Electron Energy and Electron Energy Levels in Hydrogen Atom Questions in English

(D) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$,where $R$ is the Rydberg constant and $Z$ is the atomic number.
For the transition from $M$-shell $(n_i = 3)$ to $L$-shell $(n_f = 2)$:
$\frac{1}{\lambda} = K \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = K \left( \frac{1}{4} - \frac{1}{9} \right) = K \left( \frac{9-4}{36} \right) = \frac{5K}{36}$,where $K = R Z^2$.
For the transition from $N$-shell $(n_i = 4)$ to $L$-shell $(n_f = 2)$:
$\frac{1}{\lambda'} = K \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = K \left( \frac{1}{4} - \frac{1}{16} \right) = K \left( \frac{4-1}{16} \right) = \frac{3K}{16}$.
Now,taking the ratio of the two equations:
$\frac{\lambda'}{\lambda} = \frac{5K/36}{3K/16} = \frac{5}{36} \times \frac{16}{3} = \frac{5 \times 4}{9 \times 3} = \frac{20}{27}$.
Therefore,$\lambda' = \frac{20}{27} \lambda$.

(B) The energy of an electron in the $n$th orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.
To remove an electron from the $n=2$ orbit to infinity (ionization),we need to provide energy equal to the magnitude of the binding energy at that level.
The energy of the electron in the $n=2$ state is $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \ eV$.
The energy required to remove this electron is the difference between the energy at infinity $(0 \ eV)$ and the energy at $n=2$ $(E_{\text{req}} = 0 - (-3.4) = 3.4 \ eV)$.

(A) The Rydberg formula for the wavelength of emitted photons is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the first jump from the third excited state $(n_i = 4)$ to the second excited state $(n_f = 3)$:
$\frac{1}{\lambda_1} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16 - 9}{144} \right) = R \left( \frac{7}{144} \right)$.
For the second jump from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$:
$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right)$.
Now,calculating the ratio $\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{R(5/36)}{R(7/144)} = \frac{5}{36} \times \frac{144}{7} = \frac{5 \times 4}{7} = \frac{20}{7}$.

(B) The energy of a photon emitted during a transition between two energy levels $E_i$ and $E_f$ is given by $\Delta E = E_i - E_f = \frac{hc}{\lambda}$.
For the transition from $4E$ to $E$:
$\Delta E_1 = 4E - E = 3E$
So,$3E = \frac{hc}{\lambda_1} \implies \lambda_1 = \frac{hc}{3E}$ --- $(1)$
For the transition from $\frac{7}{3}E$ to $E$:
$\Delta E_2 = \frac{7}{3}E - E = \frac{4}{3}E$
So,$\frac{4}{3}E = \frac{hc}{\lambda_2} \implies \lambda_2 = \frac{3hc}{4E}$ --- $(2)$
Taking the ratio of $\lambda_1$ to $\lambda_2$:
$\frac{\lambda_1}{\lambda_2} = \left( \frac{hc}{3E} \right) / \left( \frac{3hc}{4E} \right) = \frac{hc}{3E} \times \frac{4E}{3hc} = \frac{4}{9}$

(D) In the Bohr model of an atom,the total energy $(E)$,kinetic energy $(K)$,and potential energy $(U)$ of an electron in the $n^{th}$ orbit are related as follows:
$K = -E$
$U = 2E$
Given that the kinetic energy $K = 13.6 \frac{Z^2}{n^2} \ eV$.
Since $K = -E$,the total energy $E = -13.6 \frac{Z^2}{n^2} \ eV$.
The potential energy $U$ is given by $U = 2E$.
Substituting the value of $E$:
$U = 2 \times (-13.6 \frac{Z^2}{n^2} \ eV) = -27.2 \frac{Z^2}{n^2} \ eV$.
Therefore,the potential energy is $-27.2 \frac{Z^2}{n^2} \ eV$.

