Bohr's Model of Hydrogen Atom Questions in English

(C) The energy of an electron in the $n^{th}$ orbit of a Hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \, eV$.
Given $E_n = -1.5 \, eV$,we have $-1.5 = -\frac{13.6}{n^2}$.
$n^2 = \frac{13.6}{1.5} \approx 9.06$,which implies $n \approx 3$.
According to Bohr's postulate,the angular momentum $L$ is given by $L = n \frac{h}{2\pi}$.
Substituting $n = 3$ and $h = 6.626 \times 10^{-34} \, J \cdot s$:
$L = 3 \times \frac{6.626 \times 10^{-34}}{2 \times 3.14159} \approx 3 \times 1.054 \times 10^{-34} \approx 3.16 \times 10^{-34} \, J \cdot s$.
Rounding to the provided options,the correct value is $3.15 \times 10^{-34} \, J \cdot s$.

(B) The time period of revolution $T$ is given by $T = \frac{2\pi r}{v}$.
The radius of the $n^{th}$ Bohr orbit is $r = \frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2}$.
The velocity of the electron in the $n^{th}$ Bohr orbit is $v = \frac{Z e^2}{2 \varepsilon_0 n h}$.
Substituting these into the expression for $T$:
$T = \frac{2\pi (\frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2})}{(\frac{Z e^2}{2 \varepsilon_0 n h})} = \frac{4 \varepsilon_0^2 n^3 h^3}{m Z^2 e^4}$.
Since $h, \varepsilon_0, m,$ and $e$ are constants,we have $T \propto \frac{n^3}{Z^2}$.

(D) 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 first orbit $(n=1)$ is ${r_0}$,we have ${r_1} = {r_0}$.
For the fourth orbit $(n=4)$,the radius ${r_4}$ is calculated as:
${r_4} = {r_1} \times {n^2} = {r_0} \times {4^2}$.
${r_4} = {r_0} \times 16 = 16{r_0}$.

(C) According to the Bohr model,the radius of the $n$-th orbit is $r_n \propto n^2$ and the velocity of the electron in the $n$-th orbit is $v_n \propto 1/n$.
The time period of revolution $T$ is given by $T = \frac{2\pi r_n}{v_n}$.
Substituting the proportionalities,we get $T \propto \frac{n^2}{1/n} = n^3$.
Therefore,the ratio of the time periods for $n = 2$ and $n = 1$ is $\frac{T_2}{T_1} = \left(\frac{2}{1}\right)^3 = \frac{8}{1}$.
Thus,the ratio is $8 : 1$.

(B) The radius of the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $r_n = r_0 \frac{n^2}{Z}$, where $r_0$ is the radius of the first Bohr orbit of hydrogen $(0.5 \ \mathring{A})$, $n$ is the principal quantum number, and $Z$ is the atomic number.
For the third excited state of $He^+$, the principal quantum number is $n = 3 + 1 = 4$.
The atomic number of Helium is $Z = 2$.
Substituting these values into the formula:
$r_4 = 0.5 \times \frac{4^2}{2} = 0.5 \times \frac{16}{2} = 0.5 \times 8 = 4 \ \mathring{A}$.
Thus, the correct option is $B$.

(D) The speed of an electron in the $n^{th}$ Bohr orbit of a hydrogen-like atom is given by $v_n = \frac{2\pi Z e^2}{nh}$.
For the first Bohr orbit of hydrogen,we have $n = 1$ and $Z = 1$.
Substituting these values,the speed of the electron is $v = \frac{2\pi e^2}{h}$.
We are asked to find the ratio of this speed to the speed of light $c$.
Therefore,the ratio is $\frac{v}{c} = \frac{2\pi e^2}{hc}$.

(A) According to Bohr's postulate,the angular momentum of an electron is quantized as: $mvr = \frac{nh}{2\pi}$.
From this,the orbital velocity $v$ is given by: $v = \frac{nh}{2\pi mr}$.
The centripetal (orbital) acceleration $a$ is defined as: $a = \frac{v^2}{r}$.
Substituting the expression for $v$ into the acceleration formula:
$a = \frac{(\frac{nh}{2\pi mr})^2}{r} = \frac{n^2 h^2}{4 \pi^2 m^2 r^2 \cdot r} = \frac{n^2 h^2}{4 \pi^2 m^2 r^3}$.
Thus,the correct option is $A$.

