Bohr's Model of Hydrogen Atom Questions in English

(A) The total energy of the $n$-th orbit is given by $E_n = -\frac{m e^4}{8 \varepsilon_0^2 h^2 n^2}$.
Since $E_n \propto m$,the ratio of the energy of the $\mu^{-}$-system to the hydrogen atom is $\frac{E_{\mu}}{E_e} = \frac{m_{\mu}}{m_e} = 207$.
Thus,$E_{\mu} = 207 \times E_e = 207 \times (-13.6 \text{ eV}) = -13.6 \times 207 \text{ eV}$.
The radius of the $n$-th orbit is given by $r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m e^2}$.
Since $r_n \propto \frac{1}{m}$,the ratio of the radius of the $\mu^{-}$-system to the hydrogen atom is $\frac{r_{\mu}}{r_H} = \frac{m_e}{m_{\mu}} = \frac{1}{207}$.
Thus,$r_{\mu} = \frac{r_H}{207}$.

(B) The potential energy $U$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $U = -27.2 / n^2 \text{ eV}$.
Given $U = -6.8 \text{ eV}$,we have $-27.2 / n^2 = -6.8$,which implies $n^2 = 27.2 / 6.8 = 4$,so $n = 2$.
The radius of the $n^{th}$ orbit is given by $r_n = n^2 r_0$. For $n = 2$,$r_2 = 2^2 r_0 = 4 r_0$.
According to Bohr's quantization condition,the circumference of the orbit is an integer multiple of the de Broglie wavelength: $2 \pi r_n = n \lambda$.
Substituting the values,$2 \pi (4 r_0) = 2 \lambda$.
Solving for $\lambda$,we get $\lambda = (8 \pi r_0) / 2 = 4 \pi r_0$.

(B) According to the de-Broglie hypothesis,a revolving electron in a circular orbit exhibits wave nature.
For a stable circular orbit,the circumference of the orbit must be an integral multiple of the de-Broglie wavelength,which is given by the condition:
$2 \pi r_n = n \lambda$
where $r_n$ is the radius of the $n$th orbit,$n$ is the principal quantum number,and $\lambda$ is the de-Broglie wavelength.
For the first Bohr orbit,$n = 1$.
Substituting $n = 1$ into the equation,we get:
$2 \pi r_1 = 1 \cdot \lambda$
$\lambda = 2 \pi r_1$
Thus,the de-Broglie wavelength of the electron in the first Bohr orbit is equal to the circumference of the first orbit.

(C) The orbital angular momentum of an electron in a hydrogen atom is given by $L = \frac{nh}{2 \pi}$.
Initially,the electron is in the ground state,so $n_1 = 1$. The initial angular momentum is $L_1 = \frac{1 \cdot h}{2 \pi} = \frac{h}{2 \pi}$.
After absorbing energy $\Delta E$,the angular momentum increases by $\frac{h}{2 \pi}$.
Thus,the new angular momentum is $L_2 = L_1 + \frac{h}{2 \pi} = \frac{h}{2 \pi} + \frac{h}{2 \pi} = \frac{2h}{2 \pi} = \frac{2h}{2 \pi}$.
Comparing this with $L = \frac{nh}{2 \pi}$,we find the new principal quantum number is $n_2 = 2$.
The energy of an electron in the $n$-th state is given by $E_n = -\frac{13.6 \ eV}{n^2}$.
For $n_1 = 1$,$E_1 = -13.6 \ eV$.
For $n_2 = 2$,$E_2 = -\frac{13.6 \ eV}{2^2} = -\frac{13.6 \ eV}{4} = -3.4 \ eV$.
The energy absorbed is $\Delta E = E_2 - E_1 = -3.4 \ eV - (-13.6 \ eV) = 10.2 \ eV$.

(B) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
Here,the electron jumps from $n_i = 2$ to $n_f = 1$.
Substituting the values: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$.
$\frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$.
Therefore,$\lambda = \frac{4}{3R}$.