(D) The energy of the emitted photon is equal to the difference in energy levels.
For the transition from $2E$ to $E$:
$\Delta E_1 = 2E - E = E = \frac{hc}{\lambda} \Rightarrow E = \frac{hc}{\lambda} \quad ... (1)$
For the transition from $\frac{4E}{3}$ to $E$:
$\Delta E_2 = \frac{4E}{3} - E = \frac{E}{3} = \frac{hc}{\lambda'} \quad ... (2)$
Substituting the value of $E$ from equation $(1)$ into equation $(2)$:
$\frac{1}{3} \left( \frac{hc}{\lambda} \right) = \frac{hc}{\lambda'}$
$\frac{1}{3\lambda} = \frac{1}{\lambda'} \Rightarrow \lambda' = 3\lambda$
Therefore, the wavelength of the photon produced is $3\lambda$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
For a hydrogen atom,the atomic number $Z = 1$,so $E_n(H) = -13.6 \frac{1^2}{n^2} = E_n$.
For a singly ionized helium atom $(He^+)$,the atomic number $Z = 2$.
The energy in the $n^{th}$ orbit for $He^+$ is $E_n(He^+) = -13.6 \frac{2^2}{n^2} = 4 \times (-13.6 \frac{1^2}{n^2}) = 4E_n$.
Therefore,the energy is $4E_n$.

(A) In an absorption transition,an electron in a lower energy state absorbs a photon and jumps to a higher energy state. For a system with energy levels $A, B, C$ (where $A < B < C$),absorption can only occur from the ground state $A$ to $B$ or $A$ to $C$. Thus,there are $2$ possible absorption transitions.
In an emission transition,an electron in a higher energy state drops to a lower energy state by emitting a photon. For the same levels,emission can occur from $C \rightarrow B$,$C \rightarrow A$,and $B \rightarrow A$. Thus,there are $3$ possible emission transitions.
Since $2 < 3$,the number of absorption transitions is less than the number of emission transitions. The Reason correctly explains that absorption is restricted to starting from the lower level,whereas emission can occur between any two levels where the initial state is higher than the final state.

(C) The total energy $(TE)$ of an electron in an orbit is given as $-3.4 \; eV$.
For an electron in a hydrogen-like atom,the relationship between total energy $(TE)$,kinetic energy $(KE)$,and potential energy $(PE)$ is:
$KE = -TE$
$PE = 2 \times TE$
Substituting the given value of $TE = -3.4 \; eV$:
$KE = -(-3.4 \; eV) = +3.4 \; eV$
$PE = 2 \times (-3.4 \; eV) = -6.8 \; eV$
Therefore,the kinetic energy is $3.4 \; eV$ and the potential energy is $-6.8 \; eV$.

(B) The energy difference between the two levels is given as $E = 2.3 \; eV$.
To convert this energy into Joules,we multiply by the charge of an electron $(1.6 \times 10^{-19} \; C)$:
$E = 2.3 \times 1.6 \times 10^{-19} \; J = 3.68 \times 10^{-19} \; J$.
According to the Bohr model,the frequency $\nu$ of the emitted radiation is related to the energy difference by the equation $E = h\nu$,where $h$ is Planck's constant $(6.626 \times 10^{-34} \; Js)$.
Rearranging for frequency: $\nu = \frac{E}{h}$.
Substituting the values: $\nu = \frac{3.68 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 5.55 \times 10^{14} \; Hz$.
Rounding to two significant figures,we get $\nu \approx 5.6 \times 10^{14} \; Hz$.

For the ground level,$n_{1} = 1$.
The energy of this level is given by $E_{1} = \frac{-13.6}{n_{1}^{2}} \, eV = -13.6 \, eV$.
The atom is excited to the level $n_{2} = 4$.
The energy of this level is $E_{2} = \frac{-13.6}{n_{2}^{2}} \, eV = \frac{-13.6}{16} = -0.85 \, eV$.
The energy of the absorbed photon is $E = E_{2} - E_{1} = -0.85 - (-13.6) = 12.75 \, eV$.
Converting energy to Joules: $E = 12.75 \times 1.6 \times 10^{-19} \, J = 2.04 \times 10^{-18} \, J$.
Using the relation $E = \frac{hc}{\lambda}$,the wavelength is $\lambda = \frac{hc}{E}$.
$\lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{2.04 \times 10^{-18}} \approx 9.75 \times 10^{-8} \, m = 97.5 \, nm$.
The frequency is $\nu = \frac{c}{\lambda} = \frac{3 \times 10^{8}}{9.75 \times 10^{-8}} \approx 3.08 \times 10^{15} \, Hz$.

