current $I$ produced by a revolving charge is given by the formula $I = qf$, where $q$ is the charge of the electron and $f$ is the frequency of revolution.
Given:
Charge of electron $q = 1.6 \times 10^{-19} \, C$
Frequency $f = 6.6 \times 10^{15} \, \text{rev/s}$
Substituting the values:
$I = (1.6 \times 10^{-19} \, C) \times (6.6 \times 10^{15} \, \text{s}^{-1})$
$I = 10.56 \times 10^{-4} \, A$
$I \approx 1.056 \times 10^{-3} \, A$
$I \approx 1 \, mA$
Therefore, the equivalent current is nearly $1 \, mA$.

(A) The time period of revolution of the electron is given by $T = \frac{2\pi r}{v}$.
Since current $i$ is defined as the rate of flow of charge,$i = \frac{e}{T}$.
Substituting $T$,we get $i = \frac{ev}{2\pi r}$.
Given: $e = 1.6 \times 10^{-19} \ C$,$v = 2 \times 10^6 \ m/s$,and $r = 0.5 \ \mathring{A} = 0.5 \times 10^{-10} \ m$.
$i = \frac{1.6 \times 10^{-19} \times 2 \times 10^6}{2 \times 3.14 \times 0.5 \times 10^{-10}}$.
$i = \frac{3.2 \times 10^{-13}}{3.14 \times 10^{-10}} \approx 1.019 \times 10^{-3} \ A$.
Converting to $mA$,$i \approx 1 \ mA$.

(C) According to de Broglie's hypothesis,the circumference of the $n^{th}$ orbit is equal to an integral multiple of the de Broglie wavelength: $2\pi r_n = n\lambda$.
Given: Wavelength $\lambda = 10^{-9} \ m$.
The radius of the $n^{th}$ orbit of a hydrogen atom is given by $r_n = n^2 a_0$,where $a_0 = 0.529 \times 10^{-10} \ m$.
For $n=3$,$r_3 = 3^2 \times 0.529 \times 10^{-10} \ m = 9 \times 0.529 \times 10^{-10} \ m \approx 4.76 \times 10^{-10} \ m$.
Using the condition $n = \frac{2\pi r_n}{\lambda}$:
$n = \frac{2 \times 3.14159 \times 4.76 \times 10^{-10}}{10^{-9}} \approx 3$.
Thus,the principal quantum number is $3$.

(D) According to the Bohr model of the hydrogen atom,the radius of the $n^{th}$ stationary orbit is given by the formula:
$r_n = \frac{\varepsilon_0 n^2 h^2}{\pi Z m e^2}$
In this expression,$\varepsilon_0$ is the permittivity of free space,$h$ is Planck's constant,$Z$ is the atomic number,$m$ is the mass of the electron,and $e$ is the elementary charge.
Since all these parameters are constants for a given atom,the radius $r$ is directly proportional to the square of the principal quantum number $n$.
Therefore,$r \propto n^2$.

(D) The wavelength $\lambda$ of the spectral line emitted during a transition between energy levels is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For a transition from $n_2 = 2$ to $n_1 = 1$,the term $\left( \frac{1}{1^2} - \frac{1}{2^2} \right) = \frac{3}{4}$ is constant for all hydrogen-like species.
Therefore,$\frac{1}{\lambda} \propto Z^2$,which implies $\lambda \propto \frac{1}{Z^2}$.
To obtain the minimum wavelength $\lambda$,the atomic number $Z$ must be maximum.
- Hydrogen atom $(H)$: $Z = 1$
- Deuterium atom $(D)$: $Z = 1$
- Uni-ionized helium $(He^+)$: $Z = 2$
- Di-ionized lithium $(Li^{2+})$: $Z = 3$
Since $Li^{2+}$ has the highest atomic number $(Z = 3)$,it produces the spectral line with the minimum wavelength.