(B) The radius of the $n^{th}$ orbit of an electron in a hydrogen atom is given by the formula $r_n = n^2 r_1$,where $r_1 = 0.53 \; \mathring{A}$ is the radius of the first orbit.
For the third orbit,$n = 3$.
Substituting the value of $n$ into the formula:
$r_3 = 3^2 \times r_1$
$r_3 = 9 \times 0.53 \; \mathring{A}$
$r_3 = 4.77 \; \mathring{A}$.

(A) According to Bohr's postulate for the hydrogen atom,an electron can revolve only in those orbits for which its angular momentum $(L)$ is an integral multiple of $\frac{h}{2\pi}$.
Mathematically,this is expressed as: $L = n \cdot \frac{h}{2\pi}$,where $n = 1, 2, 3, ...$ is the principal quantum number,$h$ is Planck's constant,and $L$ is the angular momentum of the electron.
Therefore,the correct option is $A$.

(C) The orbital quantum number $l$ can take integer values ranging from $0$ to $n - 1$.
For the principal quantum number $n = 3$,the possible values of $l$ are $0, 1, 2, ..., (3 - 1)$.
Therefore,$l = 0, 1, 2$.
This corresponds to the $3s, 3p,$ and $3d$ orbitals respectively.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
Given $E_n = -3.4 \text{ eV}$,we have $-\frac{13.6}{n^2} = -3.4$,which implies $n^2 = \frac{13.6}{3.4} = 4$,so $n = 2$.
According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.
Substituting $n = 2$ and $h = 6.626 \times 10^{-34} \text{ J s}$,we get $L = \frac{2h}{2\pi} = \frac{h}{\pi}$.
$L = \frac{6.626 \times 10^{-34}}{3.1416} \approx 2.11 \times 10^{-34} \text{ J s}$.

(C) The Rydberg formula for the wavelength $\lambda$ of a transition between energy levels $n_2$ and $n_1$ is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Since the transition $2 \to 1$ is the same for all three,the term $\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ is constant.
Therefore,$\lambda \propto \frac{1}{Z^2}$.
For $Li^{++}$,$Z = 3$,so $\lambda_{Li^{++}} \propto \frac{1}{3^2} = \frac{1}{9}$.
For $He^{+}$,$Z = 2$,so $\lambda_{He^{+}} \propto \frac{1}{2^2} = \frac{1}{4}$.
For $H$,$Z = 1$,so $\lambda_{H} \propto \frac{1}{1^2} = \frac{1}{1}$.
Thus,$\lambda_{Li^{++}} : \lambda_{He^{+}} : \lambda_{H} = \frac{1}{9} : \frac{1}{4} : 1$.
Multiplying by $36$,we get $4 : 9 : 36$.

(A) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.
For the first orbit $(n_1 = 1)$,$E_1 = -13.6 \ eV$.
For the second orbit $(n_2 = 2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \ eV$.
The energy difference is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$.
Converting this to Joules: $\Delta E = 10.2 \times 1.6 \times 10^{-19} \ J = 1.632 \times 10^{-18} \ J$.
Using the relation $\Delta E = \frac{hc}{\lambda}$,where $h = 6.63 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$:
$\lambda = \frac{hc}{\Delta E} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.632 \times 10^{-18}} \ m \approx 1.215 \times 10^{-7} \ m$.

(A) The speed of an electron in the $n$-th orbit of a hydrogen atom is given by $v_n = \frac{e^2}{2\varepsilon_0 nh}$.
For the ground state,$n = 1$,so $v_1 = \frac{e^2}{2\varepsilon_0 h}$.
The ratio of this speed to the velocity of light $c$ is given by $\frac{v_1}{c} = \frac{e^2}{2\varepsilon_0 hc}$.
This ratio is also known as the fine-structure constant,denoted by $\alpha$.

(D) According to Bohr's quantization condition,the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr_n = \frac{nh}{2\pi}$.
Since the momentum $p = mv = \frac{h}{\lambda}$,we can substitute this into the equation:
$\frac{h}{\lambda} \times r_n = \frac{nh}{2\pi}$.
Rearranging for the wavelength $\lambda$,we get $\lambda = \frac{2\pi r_n}{n}$.
For the first Bohr orbit,$n = 1$,so $\lambda = 2\pi r_1$.
Since the circumference of the first orbit is $2\pi r_1$,the de-Broglie wavelength is equal to the circumference of the first orbit.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -\frac{13.6 Z^2}{n^2} \; eV$.
For a hydrogen atom,$Z = 1$ and $n = 1$,so the ground state energy is $E_1 = -13.6 \; eV$. The ionisation energy is the energy required to remove the electron,which is $13.6 \; eV$.
For a singly ionised helium atom $(He^+)$,the atomic number $Z = 2$. The ground state energy is $E_1 = -\frac{13.6 \times (2)^2}{(1)^2} = -13.6 \times 4 = -54.4 \; eV$.
Therefore,the ionisation energy required to remove the electron from the ground state of $He^+$ is $54.4 \; eV$.