(N/A) The total energy of the electron is $E = -3.4 \; eV$. The kinetic energy $(K)$ of an electron in a hydrogen atom is equal to the negative of its total energy.
$\Rightarrow K = -E$
$K = -(-3.4 \; eV) = +3.4 \; eV$
Thus,the kinetic energy of the electron in this state is $+3.4 \; eV$.
$(b)$ The potential energy $(U)$ of the electron in a hydrogen atom is equal to $-2$ times its kinetic energy.
$\Rightarrow U = -2K$
$U = -2 \times 3.4 \; eV = -6.8 \; eV$
Thus,the potential energy of the electron in this state is $-6.8 \; eV$.
$(c)$ Potential energy is defined relative to a reference point where it is assumed to be zero. If the reference point is changed,the potential energy $(U)$ and consequently the total energy $(E = K + U)$ will change. However,the kinetic energy $(K)$ remains unchanged because it depends only on the velocity of the electron.

(A) According to the Bohr model of the atom,an electron revolves around a positively charged nucleus in stable orbits. The electrostatic force of attraction $F_{e}$ between the electron and the nucleus provides the necessary centripetal force $F_{c}$ to keep the electron in its orbit.
For a stable orbit in a hydrogen atom:
$F_{e} = F_{c}$
$\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{e^{2}}{r^{2}} = \frac{m v^{2}}{r}$
From this,we get the radius relation: $r = \frac{e^{2}}{4 \pi \epsilon_{0} m v^{2}} \quad \dots (1)$
The kinetic energy $K$ of the electron is:
$K = \frac{1}{2} m v^{2} = \frac{e^{2}}{8 \pi \epsilon_{0} r} \quad \dots (2)$
The potential energy $U$ of the electron in the field of the nucleus ($Z=1$ for hydrogen) is:
$U = -\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{e^{2}}{r} \quad \dots (3)$
The total energy $E$ is the sum of kinetic and potential energy:
$E = K + U = \frac{e^{2}}{8 \pi \epsilon_{0} r} - \frac{e^{2}}{4 \pi \epsilon_{0} r} = -\frac{e^{2}}{8 \pi \epsilon_{0} r}$
The negative sign indicates that the electron is bound to the nucleus.

(A) In an atom,the total energy $E$ of an electron is the sum of its kinetic energy $K$ and potential energy $U$.
Since the electron is bound to the nucleus by the electrostatic force of attraction,the potential energy $U$ is negative.
Mathematically,$U = -\frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r}$.
The kinetic energy $K$ is positive,given by $K = \frac{1}{2}mv^2 = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{2r}$.
Thus,the total energy $E = K + U = -\frac{1}{4\pi\epsilon_0} \frac{Ze^2}{2r}$.
$A$ negative total energy indicates that the electron is in a bound state,meaning it is attracted to the nucleus and requires external energy to be removed from the atom. Therefore,the fact that the electron is attracted by the nucleus is the reason for its negative total energy.