(B) The linear momentum $p$ of an object is given by the product of its mass $m$ and its velocity $v$.
$p = m \times v$
Given:
Mass $m = 9.1 \times 10^{-31} \ kg$
Velocity $v = 2.2 \times 10^6 \ m/s$
Substituting these values into the formula:
$p = (9.1 \times 10^{-31} \ kg) \times (2.2 \times 10^6 \ m/s)$
$p = (9.1 \times 2.2) \times 10^{-31+6} \ kg \cdot m/s$
$p = 20.02 \times 10^{-25} \ kg \cdot m/s$
$p = 2.002 \times 10^{-24} \ kg \cdot m/s$
Rounding to two significant figures,we get:
$p \approx 2.0 \times 10^{-24} \ kg \cdot m/s$
Thus,the correct option is $B$.

(C) According to Bohr's model,the radius of the $n^{th}$ orbit of a hydrogen atom is given by the formula $r_n = n^2 a_0$,where $a_0$ is the Bohr radius.
For the second Bohr orbit,the principal quantum number $n = 2$.
Substituting $n = 2$ into the formula,we get $r_2 = (2)^2 a_0 = 4 a_0$.
Therefore,the radius of the second Bohr orbit is $4 a_0$.

(C) For the ionization of the second $He$ electron,the ion $He^+$ acts as a hydrogen-like atom with atomic number $Z = 2$.
The ionization energy for a hydrogen-like atom is given by the formula $E = Z^2 \times 13.6 \, eV$.
Substituting $Z = 2$ into the formula:
$E = (2)^2 \times 13.6 \, eV = 4 \times 13.6 \, eV = 54.4 \, eV$.
Therefore,the ionization potential is $54.4 \, V$.

(B) In the $CGS$ system,the electrostatic potential energy $U$ of an electron in an orbit of radius $r$ in a hydrogen atom is given by $U = -\frac{e^2}{r}$.
According to the virial theorem for a system with a $1/r$ potential,the kinetic energy $K$ is related to the potential energy $U$ by the relation $K = -\frac{1}{2}U$.
Substituting the value of $U$,we get $K = -\frac{1}{2} \left( -\frac{e^2}{r} \right) = \frac{e^2}{2r}$.

(B) Let the energies of the states $A, B,$ and $C$ be $E_A, E_B,$ and $E_C$ respectively.
From the principle of conservation of energy,the energy of the transition from $C$ to $A$ is equal to the sum of the energies of the transitions from $C$ to $B$ and $B$ to $A$.
Thus,$(E_C - E_A) = (E_C - E_B) + (E_B - E_A)$.
Using the relation $E = \frac{hc}{\lambda}$,we can write:
$\frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$
Dividing both sides by $hc$,we get:
$\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$
$\frac{1}{\lambda_3} = \frac{\lambda_1 + \lambda_2}{\lambda_1 \lambda_2}$
Therefore,$\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$.

(C) According to Bohr's second postulate,the angular momentum $L$ of an electron revolving in a stable orbit is quantized.
It is given by the relation $L = mvr = \frac{nh}{2\pi}$,where $n$ is the principal quantum number $(n = 1, 2, 3, ...)$,$h$ is Planck's constant,and $m, v, r$ are the mass,velocity,and radius of the orbit respectively.
Therefore,the correct option is $C$.

(D) The ionisation energy of a hydrogen-like atom is given by the formula $E_n = 13.6 \times Z^2 \text{ eV}$,where $Z$ is the atomic number.
For a sodium atom $(Na)$,the atomic number $Z = 11$.
$A$ $10$ times ionised sodium atom is a hydrogen-like ion with one electron remaining.
Therefore,the ionisation energy is $E = 13.6 \times Z^2 \text{ eV} = 13.6 \times (11)^2 \text{ eV}$.

(D) According to Bohr's model,the radius $r$ of the $n^{th}$ orbit is given by the formula:
$r = \frac{\varepsilon_0 n^2 h^2}{\pi Z m e^2}$
In this expression,$\varepsilon_0$,$h$,$\pi$,$m$,and $e$ are all physical constants.
Therefore,the radius $r$ is directly proportional to $n^2$ and inversely proportional to $Z$.
Thus,$r \propto \frac{n^2}{Z}$.