(B) The frequency of a spectral line in a hydrogen-like atom is given by the Rydberg formula: $\nu = R c Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the $1^{st}$ line of the Balmer series,$n_1 = 2$ and $n_2 = 3$.
Thus,$\nu \propto Z^2$.
For hydrogen $({H_2})$,$Z = 1$,so $\nu_{H} = k(1)^2 = \nu_0$.
For singly ionized helium $({He^+})$,$Z = 2$,so $\nu_{He} = k(2)^2 = 4k = 4\nu_0$.
Therefore,the frequency of the corresponding line for ${He^+}$ is $4\nu_0$.

(B) For an electron in a $3d$ orbital,the principal quantum number $n$ is $3$.
For a $d$ subshell,the azimuthal quantum number $l$ is always $2$.
The magnetic quantum number $m_l$ can take any integer value from $-l$ to $+l$,which means $m_l \in \{-2, -1, 0, +1, +2\}$.
The spin quantum number $m_s$ can be either $+1/2$ or $-1/2$.
Comparing these requirements with the given options,option $B$ $(n=3, l=2, m_l=+2, m_s=-1/2)$ satisfies all these conditions.

(A) According to the Bohr model of the hydrogen atom,the radius of the $n^{th}$ orbit is given by the formula $r_n = n^2 a_0$,where $a_0$ is the radius of the first orbit $(n=1)$.
For the second orbit,$n = 2$.
Substituting the value of $n$ into the formula: $r_2 = (2)^2 a_0 = 4a_0$.
Therefore,the radius of the second orbit is $4a_0$.

(D) The radius of the $n^{th}$ orbit of a hydrogen-like atom is given by the formula $r_n = a_0 \frac{n^2}{Z}$,where $a_0$ is the Bohr radius,$n$ is the principal quantum number,and $Z$ is the atomic number.
For the first orbit,$n = 1$,so $r_1 \propto \frac{1}{Z}$.
Comparing the atomic numbers $(Z)$ for the given systems:
- Hydrogen atom $(H)$: $Z = 1$
- Deuterium atom $(D)$: $Z = 1$
- Single ionized helium $(He^+)$: $Z = 2$
- Doubly ionized lithium $(Li^{2+})$: $Z = 3$
Since the radius $r$ is inversely proportional to $Z$,the system with the highest atomic number will have the minimum radius.
Therefore,doubly ionized lithium $(Z = 3)$ has the minimum radius.

(B) According to Bohr's postulates for the hydrogen atom,an electron can revolve only in those orbits for which its angular momentum $(L)$ is an integral multiple of $h / 2\pi$.
The condition is given by: $L = n(h / 2\pi)$,where $n = 1, 2, 3, ...$ is the principal quantum number and $h$ is Planck's constant.
This is known as the Bohr's quantization condition for angular momentum.

(A) According to Bohr's hypothesis,an electron can revolve only in those orbits in which its angular momentum is an integral multiple of $\frac{h}{2 \pi}$,where $h$ is Planck's constant.
In these orbits,the angular momentum of the electron can have magnitudes such as $\frac{h}{2 \pi}, \frac{2h}{2 \pi}, \frac{3h}{2 \pi}, \dots$ etc.,but never values like $\frac{1.5h}{2 \pi}, \frac{2.5h}{2 \pi}, \dots$ etc.
This condition is known as the quantization of angular momentum,which is a fundamental postulate of the Bohr model.

(C) The radius of the $n^{th}$ orbit for a hydrogen-like atom is given by the formula:
$r_n = r_0 \left( \frac{n^2}{Z} \right)$
where $r_0$ is the radius of the first orbit of the hydrogen atom and $Z$ is the atomic number.
For Beryllium $(Be^{+++})$,the atomic number $Z = 4$.
We want the orbital radius $r_n$ to be equal to the ground state radius of hydrogen,which is $r_0$.
Setting $r_n = r_0$,we get:
$r_0 = r_0 \left( \frac{n^2}{4} \right)$
$1 = \frac{n^2}{4}$
$n^2 = 4$
$n = 2$
Therefore,the $n = 2$ state of triply ionised Beryllium has the same orbital radius as the ground state of hydrogen.