(N/A) The total energy of an electron in an atom is given by:
$E_{n} = -\frac{m Z^{2} e^{4}}{8 n^{2} h^{2} \epsilon_{0}^{2}}$
This simplifies to:
$E_{n} = -\frac{13.6 Z^{2}}{n^{2}} \text{ eV}$
The negative sign indicates that as the value of $n$ increases,the magnitude of the negative energy decreases,meaning the energy of the electron increases as it moves to orbits further from the nucleus.
When an electron is in the orbit closest to the nucleus $(n = 1)$,it has the lowest energy (the maximum negative value). This state is called the ground state.
For a hydrogen atom $(Z = 1, n = 1)$:
$E_{1} = -13.6 \text{ eV}$
The ionization energy is the minimum energy required to remove an electron from the ground state to infinity. For hydrogen,this is $13.6 \text{ eV}$.
Excitation energy is the energy required to move an electron from the ground state to a higher energy state. For the first excited state $(n = 2)$:
$E_{2} = -\frac{13.6}{2^{2}} = -3.4 \text{ eV}$
$\Delta E = E_{2} - E_{1} = -3.4 - (-13.6) = 10.2 \text{ eV}$

(A) The ground state of an atom is its lowest energy state. When an atom absorbs energy,its electrons jump from a lower energy level to a higher energy level. This state,which has higher energy than the ground state,is called the excited state of the atom.

(C) In an atom,the total energy of an electron in an orbit is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
As the distance from the nucleus increases,the principal quantum number $n$ increases.
At an infinite distance,$n \to \infty$.
Therefore,the total energy $E_{\infty} = \lim_{n \to \infty} \left( -\frac{13.6}{n^2} \right) = 0 \text{ eV}$.
Thus,the total energy of an electron at an infinite distance from the nucleus is zero.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \ eV$.
For the ground state,$n = 1$.
The first excited state corresponds to $n = 2$.
The second excited state corresponds to $n = 3$.
Substituting $n = 3$ into the formula,we get $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \ eV$.
To release the electron (ionize the atom),we need to provide enough energy to bring its total energy to $0 \ eV$.
Therefore,the energy required is $E_{\text{ionization}} = 0 - E_3 = 0 - (-1.51 \ eV) = 1.51 \ eV$.

(N/A) $1$. Emission Lines: When an atom in an excited state transitions to a lower energy state,it emits a photon with energy equal to the difference between the two energy levels $(E_2 - E_1 = h
u)$. This results in a bright line in the spectrum at a specific wavelength,known as an emission line.
$2$. Absorption Lines: When an atom in a lower energy state absorbs a photon of energy exactly equal to the difference between its current state and a higher energy state $(h
u = E_2 - E_1)$,it transitions to the higher state. This removal of specific wavelengths from a continuous spectrum creates dark lines,known as absorption lines.

(B) An emission line is a spectral line that is formed when an electron in an atom or molecule transitions from a higher energy state $(E_2)$ to a lower energy state $(E_1)$.
During this transition,the excess energy is released in the form of a photon.
The energy of the emitted photon is given by the equation: $E = E_2 - E_1 = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency of the emitted radiation.
This results in a bright line at a specific wavelength in the emission spectrum.

(B) In a hydrogen atom,the energy of the $n^{\text{th}}$ level is given by $E_n = -\frac{E_0}{n^2}$,where $E_0$ is the ionization energy of hydrogen.
For a transition from $(n+1)$ to $n$,the energy of the emitted radiation $\Delta E$ is the difference between the energy levels:
$\Delta E = E_{n+1} - E_n = E_0 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$
Using the relation $\Delta E = h\nu$,we have:
$h\nu = E_0 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right) = E_0 \left( \frac{2n+1}{n^2(n+1)^2} \right)$
For $n >> 1$,we can approximate $(n+1) \approx n$ and $(2n+1) \approx 2n$:
$h\nu \approx E_0 \left( \frac{2n}{n^2 \cdot n^2} \right) = E_0 \left( \frac{2n}{n^4} \right) = \frac{2E_0}{n^3}$
Thus,the frequency $\nu$ is proportional to $\frac{1}{n^3}$.

(C) According to the Bohr model of the hydrogen atom,the total energy of an electron in the $n^{th}$ stationary orbit is given by the formula:
$E_{n} = -\frac{13.6}{n^{2}} \; eV$
where $n$ is the principal quantum number representing the orbit number $(n = 1, 2, 3, \dots)$.