(D) In the given hypothetical world,the angular momentum is $L = 2n' \frac{h}{2\pi} = n' \frac{h}{\pi}$,where $n' = 1, 2, 3, \dots$.
Comparing this with the standard Bohr quantization $L = n \frac{h}{2\pi}$,we see that the allowed orbits correspond to $n = 2n'$.
The energy of an orbit is $E_n = -\frac{13.6}{n^2} \text{ eV}$. Substituting $n = 2n'$,we get $E_{n'} = -\frac{13.6}{(2n')^2} = -\frac{13.6}{4n'^2} = -\frac{3.4}{n'^2} \text{ eV}$.
For the visible range,the transition must end at the first excited state of this system. The ground state is $n'=1$ $(n=2)$,and the first excited state is $n'=2$ $(n=4)$.
The transition for the largest wavelength (smallest energy) in the visible range is from $n'=2$ to $n'=1$.
The energy difference is $\Delta E = E_2 - E_1 = -\frac{3.4}{2^2} - (-\frac{3.4}{1^2}) = -0.85 + 3.4 = 2.55 \text{ eV}$.
The wavelength is $\lambda = \frac{hc}{\Delta E} = \frac{1242 \text{ eV-nm}}{2.55 \text{ eV}} \approx 487 \text{ nm}$.

(A, B) The energy required for a transition in $He^{+}$ is given by $E = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. For $He^{+}$,$Z=2$,so $E = 13.6 \times 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 54.4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) eV$.
For a transition from $n=2$ to $n=4$,$E = 54.4 \left( \frac{1}{4} - \frac{1}{16} \right) = 54.4 \left( \frac{3}{16} \right) = 10.2 eV$.
Since the incident photon energy is $10.2 eV$,the $He^{+}$ electron can be excited from $n=2$ to $n=4$.
Once in the $n=4$ state,the number of spectral lines emitted is given by $\frac{n(n-1)}{2} = \frac{4(4-1)}{2} = 6$.
Therefore,statements $A$ and $B$ are correct.

(A) According to Bohr's postulate of quantization of angular momentum,the angular momentum $L$ is given by:
$L = mvr = \frac{nh}{2\pi}$
From this equation,we can express the quantum number $n$ as:
$n = \frac{2\pi m}{h} \cdot (vr)$
Since $m$,$h$,and $\pi$ are constants,we have:
$n \propto vr$
Therefore,the quantity $vr$ is proportional to the quantum number $n$.

(D) According to the Bohr's atomic model,the angular momentum $L$ is given by:
$L = mvr = \frac{nh}{2\pi} \dots (i)$
Since angular velocity $\omega = \frac{v}{r}$,we have $v = r\omega$.
Substituting $v = r\omega$ into equation $(i)$:
$m(r\omega)r = \frac{nh}{2\pi} \Rightarrow m\omega r^2 = \frac{nh}{2\pi} \Rightarrow \omega = \frac{nh}{2\pi mr^2} \dots (ii)$
The radius of the electron in the $n$-th orbit is given by:
$r = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2} \dots (iii)$
Substituting the expression for $r$ from equation $(iii)$ into equation $(ii)$:
$\omega = \frac{nh}{2\pi m} \left( \frac{\pi m Z e^2}{n^2 h^2 \epsilon_0} \right)^2$
$\omega = \frac{nh}{2\pi m} \cdot \frac{\pi^2 m^2 Z^2 e^4}{n^4 h^4 \epsilon_0^2}$
$\omega = \frac{\pi m Z^2 e^4}{2 h^3 \epsilon_0^2} \cdot \frac{1}{n^3}$
Therefore,$\omega \propto \frac{1}{n^3}$.