(B) According to Bohr's model,the radius of the $n^{th}$ orbit is given by the relation $r_n \propto n^2$.
Given that the radius of the second orbit $(n=2)$ is $R$,we have $r_2 = k(2)^2 = 4k = R$,where $k$ is a constant.
Therefore,$k = \frac{R}{4}$.
For the third orbit $(n=3)$,the radius $r_3$ is given by $r_3 = k(3)^2 = 9k$.
Substituting the value of $k$,we get $r_3 = 9 \times \frac{R}{4} = 2.25 R$.

(A) In the $C.G.S.$ system,the electrostatic force of attraction between the nucleus (charge $Ze$) and the electron (charge $e$) provides the necessary centripetal force for circular motion.
The electrostatic force is given by $F_e = \frac{(Ze)(e)}{r^2} = \frac{Ze^2}{r^2}$.
The centripetal force required for circular motion is $F_c = \frac{mv^2}{r}$.
Equating the two forces: $\frac{mv^2}{r} = \frac{Ze^2}{r^2}$.
Multiplying both sides by $r$,we get $mv^2 = \frac{Ze^2}{r}$.
The kinetic energy $K$ is defined as $K = \frac{1}{2}mv^2$.
Substituting the expression for $mv^2$: $K = \frac{1}{2} \left( \frac{Ze^2}{r} \right) = \frac{Ze^2}{2r}$.

(D) According to Bohr's postulate for the quantization of angular momentum,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $mvr = n\frac{h}{2\pi}$.
Rearranging this equation,we get $2\pi r = n\left(\frac{h}{mv}\right)$.
Since the de Broglie wavelength $\lambda$ is defined as $\lambda = \frac{h}{p} = \frac{h}{mv}$,we can substitute this into the equation.
Therefore,the circumference of the orbit is $2\pi r = n\lambda$.

(C) According to Bohr's model for a hydrogen-like atom,the kinetic energy $(K.E.)$ of an electron in an orbit of radius $r$ is given by $K.E. = \frac{kZe^2}{2r}$.
The potential energy $(P.E.)$ of the electron in the same orbit is given by $P.E. = -\frac{kZe^2}{r}$.
To find the ratio of kinetic energy to potential energy,we divide the two expressions:
$\frac{K.E.}{P.E.} = \frac{\frac{kZe^2}{2r}}{-\frac{kZe^2}{r}} = -\frac{1}{2} = -0.5$.
Thus,the ratio is $-0.5$.

(C) The wave number $\bar{\nu}$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
Here,the transition is from $n_2 = 4$ to $n_1 = 2$.
Substituting the values: $\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right]$.
$\frac{1}{\lambda} = R \left[ \frac{1}{4} - \frac{1}{16} \right]$.
$\frac{1}{\lambda} = R \left[ \frac{4 - 1}{16} \right] = \frac{3R}{16}$.

(A) In the Bohr model of the hydrogen atom,the kinetic energy $(K.E.)$ of an electron in an orbit is given by $K.E. = \frac{kZe^2}{2r}$.
The total energy $(T.E.)$ of the electron in the same orbit is given by $T.E. = -\frac{kZe^2}{2r}$.
Comparing these two expressions,we can see that $K.E. = -(T.E.)$.
Therefore,the ratio of the kinetic energy to the total energy is $\frac{K.E.}{T.E.} = \frac{K.E.}{-K.E.} = -1$.

(A) According to Bohr's model,the energy of an electron in an orbit is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
As the principal quantum number $n$ increases,the energy of the orbit increases.
When an electron transitions from a lower energy level $(n=1)$ to a higher energy level $(n=3)$,it must gain energy from an external source to overcome the energy difference.
Therefore,the electron absorbs energy equal to $\Delta E = E_3 - E_1$.