(A) The energy required to excite an electron in a hydrogen atom from state $n_1$ to $n_2$ is given by the formula:
$\Delta E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) eV$
For a hydrogen atom,$Z = 1$. The electron is excited from the ground state $(n_1 = 1)$ to the state $n_2 = 3$.
Substituting these values into the formula:
$\Delta E = 13.6 \times (1)^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) eV$
$\Delta E = 13.6 \left( 1 - \frac{1}{9} \right) eV$
$\Delta E = 13.6 \left( \frac{8}{9} \right) eV$
$\Delta E = 13.6 \times 0.888... eV$
$\Delta E \approx 12.088 eV \approx 12.1 eV$
Thus,the energy transferred to the atom is $12.1 eV$.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like ion is given by the formula:
$E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$
For a $He^{+}$ ion,the atomic number $Z = 2$.
For the $2^{nd}$ orbit,the principal quantum number $n = 2$.
Substituting these values into the formula:
$E_2 = -13.6 \times \frac{2^2}{2^2} \ eV$
$E_2 = -13.6 \times \frac{4}{4} \ eV$
$E_2 = -13.6 \ eV$

(A) The energy of an electron in the $n^{th}$ orbit of a Hydrogen atom is given by the formula:
$E_n = -\frac{13.6}{n^2} \ eV$
For the $4^{th}$ orbit $(n=4)$:
$E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \ eV$
When an external energy of $15 \ eV$ is supplied to the electron,the final energy $(E_f)$ of the electron after it is ejected from the atom is given by:
$E_f = E_{supplied} + E_4$
$E_f = 15 \ eV + (-0.85 \ eV)$
$E_f = 14.15 \ eV$
Thus,the final kinetic energy of the electron is $14.15 \ eV$.

(B) The energy of an electron in a hydrogen-like atom is given by the formula $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
For the hydrogen atom,the atomic number $Z = 1$ and for the ground state,$n = 1$.
Thus,the ground state energy of hydrogen is $E_H = -13.6 \times \frac{1^2}{1^2} = -13.6 \text{ eV}$.
For a singly ionized carbon atom $(C^+)$,the atomic number $Z = 6$.
We want to find the energy level $n$ such that the energy $E_C$ equals the ground state energy of hydrogen:
$-13.6 \frac{6^2}{n^2} = -13.6$.
Dividing both sides by $-13.6$,we get $\frac{36}{n^2} = 1$.
Therefore,$n^2 = 36$,which gives $n = 6$.

(B) The energy of a photon emitted during a transition is given by $\Delta E = h\nu = 13.6 \text{ eV} \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
Since $\nu = \frac{\Delta E}{h}$,the frequency $\nu$ is maximum when the energy difference $\Delta E$ is maximum.
Comparing the energy gaps:
For $n = 4$ to $n = 3$: $\Delta E = 13.6 \left( \frac{1}{9} - \frac{1}{16} \right) \approx 0.66 \text{ eV}$.
For $n = 2$ to $n = 1$: $\Delta E = 13.6 \left( \frac{1}{1} - \frac{1}{4} \right) = 13.6 \times 0.75 = 10.2 \text{ eV}$.
For $n = 5$ to $n = 4$: $\Delta E = 13.6 \left( \frac{1}{16} - \frac{1}{25} \right) \approx 0.31 \text{ eV}$.
For $n = 3$ to $n = 2$: $\Delta E = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) \approx 1.89 \text{ eV}$.
The transition from $n = 2$ to $n = 1$ results in the largest energy change,hence the maximum frequency.

(A) The energy released when an electron transitions from $n = 5$ to $n = 1$ is given by:
$\Delta E = E_5 - E_1 = -0.54 \, eV - (-13.6 \, eV) = 13.06 \, eV$.
By the law of conservation of linear momentum,the initial momentum of the atom is zero,so the final momentum of the atom must be equal and opposite to the momentum of the emitted photon:
$P_{\text{atom}} = P_{\text{photon}} = \frac{h}{\lambda}$.
Since the energy of the photon is $\Delta E = \frac{hc}{\lambda}$,we have $\frac{1}{\lambda} = \frac{\Delta E}{hc}$.
Substituting this into the momentum equation,the recoil momentum of the atom is $Mv = \frac{\Delta E}{c}$,where $M$ is the mass of the hydrogen atom $(M \approx 1.67 \times 10^{-27} \, kg)$ and $c$ is the speed of light $(3 \times 10^8 \, m/s)$.
The recoil speed $v$ is given by:
$v = \frac{\Delta E}{Mc} = \frac{13.06 \times 1.6 \times 10^{-19} \, J}{1.67 \times 10^{-27} \, kg \times 3 \times 10^8 \, m/s} \approx 4.17 \, m/s$.