(D) The radius of the $n^{th}$ orbit in a hydrogen atom is given by $r_n = a_0 n^2$, where $a_0$ is the Bohr radius and $n$ is the principal quantum number.
The area $A$ of the orbit is given by $A = \pi r_n^2 = \pi (a_0 n^2)^2 = \pi a_0^2 n^4$.
Thus, the area is proportional to the fourth power of the principal quantum number: $A_n \propto n^4$.
For the ground state, $n_1 = 1$. For the first excited state, $n_2 = 2$.
The ratio of the area of the first excited state $(A_2)$ to the ground state $(A_1)$ is:
$\frac{A_2}{A_1} = \left( \frac{n_2}{n_1} \right)^4 = \left( \frac{2}{1} \right)^4 = \frac{16}{1}$.
Therefore, the ratio is $16 : 1$.

(D) For an electron revolving in an orbit around a nucleus,the total energy $E$,kinetic energy $K.E.$,and potential energy $P.E.$ are related as follows:
$K.E. = -E$
$P.E. = 2E$
From these relations,we can see that $K.E. = -\frac{1}{2} P.E.$
In terms of magnitude,the kinetic energy is half of the potential energy.
$|K.E.| = \frac{1}{2} |P.E.|$
Therefore,the kinetic energy is half of its $P.E.$

(C) The total energy $(E_n)$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 / n^2 \, eV$.
For the first Bohr orbit,$n = 1$,so $E_1 = -13.6 \, eV$.
The kinetic energy $(K.E.)$ of an electron is equal to the negative of its total energy,i.e.,$K.E. = -E_n$.
Therefore,$K.E. = -(-13.6 \, eV) = +13.6 \, eV$.
Thus,the correct option is $C$.

(C) According to Bohr's theory of the atom,electrons can revolve only in those orbits in which their angular momentum is an integral multiple of $\frac{h}{2 \pi}$ (where $h$ is Planck's constant). Hence,angular momentum is quantized: $L = n \frac{h}{2 \pi}$.
Also,the energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E = -\frac{13.6}{n^2} \, \text{eV}$.
Since $n$ can only take integer values $(n = 1, 2, 3, ...)$,the energy levels are also quantized. Therefore,both energy and angular momentum are quantized.

(C) According to Bohr's theory,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$ and the velocity of the electron in the $n^{th}$ orbit is given by $v_n \propto \frac{1}{n}$.
The time period $(T)$ is defined as the ratio of the circumference of the orbit to the velocity of the electron:
$T = \frac{2\pi r_n}{v_n}$
Substituting the proportionalities:
$T \propto \frac{n^2}{1/n}$
$T \propto n^3$
Therefore,the correct relation is $T \propto n^3$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: ${E_n} = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For a hydrogen atom,the atomic number $Z_H = 1$,so the energy is ${E_n} = -13.6 \times \frac{1^2}{n^2} = -\frac{13.6}{n^2} \text{ eV}$.
For a singly ionized helium atom $(He^+)$,the atomic number $Z_{He} = 2$. The energy in the $n^{th}$ orbit is given by: ${E'_{n}} = -13.6 \times \frac{Z_{He}^2}{n^2} = -13.6 \times \frac{2^2}{n^2} = 4 \times \left( -\frac{13.6}{n^2} \right)$.
Substituting the expression for the hydrogen atom energy ${E_n}$,we get: ${E'_{n}} = 4{E_n}$.

(A) The total energy of an electron in the $n$-th orbit is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
For $n = 2$,$E_2 = -\frac{13.6}{4} = -3.4 \text{ eV}$.
For $n = 1$,$E_1 = -\frac{13.6}{1} = -13.6 \text{ eV}$.
We know that kinetic energy $K = -E$ and potential energy $U = 2E$.
At $n = 2$: $K_2 = 3.4 \text{ eV}$ and $U_2 = -6.8 \text{ eV}$.
At $n = 1$: $K_1 = 13.6 \text{ eV}$ and $U_1 = -27.2 \text{ eV}$.
Ratio of kinetic energies: $\frac{K_1}{K_2} = \frac{13.6}{3.4} = 4$.
Ratio of potential energies: $\frac{U_1}{U_2} = \frac{-27.2}{-6.8} = 4$.
Thus,both kinetic energy and potential energy increase by a factor of $4$.