(A) The electrostatic potential energy $(P.E.)$ of an electron at a distance $r$ from the nucleus is given by $P.E. = \frac{1}{4\pi \varepsilon_0} \frac{(Ze)( -e)}{r}$. For a hydrogen atom $(Z=1)$,this becomes $P.E. = - \frac{e^2}{4\pi \varepsilon_0 r}$.
According to the virial theorem for a stable orbit,the kinetic energy $(K.E.)$ is related to the potential energy by $K.E. = -\frac{1}{2} (P.E.)$.
Substituting the expression for $P.E.$,we get $K.E. = -\frac{1}{2} \left( - \frac{e^2}{4\pi \varepsilon_0 r} \right) = + \frac{e^2}{8\pi \varepsilon_0 r}$.
Thus,the kinetic energy is $+ \frac{e^2}{8\pi \varepsilon_0 r}$ and the potential energy is $- \frac{e^2}{4\pi \varepsilon_0 r}$.

(C) According to Bohr's postulate,the angular momentum $L$ of an electron in a stationary orbit is given by the quantization condition:
$L = mvr = \frac{nh}{2\pi}$
where $n$ is the principal quantum number,$h$ is Planck's constant,$m$ is the mass of the electron,$v$ is its velocity,and $r$ is the radius of the orbit.
For the lowest energy level (ground state) of a hydrogen atom,$n = 1$.
Substituting $n = 1$ into the formula,we get:
$L = \frac{1 \cdot h}{2\pi} = \frac{h}{2\pi}$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{m e^4}{8 \varepsilon_0^2 n^2 h^2}$.
Using the Bohr model,the wavenumber $\bar{\nu}$ for a transition from $n_2$ to $n_1$ is $\bar{\nu} = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
From the energy difference $\Delta E = h c \bar{\nu} = E_{n_2} - E_{n_1}$,we get $R = \frac{m e^4}{8 \varepsilon_0^2 c h^3}$.
Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we have $\varepsilon_0 = \frac{1}{4 \pi k}$,so $\varepsilon_0^2 = \frac{1}{16 \pi^2 k^2}$.
Thus,$R = \frac{m e^4}{8 (\frac{1}{16 \pi^2 k^2}) c h^3} = \frac{16 \pi^2 k^2 m e^4}{8 c h^3} = \frac{2 \pi^2 k^2 m e^4}{c h^3}$.
Substituting $k = \frac{1}{4 \pi \varepsilon_0}$,we get $R = \left( \frac{1}{4 \pi \varepsilon_0} \right)^2 \frac{2 \pi^2 m e^4}{c h^3}$.

(C) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by the formula: $L = n \left( \frac{h}{2\pi} \right)$.
For the second orbit,we have $n = 2$.
Substituting the value of $n$ into the formula: $L = 2 \times \left( \frac{h}{2\pi} \right) = \frac{h}{\pi}$.
Therefore,the angular momentum of the electron in the second orbit is $\frac{h}{\pi}$.

(D) 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 $Z$ is the atomic number and $n$ is the orbit number.
From this relation,we can see that for a given atom (constant $Z$),the velocity is inversely proportional to the orbit number: $v_n \propto \frac{1}{n}$.
Given that the velocity in the second orbit $(n_2 = 2)$ is $v$,we have $v_2 = v$.
We need to find the velocity in the fifth orbit $(n_5 = 5)$,denoted as $v_5$.
Using the proportionality $v_n \propto \frac{1}{n}$,we can write the ratio: $\frac{v_5}{v_2} = \frac{n_2}{n_5}$.
Substituting the known values: $\frac{v_5}{v} = \frac{2}{5}$.
Therefore,$v_5 = \frac{2}{5}v$.

(D) The number of possible emission transitions from an excited state $n$ to the ground state is given by the formula $N_E = \frac{n(n - 1)}{2}$.
Here,the electron is in the fourth energy state,so $n = 4$.
Substituting the value of $n$ into the formula:
$N_E = \frac{4(4 - 1)}{2} = \frac{4 \times 3}{2} = \frac{12}{2} = 6$.
Therefore,there are $6$ possible emission transitions.