(C) The total energy of the system before collision is the kinetic energy of the free electron,as the potential energy at a large distance is $0$. Total energy $E_i = 2.6 \, eV + 0 = 2.6 \, eV$.
The hydrogen atom is formed in the first excited state $(n=2)$. The energy of the hydrogen atom in the $n$-th state is given by $E_n = -13.6/n^2 \, eV$. For $n=2$,$E_f = -13.6/4 = -3.4 \, eV$.
The energy released as a photon is the difference between the initial and final total energy: $\Delta E = E_i - E_f = 2.6 - (-3.4) = 6.0 \, eV$.
Convert the energy to Joules: $\Delta E = 6.0 \times 1.6 \times 10^{-19} \, J = 9.6 \times 10^{-19} \, J$.
Using the relation $\Delta E = hf$,the frequency $f = \Delta E / h = (9.6 \times 10^{-19}) / (6.6 \times 10^{-34}) \approx 1.45 \times 10^{15} \, Hz$.
Since $1 \, MHz = 10^6 \, Hz$,the frequency in $MHz$ is $1.45 \times 10^{15} / 10^6 = 1.45 \times 10^9 \, MHz$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
The first excited state corresponds to $n = 2$.
Therefore,$T_1 = -\frac{13.6}{2^2} = -\frac{13.6}{4} \text{ eV}$.
The second excited state corresponds to $n = 3$.
Therefore,$T_2 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \text{ eV}$.
The ratio $T_1 : T_2$ is calculated as:
$\frac{T_1}{T_2} = \frac{-13.6/4}{-13.6/9} = \frac{9}{4}$.
Thus,the ratio is $9:4$.

(B) In Bohr's model for a hydrogen atom,the energy levels are given by $E_n = -\frac{13.6 \text{ eV}}{n^2}$. As the electron transitions to a higher level ($n$ increases),$E$ increases (becomes less negative).
For a hydrogen-like atom,the kinetic energy is $K = -E = \frac{13.6 \text{ eV}}{n^2}$. As $n$ increases,$K$ decreases.
The potential energy is $P = 2E = -\frac{27.2 \text{ eV}}{n^2}$. As $n$ increases,the magnitude of $P$ decreases,meaning $P$ becomes less negative,so $P$ increases.
Therefore,as the electron moves to a higher energy level,$K$ decreases,while $P$ and $E$ increase.

(B) The energy of a photon emitted during a transition from level $n_2$ to $n_1$ is given by $E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
For transition $(i)$ from $n=3$ to $n=2$: $E_1 = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{5}{36} \right)$.
For transition $(ii)$ from $n=\infty$ to $n=2$: $E_2 = 13.6 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = 13.6 \left( \frac{1}{4} - 0 \right) = 13.6 \left( \frac{1}{4} \right)$.
The ratio is $\frac{E_1}{E_2} = \frac{13.6(5/36)}{13.6(1/4)} = \frac{5/36}{1/4} = \frac{5}{9}$.
Given the ratio is $\frac{x}{x+4}$,we have $\frac{x}{x+4} = \frac{5}{9}$.
Cross-multiplying gives $9x = 5x + 20$,which simplifies to $4x = 20$,so $x = 5$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
For the transition from the second energy level $(n=2)$ to the first energy level $(n=1)$,the energy of the emitted photon is:
$\Delta E_1 = E_2 - E_1 = -\frac{13.6}{2^2} - (-\frac{13.6}{1^2}) = 13.6(1 - \frac{1}{4}) = 13.6 \times \frac{3}{4} \text{ eV}$.
For the transition from the highest permitted energy level $(n=\infty)$ to the first energy level $(n=1)$,the energy of the emitted photon is:
$\Delta E_2 = E_{\infty} - E_1 = 0 - (-\frac{13.6}{1^2}) = 13.6 \text{ eV}$.
The ratio of the energies is:
$\frac{\Delta E_1}{\Delta E_2} = \frac{13.6 \times \frac{3}{4}}{13.6} = \frac{3}{4}$.