(D) In $1913$,Niels Bohr proposed a model for the hydrogen atom. The Bohr model is based on the following assumptions:
$1$. The nucleus is of infinite mass and is at rest.
$2$. Electrons in a quantized orbit will not radiate energy.
$3$. The mass of the electron remains constant.
Since all these statements are fundamental postulates of Bohr's model,the correct option is $(d)$.

(C) The radius of the $n^{th}$ Bohr orbit is given by the formula $r_n = r_0 n^2$,where $r_0$ is the radius of the first orbit $(n=1)$.
Given that the radius of the first orbit $(n=1)$ is $r$,we have $r = r_0(1)^2$,so $r_0 = r$.
For the $2^{nd}$ Bohr orbit $(n=2)$,the radius $r_2$ is given by $r_2 = r_0(2)^2 = 4r_0$.
Substituting $r_0 = r$,we get $r_2 = 4r$.

(C) The speed of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $v = \frac{e^2}{2\varepsilon_0 nh}$.
For the ground state,the principal quantum number $n = 1$.
Substituting $n = 1$,we get the speed $v = \frac{e^2}{2\varepsilon_0 h}$.
To find the ratio of the speed of the electron $(v)$ to the speed of light $(c)$,we calculate $\frac{v}{c} = \frac{e^2}{2\varepsilon_0 ch}$.
Substituting the physical constants: $e = 1.6 \times 10^{-19} \ C$,$\varepsilon_0 = 8.85 \times 10^{-12} \ F/m$,$c = 3 \times 10^8 \ m/s$,and $h = 6.63 \times 10^{-34} \ J\cdot s$.
$\frac{v}{c} = \frac{(1.6 \times 10^{-19})^2}{2 \times 8.85 \times 10^{-12} \times 3 \times 10^8 \times 6.63 \times 10^{-34}} \approx \frac{1}{137}$.
Thus,the ratio is $1/137$.

(B) The recoil momentum of the $H$ atom is equal to the momentum of the emitted photon.
The momentum of a photon is given by $p = \frac{h}{\lambda}$.
Using the Rydberg formula for the wavenumber: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Substituting this into the momentum equation: $p = hR \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Given $n_1 = 1$ and $n_2 = 4$,we have:
$p = hR \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = hR \left( 1 - \frac{1}{16} \right) = \frac{15hR}{16}$.
Using $h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s}$ and $R = 1.097 \times 10^7 \text{ m}^{-1}$:
$p = \frac{15 \times (6.63 \times 10^{-34}) \times (1.097 \times 10^7)}{16} \approx 6.8 \times 10^{-27} \text{ kg} \cdot \text{m/s}$ (or $\text{N} \cdot \text{s}$).

(A) The energy of the ground state of a hydrogen atom is $E_1 = -13.6 \ eV$.
When it absorbs $10.2 \ eV$ of energy,its new energy becomes $E = -13.6 \ eV + 10.2 \ eV = -3.4 \ eV$.
This corresponds to the first excited state,where the principal quantum number $n = 2$.
The orbital angular momentum $L$ of an electron in a hydrogen atom is given by $L = \frac{nh}{2\pi}$.
In the ground state $(n_1 = 1)$,the angular momentum is $L_1 = \frac{1 \cdot h}{2\pi}$.
In the first excited state $(n_2 = 2)$,the angular momentum is $L_2 = \frac{2 \cdot h}{2\pi} = \frac{h}{\pi}$.
The increase in orbital angular momentum is $\Delta L = L_2 - L_1 = \frac{2h}{2\pi} - \frac{h}{2\pi} = \frac{h}{2\pi}$.
Substituting the value of $h = 6.6 \times 10^{-34} \ J \cdot s$:
$\Delta L = \frac{6.6 \times 10^{-34}}{2 \times 3.14} \approx \frac{6.6 \times 10^{-34}}{6.28} \approx 1.05 \times 10^{-34} \ J \cdot s$.

(A) The energy of a photon emitted during a transition from $n_2$ to $n_1$ is given by $\Delta E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Since the transition $n=2$ to $n=1$ is the same for all,$\Delta E \propto Z^2$.
We know that $\Delta E = \frac{hc}{\lambda}$,therefore $\frac{hc}{\lambda} \propto Z^2$,which implies $\lambda Z^2 = \text{constant}$.
For Hydrogen $(Z=1)$,$\lambda_1 Z_1^2 = \lambda_1 (1)^2 = \lambda_1$.
For Deuterium $(Z=1)$,$\lambda_2 Z_2^2 = \lambda_2 (1)^2 = \lambda_2$.
For Helium $(Z=2)$,$\lambda_3 Z_3^2 = \lambda_3 (2)^2 = 4\lambda_3$.
For Lithium $(Z=3)$,$\lambda_4 Z_4^2 = \lambda_4 (3)^2 = 9\lambda_4$.
Equating these,we get $\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$.