(A) The linear velocity $v$ of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:
$v = \frac{Z e^2}{2 \epsilon_0 n h}$
Where $Z$ is the atomic number,$e$ is the charge of the electron,$\epsilon_0$ is the permittivity of free space,$n$ is the principal quantum number,and $h$ is Planck's constant.
From this expression,it is clear that all terms except $n$ are constants for a given atom.
Therefore,the relationship is $v \propto \frac{1}{n}$.

(A-D) According to Bohr's model for a hydrogen atom:
$v_{n} \propto \frac{1}{n}$
$E_{n} \propto \frac{1}{n^{2}}$
$r_{n} \propto n^{2}$
For option $A$: $\frac{E_{n} r_{n}}{E_{1} r_{1}} \propto \frac{(1/n^{2}) \cdot n^{2}}{1} = 1$. This is a constant,so the slope is $0$.
For option $B$: $\frac{r_{n} v_{n}}{r_{1} v_{1}} \propto \frac{n^{2} \cdot (1/n)}{1} = n$. This is a straight line with slope $1$.
For option $C$: $\frac{r_{n}}{r_{1}} = n^{2}$. Taking natural log on both sides: $\ln \left(\frac{r_{n}}{r_{1}}\right) = 2 \ln(n)$. This is a straight line with slope $2$.
For option $D$: $\frac{r_{n}}{E_{n}} \propto \frac{n^{2}}{1/n^{2}} = n^{4}$. Thus,$\frac{r_{n} E_{1}}{E_{n} r_{1}} = n^{4}$. Taking natural log: $\ln \left(\frac{r_{n} E_{1}}{E_{n} r_{1}}\right) = 4 \ln(n)$. This is a straight line with slope $4$.
All options $A, B, C,$ and $D$ are mathematically correct based on Bohr's model.

(B) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by $L = mvr = \frac{nh}{2\pi}$.
From the de-Broglie hypothesis,the wavelength of an electron is $\lambda = \frac{h}{mv}$.
Substituting $mv = \frac{h}{\lambda}$ into the quantization condition,we get $\frac{h}{\lambda} r = \frac{nh}{2\pi}$.
This simplifies to $2\pi r = n\lambda$,where $2\pi r$ is the circumference of the $n^{th}$ orbit.
For the second Bohr orbit,$n = 2$.
Therefore,the circumference of the second orbit is $2\pi r = 2\lambda$.
This implies that the number of de-Broglie wavelengths contained in the second Bohr orbit is $2$.

(D) The energy of the photon is given by $E = \frac{hc}{\lambda}$.
Substituting the given values:
$E = \frac{4 \times 10^{-15} \ eV \cdot s \times 3 \times 10^{8} \ m/s}{300 \times 10^{-9} \ m} = \frac{12 \times 10^{-7}}{300 \times 10^{-9}} \ eV = \frac{1200}{300} \ eV = 4 \ eV$.
The energy required to excite a hydrogen atom from the ground state $(n=1)$ to the first excited state $(n=2)$ is $\Delta E = 13.6 \ eV \times (1 - \frac{1}{4}) = 13.6 \times 0.75 = 10.2 \ eV$.
Since the energy of the photon $(4 \ eV)$ is less than the energy required for the first excitation $(10.2 \ eV)$ and also less than the ionization energy $(13.6 \ eV)$,the electron cannot absorb this energy to transition to a higher state.
Therefore,the electron will remain in the ground state.

(A) For a hydrogen-like atom,the velocity of an electron in the $n^{th}$ orbit is given by $v \propto \frac{Z}{n}$.
The radius of the $n^{th}$ orbit is given by $r \propto \frac{n^2}{Z}$.
From these two relations,we can write $r \propto \frac{n}{v}$ (since $Z \propto \frac{n}{v} \cdot n = \frac{n^2}{r}$,substituting $Z$ into the velocity relation).
Given that the radii are nearly the same $(r_1 \approx r_2)$,we have $\frac{n_1}{v_1} = \frac{n_2}{v_2}$.
Therefore,$\frac{n_1}{n_2} = \frac{v_1}{v_2} = \frac{3 \times 10^5}{2.5 \times 10^5} = \frac{3}{2.5} = \frac{6}{5}$.
Thus,the possible order of energy states ($n_1$ and $n_2$) is $6$ and $5$.