(D) The energy of a photon with wavelength $\lambda$ is given by $E = \frac{12400 \, \text{eV} \cdot \mathring{A}}{\lambda (\mathring{A})}$.
Calculating the energies of the given wavelengths:
$E_1 = \frac{12400}{912} \approx 13.6 \, \text{eV}$
$E_2 = \frac{12400}{1026} \approx 12.08 \, \text{eV}$
$E_3 = \frac{12400}{1216} \approx 10.2 \, \text{eV}$
$E_4 = \frac{12400}{3646} \approx 3.4 \, \text{eV}$
At room temperature,hydrogen atoms are in the ground state $(n=1)$.
For absorption to occur,the photon energy must match the energy difference between the ground state and an excited state $(n > 1)$. The minimum energy required for excitation is $E_2 - E_1 = -3.4 - (-13.6) = 10.2 \, \text{eV}$.
Since $3.4 \, \text{eV}$ is less than the required $10.2 \, \text{eV}$ for excitation from the ground state,the wavelength $3646 \, \mathring{A}$ will not be strongly absorbed.

(B) The first excitation energy of a hydrogen atom in the ground state $(n=1)$ to the first excited state $(n=2)$ is $E_2 - E_1 = -3.4 \,eV - (-13.6 \,eV) = 10.2 \,eV$.
If the kinetic energy $E$ of the incident electron is less than $10.2 \,eV$,the hydrogen atom cannot absorb any energy because there are no energy levels available between $n=1$ and $n=2$. Consequently,no internal energy change occurs,and the collision is elastic.
If the kinetic energy $E$ is greater than or equal to $10.2 \,eV$,the hydrogen atom can absorb $10.2 \,eV$ of energy to transition to the $n=2$ state. In this case,the collision is inelastic.
Therefore,the collision is elastic for $E < 10.2 \,eV$.

(C) For a particle of mass $m$ confined to a one-dimensional box of length $L$,the energy levels are given by $E_k = \frac{k^2 h^2}{8 m L^2}$.
Given $L = \sqrt{\frac{35 h \lambda}{8 m c}}$,we have $L^2 = \frac{35 h \lambda}{8 m c}$.
Substituting $L^2$ into the energy expression:
$E_k = \frac{k^2 h^2}{8 m (35 h \lambda / 8 m c)} = \frac{k^2 h c}{35 \lambda}$.
When the electron absorbs a photon of wavelength $\lambda$,it transitions from $k=1$ to $n$:
$E_n - E_1 = \frac{h c}{\lambda}$
$\frac{n^2 h c}{35 \lambda} - \frac{1^2 h c}{35 \lambda} = \frac{h c}{\lambda}$
$\frac{n^2 - 1}{35} = 1 \Rightarrow n^2 - 1 = 35 \Rightarrow n^2 = 36 \Rightarrow n = 6$.
Next,the electron transitions from $n=6$ to $m$ by emitting a photon of wavelength $\lambda^{\prime} = 1.75 \lambda = \frac{7}{4} \lambda$:
$E_6 - E_m = \frac{h c}{\lambda^{\prime}}$
$\frac{6^2 h c}{35 \lambda} - \frac{m^2 h c}{35 \lambda} = \frac{h c}{1.75 \lambda}$
$\frac{36 - m^2}{35} = \frac{1}{1.75} = \frac{1}{7/4} = \frac{4}{7}$
$36 - m^2 = 35 \times \frac{4}{7} = 5 \times 4 = 20$
$m^2 = 36 - 20 = 16 \Rightarrow m = 4$.
Thus,$n=6$ and $m=4$.