(D) According to Bohr's quantization condition,the angular momentum $L$ is given by $mvr = \frac{h}{2\pi}$.
Since $v = r\omega = r(2\pi\nu)$,where $\nu$ is the frequency of revolution:
$m(r \cdot 2\pi\nu)r = \frac{h}{2\pi}$
$\Rightarrow \nu = \frac{h}{4\pi^2 mr^2}$
Substituting the values $h = 6.6 \times 10^{-34} \ J \cdot s$,$m = 9.1 \times 10^{-31} \ kg$,and $r = 0.53 \times 10^{-10} \ m$:
$\nu = \frac{6.6 \times 10^{-34}}{4 \times (3.14)^2 \times 9.1 \times 10^{-31} \times (0.53 \times 10^{-10})^2}$
$\nu \approx 6.5 \times 10^{15} \ \text{rev/sec}$.
Thus,the order of magnitude is $10^{15}$.

(A) The radius of the $n$-th orbit in Bohr's model is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2}$. Since $r \propto 1/m$,if the mass $m$ is doubled,the new radius $r_0$ becomes $r_0 = a_0/2$.
The energy of the $n$-th orbit is given by $E_n = -\frac{m e^4}{8 \epsilon_0^2 n^2 h^2}$. Since $E \propto m$,if the mass $m$ is doubled,the new energy $E_0$ becomes $E_0 = 2 \times (-13.6 \text{ eV}) = -27.2 \text{ eV}$.
Thus,the correct option is $A$.

(A) The electronic configuration of iodine $(Z=53)$ is $2, 8, 18, 18, 7$. The valence shell is the $n=5$ shell.
Using the Bohr radius formula for a hydrogen-like atom,the radius is given by $r_n = (0.053 \times 10^{-9} \ m) \frac{n^2}{Z}$.
Substituting $n=5$ and $Z=53$:
$r_5 = (0.053 \times 10^{-9} \ m) \times \frac{5^2}{53}$
$r_5 = (0.053 \times 10^{-9} \ m) \times \frac{25}{53}$
$r_5 \approx 0.025 \times 10^{-9} \ m = 2.5 \times 10^{-11} \ m$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2} \, eV$.
For $Li^{++}$,the atomic number $Z = 3$.
The energy required for the transition from $n_1 = 1$ to $n_2 = 3$ is:
$\Delta E = E_3 - E_1 = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
$\Delta E = 13.6 \times 3^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 13.6 \times 9 \times \left( 1 - \frac{1}{9} \right) = 13.6 \times 9 \times \frac{8}{9} = 108.8 \, eV$.
Using the relation $\Delta E = \frac{hc}{\lambda}$,where $h = 6.626 \times 10^{-34} \, J \cdot s$,$c = 3 \times 10^8 \, m/s$,and $1 \, eV = 1.602 \times 10^{-19} \, J$:
$\lambda = \frac{hc}{\Delta E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{108.8 \times 1.602 \times 10^{-19}} \approx 1.1374 \times 10^{-8} \, m$.
$\lambda = 113.74 \, \mathring{A}$.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Substituting the values: $E = \frac{12400 \, eV \cdot \mathring{A}}{970.6 \, \mathring{A}} \approx 12.77 \, eV$.
The energy of the $n$-th state in a Hydrogen atom is $E_n = -\frac{13.6}{n^2} \, eV$.
The energy required to excite from ground state $(n=1)$ to state $n$ is $\Delta E = 13.6 \left( 1 - \frac{1}{n^2} \right)$.
Setting $\Delta E = 12.77 \, eV$,we get $1 - \frac{1}{n^2} = \frac{12.77}{13.6} \approx 0.939$.
$\frac{1}{n^2} = 1 - 0.939 = 0.061 \Rightarrow n^2 \approx 16 \Rightarrow n = 4$.
The number of emission lines for an electron transitioning from state $n$ to lower states is given by $N = \frac{n(n-1)}{2}$.
For $n=4$,$N = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6$.