(C) The centripetal force required for the circular motion of the electron is provided by the electrostatic Coulomb force between the proton and the electron.
Equating the centripetal force to the Coulomb force:
$\frac{m{v^2}}{{a_0}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{{a_0^2}}$
Canceling $a_0$ from both sides:
$m{v^2} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{{e^2}}}{{a_0}}$
Solving for $v^2$:
${v^2} = \frac{{{e^2}}}{{4\pi {\varepsilon _0}{a_0}m}}$
Taking the square root of both sides,we get the speed of the electron:
$v = \frac{e}{{\sqrt {4\pi {\varepsilon _0}{a_0}m} }}$
Thus,the correct option is $C$.

(D) According to the Bohr model,the time period $T$ of an electron in an orbit with principal quantum number $n$ is given by $T = \frac{2\pi r}{v}$.
Since $r \propto n^2$ and $v \propto \frac{1}{n}$,we have $T \propto n^3$.
Given that the time period in the initial state is eight times that in the final state: $T_{n_1} = 8 T_{n_2}$.
Substituting the proportionality: $n_1^3 = 8 n_2^3$.
Taking the cube root on both sides: $n_1 = 2 n_2$.
Checking the options:
For option $(a)$: $n_1 = 4, n_2 = 2$. Here $4 = 2(2)$,which is correct.
For option $(b)$: $n_1 = 6, n_2 = 3$. Here $6 = 2(3)$,which is correct.
Thus,both $(a)$ and $(b)$ are valid possibilities.

(D) According to the Bohr model,the energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \; eV$.
For a doubly ionized lithium atom $(Li^{2+})$,the atomic number $Z = 3$.
For the ground state,the principal quantum number $n = 1$.
Substituting these values,the energy of the ground state is $E_1 = -13.6 \times \frac{3^2}{1^2} \; eV = -13.6 \times 9 \; eV = -122.4 \; eV$.
The energy required to remove the electron (ionization energy) is the energy needed to bring the electron from the ground state to infinity $(E_{\infty} = 0)$.
Therefore,Ionization Energy $= E_{\infty} - E_1 = 0 - (-122.4 \; eV) = 122.4 \; eV$.

(B) In Bohr's model,the potential energy of an electron in the $n^{th}$ orbit is given by $PE = -\frac{ke^2}{r_n}$,where $r_n \propto n^2$. As the principal quantum number $n$ increases,the radius $r_n$ increases,making the potential energy $PE$ less negative (i.e.,it increases).
The total energy is given by $TE = -\frac{13.6}{n^2} \text{ eV}$. As $n$ increases,the denominator $n^2$ increases,making the total energy $TE$ less negative (i.e.,it increases towards zero).
Therefore,both $PE$ and $TE$ increase as the electron moves to a higher energy level.

(B) The radius of the $n^{th}$ orbit in a hydrogen-like atom is given by the formula: $r_n = r_0 \times \frac{n^2}{Z}$, where $r_0 = 0.53 \ \mathring{A}$ is the radius of the first orbit of the hydrogen atom.
For a helium ion $(He^+)$, the atomic number $Z = 2$.
For the second orbit, $n = 2$.
Substituting these values into the formula:
$r_2 = 0.53 \times \frac{2^2}{2} \ \mathring{A}$
$r_2 = 0.53 \times \frac{4}{2} \ \mathring{A}$
$r_2 = 0.53 \times 2 \ \mathring{A} = 1.06 \ \mathring{A}$.
Thus, the correct option is $B$.

(A) According to the Rydberg formula for hydrogen-like ions,the wavelength $\lambda$ of a spectral line is given by: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For a hydrogen atom,$Z = 1$,so $\frac{1}{\lambda_H} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the ionic atom,$\frac{1}{\lambda_{ion}} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Given that $\lambda_{ion} = \frac{1}{4} \lambda_H$,we have $\frac{1}{\lambda_{ion}} = 4 \frac{1}{\lambda_H}$.
Substituting the expressions,we get $R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 4 R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
This simplifies to $Z^2 = 4$,which gives $Z = 2$.
The element with atomic number $Z = 2$ is Helium $(He)$. Since it is an ion,it is $He^+$. Therefore,the correct option is $A$.