(C) According to Bohr's postulate,the angular momentum $L$ of an electron in an orbit is given by $L = \frac{nh}{2\pi}$,where $n$ is the principal quantum number.
Rearranging this,we get $\frac{2\pi L}{h} = n$.
The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{E_0}{n^2}$,where $E_0$ is the magnitude of the ground state energy.
Given that $E_n = -0.04 E_0$,we have $-\frac{E_0}{n^2} = -0.04 E_0$.
Dividing both sides by $-E_0$,we get $\frac{1}{n^2} = 0.04$.
$n^2 = \frac{1}{0.04} = \frac{100}{4} = 25$.
Therefore,$n = 5$.
Since $\frac{2\pi L}{h} = n$,the value is $5$.

(A) According to Bohr's model,the orbital angular momentum $L$ is given by the formula $L = n \frac{h}{2\pi}$.
The ground state corresponds to $n = 1$.
The first excited state is $n = 2$,the second excited state is $n = 3$,and the third excited state is $n = 4$.
Substituting $n = 4$ into the formula:
$L = 4 \times \frac{h}{2\pi} = 2 \times \frac{h}{\pi}$.
Given $h = 6.63 \times 10^{-34} \text{ Js}$ and $\pi \approx 3.14$:
$L = 2 \times \frac{6.63 \times 10^{-34}}{3.14} \approx 2 \times 2.111 \times 10^{-34} = 4.222 \times 10^{-34} \text{ kg m}^2\text{s}^{-1}$.

(D) For a hydrogen atom in the ground state $(n=1)$,Bohr's quantization condition for angular momentum is given by $mvr = \frac{nh}{2\pi}$.
Here,$m$ is the mass of the electron $(9.1 \times 10^{-31} \text{ kg})$,$v$ is the velocity,$r$ is the orbital radius $(5.3 \times 10^{-11} \text{ m})$,and $h$ is Planck's constant $(6.63 \times 10^{-34} \text{ Js})$.
Rearranging the formula for velocity $v$,we get $v = \frac{h}{2\pi mr}$.
Substituting the values: $v = \frac{6.63 \times 10^{-34}}{2 \times 3.14159 \times 9.1 \times 10^{-31} \times 5.3 \times 10^{-11}}$.
Calculating this,$v \approx 2.18 \times 10^6 \text{ ms}^{-1}$.
Rounding to two significant figures,we get $v \approx 2.2 \times 10^6 \text{ ms}^{-1}$.

(C) In the Bohr model,the Total Energy $(TE)$ of an electron in the ground state is $-13.6 \text{ eV}$.
Kinetic Energy $(KE)$ is given by the relation $KE = -TE$. Therefore,$KE = -(-13.6 \text{ eV}) = +13.6 \text{ eV}$.
Potential Energy $(PE)$ is given by the relation $PE = 2 \times TE$. Therefore,$PE = 2 \times (-13.6 \text{ eV}) = -27.2 \text{ eV}$.
Thus,the potential energy is $-27.2 \text{ eV}$ and the kinetic energy is $+13.6 \text{ eV}$.

(A) In a hydrogen-like atom,the total energy $E$ of an electron is related to its kinetic energy $K$ and potential energy $U$ by the following relations:
$E = -K$
$U = 2E$
Given that the total energy $E = -13.6 \text{ eV}$,we can find $K$ and $U$:
$K = -E = -(-13.6 \text{ eV}) = 13.6 \text{ eV}$
$U = 2E = 2 \times (-13.6 \text{ eV}) = -27.2 \text{ eV}$
The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{13.6 \text{ eV}}{-27.2 \text{ eV}} = -\frac{1}{2}$.
Thus,the correct option is $A$.