(C) In a hydrogen atom,the energy of the $n^{th}$ level is given by $E_n = -\frac{Rhc}{n^2}$.
Since the energy $E_n$ is directly proportional to the mass of the orbiting particle $(E_n \propto m)$,and the new particle has a mass $m' = 2m_e$,the energy levels for this hypothetical atom become $E'_n = -\frac{2Rhc}{n^2}$.
The longest wavelength photon corresponds to the minimum energy transition,which occurs between $n = 3$ and $n = 2$.
The energy difference is $\Delta E = E'_3 - E'_2 = -\frac{2Rhc}{3^2} - (-\frac{2Rhc}{2^2}) = 2Rhc \left( \frac{1}{4} - \frac{1}{9} \right)$.
Simplifying this,$\Delta E = 2Rhc \left( \frac{5}{36} \right) = \frac{5Rhc}{18}$.
Since $\Delta E = \frac{hc}{\lambda}$,we have $\frac{hc}{\lambda} = \frac{5Rhc}{18}$.
Therefore,$\lambda = \frac{18}{5R}$.

(D) According to the Bohr model for a hydrogen atom:
$1$. The radius of the $n^{th}$ orbit is given by $R_n = \frac{\varepsilon_0 n^2 h^2}{\pi m Z e^2}$,which implies $R \propto n^2$.
$2$. The speed of the electron in the $n^{th}$ orbit is given by $v_n = \frac{Z e^2}{2 \varepsilon_0 n h}$,which implies $v \propto \frac{1}{n}$.
$3$. The total energy of the electron in the $n^{th}$ orbit is given by $E_n = -\frac{m Z^2 e^4}{8 \varepsilon_0^2 n^2 h^2}$,which implies $E \propto \frac{1}{n^2}$.
Now,let us check the proportionality for the given options:
- For $vR$: $v \cdot R \propto (\frac{1}{n}) \cdot (n^2) = n$.
Therefore,the quantity $vR$ is proportional to the quantum number $n$.

(B) The second excited state corresponds to the principal quantum number $n = 3$.
According to Bohr's quantization condition,the angular momentum $l$ is given by $l = n \left( \frac{h}{2\pi} \right)$.
Since both are in the same state $n = 3$,their angular momenta are equal: $l_H = l_{Li} = 3 \left( \frac{h}{2\pi} \right)$.
The energy of an electron in a hydrogen-like atom is given by $E = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
For Hydrogen $(Z_H = 1)$ and $Li^{++}$ $(Z_{Li} = 3)$,the energies are $E_H = -13.6 \frac{1^2}{3^2}$ and $E_{Li} = -13.6 \frac{3^2}{3^2}$.
Taking the magnitude,$|E_H| = \frac{13.6}{9} \text{ eV}$ and $|E_{Li}| = 13.6 \text{ eV}$.
Clearly,$|E_H| < |E_{Li}|$. Thus,option $(b)$ is correct.

(D) The wavelength of the spectral lines is given by the Rydberg formula:
$\frac{1}{\lambda} = R_M \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
where $R_M = \frac{R_{\infty}}{1 + \frac{m}{M}}$,$m$ is the mass of the electron,and $M$ is the mass of the nucleus.
Since the mass of the deuterium nucleus $(M_D \approx 2M_H)$ is different from the mass of the hydrogen nucleus $(M_H)$,the Rydberg constant $R_M$ for deuterium is slightly different from that of hydrogen.
Consequently,the wavelengths of the spectral lines for deuterium differ from those of hydrogen.
Therefore,the correct option is $(d)$.

(A) The potential energy $U$ is given by $U = eV = eV_0 \ln(r/r_0)$.
The force $F$ is given by $F = -dU/dr = -d/dr(eV_0 \ln(r/r_0)) = -eV_0/r$. The magnitude of the force is $F = eV_0/r$.
This force provides the necessary centripetal force for circular motion: $mv^2/r = eV_0/r$.
Simplifying this,we get $mv^2 = eV_0$,which implies $v = \sqrt{eV_0/m}$. Note that $v$ is independent of $r$ and $n$.
According to Bohr's quantization condition,the angular momentum is $mvr = nh/(2\pi)$.
Substituting $v = \sqrt{eV_0/m}$ into the quantization condition,we get $m(\sqrt{eV_0/m})r_n = nh/(2\pi)$.
Solving for $r_n$,we get $r_n = (nh / (2\pi)) \cdot \sqrt{1/(meV_0)}$.
Since $h, m, e, V_0$ are constants,we conclude that $r_n \propto n$.