(C) For the third line of the Balmer series,the transition is from $n_2 = 5$ to $n_1 = 2$.
Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
Given $\lambda = 108.5 \ nm = 108.5 \times 10^{-9} \ m$ and $R \approx 1.097 \times 10^7 \ m^{-1}$.
Substituting the values: $\frac{1}{108.5 \times 10^{-9}} = (1.097 \times 10^7) Z^2 \left[ \frac{1}{2^2} - \frac{1}{5^2} \right]$.
$\frac{1}{108.5 \times 10^{-9}} = (1.097 \times 10^7) Z^2 \left[ \frac{1}{4} - \frac{1}{25} \right] = (1.097 \times 10^7) Z^2 \left[ \frac{21}{100} \right]$.
Solving for $Z^2$: $Z^2 \approx \frac{100}{108.5 \times 10^{-9} \times 1.097 \times 10^7 \times 21} \approx \frac{100}{25} = 4$.
Thus,$Z = 2$.
The ground state energy of a hydrogen-like ion is given by $E = -13.6 \frac{Z^2}{n^2} \ eV$.
For the ground state,$n = 1$,so $E = -13.6 \times \frac{2^2}{1^2} = -13.6 \times 4 = -54.4 \ eV$.
The magnitude of the ground state energy is $54.4 \ eV$.

(A) The binding energy of an electron in the $n = 1$ orbit of a hydrogen atom is the energy required to remove the electron from the atom to infinity,which is $13.6\, eV$.
By definition,the ionization energy is the minimum energy required to remove an electron from its ground state to an infinite distance from the nucleus.
Therefore,the ionization energy is equal to the magnitude of the binding energy.
Ionization energy = $13.6\, eV$.

(B) The ionization energy $E$ of a hydrogen atom is related to the Rydberg constant $R$ by the formula $E = Rch$,where $c$ is the speed of light and $h$ is Planck's constant.
Rearranging for $R$,we get $R = \frac{E}{ch}$.
Given $E = 13.6 \, eV = 13.6 \times 1.6 \times 10^{-19} \, J$,$c = 3 \times 10^8 \, m/s$,and $h = 6.6 \times 10^{-34} \, J \cdot s$.
Substituting these values:
$R = \frac{13.6 \times 1.6 \times 10^{-19}}{3 \times 10^8 \times 6.6 \times 10^{-34}}$
$R = \frac{21.76 \times 10^{-19}}{19.8 \times 10^{-26}}$
$R \approx 1.098 \times 10^7 \, m^{-1}$.
Thus,the order of magnitude of $R$ is $10^7 \, m^{-1}$.

(B) Bohr's model of the atom is based on several postulates. One of the fundamental postulates is the quantization of angular momentum. Bohr proposed that an electron can revolve only in those orbits for which its angular momentum is an integral multiple of $h / (2\pi)$,where $h$ is Planck's constant. This is expressed as $L = mvr = n(h / 2\pi)$,where $n = 1, 2, 3, ...$. Thus,Bohr used the principle of quantization of angular momentum to explain the stability of atoms and the observed spectral lines.

(B) The radius of an orbit in a hydrogen atom is given by $r_n = r_0 n^2$,where $r_0$ is the radius of the ground state $(n=1)$.
Therefore,the ratio of the final radius $r_f$ to the initial radius $r_i$ is given by $\frac{r_f}{r_i} = \left( \frac{n_f}{n_i} \right)^2$.
Given $r_i = 5.3 \times 10^{-11} \ m$ at $n_i = 1$ and $r_f = 21.2 \times 10^{-11} \ m$ at $n_f = n$.
Substituting the values: $\frac{21.2 \times 10^{-11}}{5.3 \times 10^{-11}} = \left( \frac{n}{1} \right)^2$.
$4 = n^2$.
Taking the square root,we get $n = 2$.

(B) The radius of an electron orbit in a hydrogen atom is given by the formula $r_n = n^2 a_0$,where $n$ is the principal quantum number and $a_0$ is the Bohr radius.
For the ground state,$n = 1$.
For the first excited state,$n = 2$.
Substituting $n = 2$ into the formula,we get $r_2 = (2)^2 a_0 = 4 a_0$.
Therefore,the radius of the first excited state is $4$ times the Bohr radius.