(D) The radius of the $n$-th orbit in a hydrogen atom is given by the formula $r_n = n^2 r_1$,where $r_1$ is the radius of the first orbit (Bohr radius).
Given $r_1 = 5.3 \times 10^{-11} \text{ m}$ and $n = 3$.
Substituting these values into the formula:
$r_3 = 3^2 \times (5.3 \times 10^{-11} \text{ m})$
$r_3 = 9 \times 5.3 \times 10^{-11} \text{ m}$
$r_3 = 47.7 \times 10^{-11} \text{ m}$
$r_3 = 4.77 \times 10^{-10} \text{ m}$.
Thus,the radius of the $n = 3$ orbit is $4.77 \times 10^{-10} \text{ m}$.
Therefore,the correct option is $D$.

(C) According to Bohr's quantization condition,the angular momentum $L$ of an electron in an orbit is given by $L = \frac{nh}{2\pi}$.
Given that $L = \frac{3h}{\pi}$,we equate the two expressions: $\frac{nh}{2\pi} = \frac{3h}{\pi}$.
Solving for $n$,we get $n = 6$.
The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \text{ eV}$.
Substituting $n = 6$ into the formula: $E_6 = -\frac{13.6}{6^2} = -\frac{13.6}{36}$.
Calculating the value: $E_6 \approx -0.377 \text{ eV}$,which rounds to $-0.38 \text{ eV}$.

(C) 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.
Therefore,the ratio of the radii is $r_i / r_f = n_i^2 / n_f^2 = 16 / 4 = 4$.
Taking the square root,we get $n_i / n_f = 2$,which implies $n_i = 2n_f$.
For the simplest transition,we take $n_f = 2$ and $n_i = 4$.
Using the Rydberg formula for the wavelength $\lambda$:
$1/\lambda = R(1/n_f^2 - 1/n_i^2)$
$1/\lambda = R(1/2^2 - 1/4^2) = R(1/4 - 1/16) = R(3/16)$.
Substituting $R = 1.0973 \times 10^7 \text{ m}^{-1}$:
$1/\lambda = 1.0973 \times 10^7 \times (3/16) \approx 0.20574 \times 10^7 \text{ m}^{-1}$.
$\lambda = 1 / (0.20574 \times 10^7) \approx 4.86 \times 10^{-7} \text{ m} = 486 \text{ nm}$.

(C) The magnetic field $B$ at the center of the orbit due to an electron moving 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.
According to Bohr's model,$v \propto \frac{1}{n}$ and $r \propto n^2$.
Substituting these into the expression for current: $I \propto \frac{1/n}{n^2} = \frac{1}{n^3}$.
Now,substituting $I$ and $r$ into the expression for $B$: $B \propto \frac{I}{r} \propto \frac{1/n^3}{n^2} = \frac{1}{n^5}$.
Therefore,the ratio of the magnetic fields for the $2^{nd}$ and $4^{th}$ orbits is $\frac{B_2}{B_4} = \left( \frac{4}{2} \right)^5 = 2^5 = 32$.
Thus,the ratio is $32:1$.

(B) For a hydrogen atom,the radius of the $n^{th}$ orbit is given by the formula $r_n = a_0 \times n^2$,where $a_0 = 0.529 \text{ Å}$ is the Bohr radius.
For the first excited state,the principal quantum number is $n = 2$.
Substituting the value of $n$ into the formula,we get $r_2 = 0.529 \times (2)^2 \text{ Å}$.
$r_2 = 0.529 \times 4 \text{ Å} = 2.116 \text{ Å}$.
Since $1 \text{ Å} = 10^{-10} \text{ m}$,we have $r_2 = 2.116 \times 10^{-10} \text{ m}$.
Rounding to two significant figures,the radial distance is approximately $2.1 \times 10^{-10} \text{ m}$.