(D) The radius of an orbit in the Bohr model is given by $r_n = n^2 \frac{a_0}{Z}$,where $a_0$ is the Bohr radius $(0.529 \ \mathring{A})$,$n$ is the principal quantum number,and $Z$ is the atomic number.
Given the atom is $_{100}Fm^{257}$,the atomic number $Z = 100$.
The outermost shell for Fermium $(Z=100)$ is the $5f$ shell,so the principal quantum number $n_{shell} = 5$.
The radius of the outermost shell is $r = (5)^2 \frac{a_0}{100} = \frac{25}{100} a_0 = 0.25 a_0$.
Thus,the radius is $0.25$ times the Bohr radius,so $n = 0.25$.

(A) The energy of an electron in the $n$-th orbit of a hydrogen-like ion is given by $E_n = -13.6 \frac{Z^2}{n^2} \, eV$.
The first excitation state corresponds to $n=2$,and the ground state corresponds to $n=1$.
The excitation energy is $\Delta E = E_2 - E_1 = -13.6 Z^2 \left( \frac{1}{2^2} - \frac{1}{1^2} \right) = -13.6 Z^2 \left( \frac{1}{4} - 1 \right) = 13.6 Z^2 \times \frac{3}{4}$.
Given $\Delta E = 40.8 \, eV$,we have $40.8 = 13.6 \times Z^2 \times 0.75$.
$40.8 = 10.2 \times Z^2 \Rightarrow Z^2 = 4 \Rightarrow Z = 2$.
The energy required to remove the electron from the ground state is the ionization energy,$E_{ion} = |E_1| = 13.6 \times Z^2 \, eV$.
$E_{ion} = 13.6 \times (2)^2 = 13.6 \times 4 = 54.4 \, eV$.

(B) Let the ground state energy of the hydrogen-like atom be $E_0 = -13.6 Z^2 \ eV$. The energy of an electron in state $k$ is given by $E_k = \frac{E_0}{k^2}$.
Given that the atom is in state $2n$ and emits a maximum energy photon of $204 \ eV$ to reach the ground state $(k=1)$:
$E_{2n} - E_1 = 204 \ eV$
$\frac{E_0}{(2n)^2} - E_0 = 204 \ eV$
$E_0 \left( \frac{1}{4n^2} - 1 \right) = 204 \ eV$ --- $(i)$
Given that the atom makes a transition from state $2n$ to state $n$ emitting a photon of $40.8 \ eV$:
$E_{2n} - E_n = 40.8 \ eV$
$\frac{E_0}{4n^2} - \frac{E_0}{n^2} = 40.8 \ eV$
$E_0 \left( \frac{1 - 4}{4n^2} \right) = 40.8 \ eV$
$E_0 \left( -\frac{3}{4n^2} \right) = 40.8 \ eV$ --- $(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{E_0 \left( \frac{1 - 4n^2}{4n^2} \right)}{E_0 \left( -\frac{3}{4n^2} \right)} = \frac{204}{40.8}$
$\frac{1 - 4n^2}{-3} = 5$
$1 - 4n^2 = -15$
$4n^2 = 16$
$n^2 = 4$
$n = 2$

(A) The energy of a state is given by $E_n = -\frac{13.6 Z^2}{n^2}$. Let $E_1 = -13.6 Z^2$.
The most energetic photon corresponds to the transition from the $n^{th}$ state to the ground state $(n=1)$:
$E_{max} = E_n - E_1 = -\frac{E_1}{n^2} - E_1 = -E_1 \left( \frac{1}{n^2} - 1 \right) = 52.224 \ eV$ ... $(i)$
The least energetic photon corresponds to the transition between adjacent states ($n$ to $n-1$):
$E_{min} = E_n - E_{n-1} = -\frac{E_1}{n^2} - \left( -\frac{E_1}{(n-1)^2} \right) = E_1 \left( \frac{1}{(n-1)^2} - \frac{1}{n^2} \right) = 1.224 \ eV$ ... $(ii)$
Dividing $(i)$ by $(ii)$:
$\frac{1 - 1/n^2}{1/(n-1)^2 - 1/n^2} = \frac{52.224}{1.224} = 42.666... \approx 42.67$
Solving for $n$, we find $n=5$.
Substituting $n=5$ into $(i)$:
$-E_1 (1/25 - 1) = 52.224 \implies E_1 (24/25) = 52.224 \implies E_1 = 54.4 \ eV$.
Since $E_1 = -13.6 Z^2 = -54.4$, we get $Z^2 = 4$, so $Z = 2$.