(C) The radius of the $n^{th}$ orbit of an electron in a hydrogen-like atom is given by the formula $r_n = r_0 n^2 / Z$.
For a hydrogen atom,the atomic number $Z = 1$,which simplifies the expression to $r_n = r_0 n^2$.
For the third orbit,we substitute $n = 3$ into the formula.
Therefore,the radius of the third orbit is $r_3 = r_0 \times (3)^2 = 9r_0$.

(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For a fixed transition ($n_i = 4$ to $n_f = 2$),the wavelength $\lambda$ is inversely proportional to the square of the atomic number $Z$,i.e.,$\lambda \propto \frac{1}{Z^2}$.
For hydrogen $(H)$,$Z = 1$,so $\lambda_H = 20.397 \, cm$.
For helium ion $(He^+)$,$Z = 2$,so $\lambda_{He^+} = \frac{\lambda_H}{Z^2} = \frac{20.397}{2^2} = \frac{20.397}{4} = 5.099 \, cm$.

(C) The excitation potential is defined as the excitation energy divided by the elementary charge $e$.
The minimum excitation energy corresponds to the transition of an electron from the ground state $(n = 1)$ to the first excited state $(n = 2)$.
The energy of the $n$-th orbit in a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \, eV$.
For $n = 1$,$E_1 = -13.6 \, eV$.
For $n = 2$,$E_2 = -\frac{13.6}{4} = -3.4 \, eV$.
The minimum excitation energy is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \, eV$.
Therefore,the minimum excitation potential is $10.2 \, V$.

(A) According to Bohr's model for a hydrogen atom:
$(I)$ The orbital speed $v_n$ is given by $v_n \propto \frac{1}{n}$. As $n$ increases (moving away from the nucleus),$v_n$ decreases. Thus,statement $(I)$ is true.
$(II)$ The radius of the $n^{th}$ orbit is $r_n \propto n^2$. Statement $(II)$ is false because it is proportional to the square of the principal quantum number.
$(III)$ The frequency of revolution $f$ is given by $f = \frac{v}{2\pi r}$. Since $v \propto \frac{1}{n}$ and $r \propto n^2$,we have $f \propto \frac{1/n}{n^2} = \frac{1}{n^3}$. Thus,statement $(III)$ is true.
$(IV)$ The electrostatic binding force is $F = \frac{ke^2}{r^2}$. Since $r$ increases as $n$ increases,the force $F$ decreases. Thus,statement $(IV)$ is false.
Therefore,statements $(I)$ and $(III)$ are correct.

(D) The radius of the $n^{th}$ orbit is given by $r_n \propto n^2$.
The velocity of an electron in the $n^{th}$ orbit is given by $v_n \propto 1/n$.
Angular velocity $\omega_n$ is defined as $\omega_n = v_n / r_n$.
Substituting the proportionalities: $\omega_n \propto (1/n) / n^2$.
Therefore,$\omega_n \propto 1/n^3$.

(C) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.
The change in angular momentum $\Delta L$ when the electron moves from orbit $n_1 = 4$ to $n_2 = 5$ is:
$\Delta L = L_2 - L_1 = \frac{n_2 h}{2\pi} - \frac{n_1 h}{2\pi} = \frac{h}{2\pi}(n_2 - n_1)$.
Substituting the given values:
$\Delta L = \frac{6.6 \times 10^{-34}}{2 \times 3.14} \times (5 - 4)$.
$\Delta L = \frac{6.6 \times 10^{-34}}{6.28} \times 1$.
$\Delta L \approx 1.05 \times 10^{-34} \ J \cdot s$.

(A) The concept of stationary orbits was proposed by Niels Bohr in $1913$,which is known as the Bohr model of the atom. According to this model,electrons revolve around the nucleus only in certain stable orbits without radiating energy. These orbits are called stationary orbits.