(B) The energy of a hydrogen-like atom in the $n$th state is given by $E_n = -13.6 \frac{Z^2}{n^2} \,eV$.
For a singly ionised helium atom,the atomic number $Z = 2$.
Therefore,the energy of the state $n=4$ is:
$E_4 = -13.6 \times \frac{2^2}{4^2} = -13.6 \times \frac{4}{16} = -3.4 \,eV$.
When the atom emits a photon of energy $2.6 \,eV$,its final energy $E_f$ will be:
$E_f = E_4 - 2.6 \,eV = -3.4 \,eV - 2.6 \,eV = -6.0 \,eV$.
Now,we find the quantum number $n$ for this energy level using $E_n = -13.6 \frac{Z^2}{n^2}$:
$-6.0 = -13.6 \times \frac{4}{n^2} \implies n^2 = \frac{13.6 \times 4}{6.0} \approx 9.06 \approx 9$.
Thus,$n = 3$.
The final energy is $-6.0 \,eV$ and the quantum number is $n=3$.

(D) The energy of a photon emitted during a transition is given by $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
For the first three energy levels $(n=1, 2, 3)$,the possible transitions are:
$1$. $n=3 \rightarrow n=1$ (Energy $E_1$,wavelength $\lambda_1$)
$2$. $n=2 \rightarrow n=1$ (Energy $E_2$,wavelength $\lambda_2$)
$3$. $n=3 \rightarrow n=2$ (Energy $E_3$,wavelength $\lambda_3$)
Since energy is inversely proportional to wavelength $(E \propto \frac{1}{\lambda})$,the transition with the largest energy change corresponds to the shortest wavelength. Thus,$\lambda_1$ is the shortest wavelength $(n=3 \rightarrow n=1)$,and $\lambda_3$ is the longest wavelength $(n=3 \rightarrow n=2)$.
From the conservation of energy,the energy of the transition from $n=3$ to $n=1$ is equal to the sum of the energies of the transitions from $n=3$ to $n=2$ and $n=2$ to $n=1$:
$E_{3 \rightarrow 1} = E_{3 \rightarrow 2} + E_{2 \rightarrow 1}$
Substituting $E = \frac{hc}{\lambda}$:
$\frac{hc}{\lambda_1} = \frac{hc}{\lambda_3} + \frac{hc}{\lambda_2}$
Dividing by $hc$ on both sides,we get:
$\frac{1}{\lambda_1} = \frac{1}{\lambda_2} + \frac{1}{\lambda_3}$

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.
The energy difference between two consecutive orbits $n$ and $n+1$ is $\Delta E = E_{n+1} - E_n = -13.6 \left( \frac{1}{(n+1)^2} - \frac{1}{n^2} \right) = 13.6 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) = 13.6 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right) = 13.6 \left( \frac{2n+1}{n^2(n+1)^2} \right)$.
As $n$ increases,the denominator $n^2(n+1)^2$ grows much faster than the numerator $(2n+1)$,causing the value of $\Delta E$ to decrease.
Therefore,the energy difference between consecutive energy levels decreases as $n$ increases.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula:
$E_n = -\frac{13.6 \,eV}{n^2}$
For the energy level corresponding to $n=7$:
$E_7 = -\frac{13.6}{7^2} \,eV$
$E_7 = -\frac{13.6}{49} \,eV$
$E_7 \approx -0.27755 \,eV$
Rounding to two decimal places,we get:
$E_7 \approx -0.28 \,eV$
Therefore,the correct option is $D$.

(A) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \, eV$.
For the transition from $n = 4$ to $n = 1$,$\Delta E = 13.6 \left( 1 - \frac{1}{16} \right) = 13.6 \times \frac{15}{16} = 12.75 \, eV$.
The energy of the photon is also given by $E = \frac{hc}{\lambda}$.
Given $h = 4 \times 10^{-15} \, eV \cdot s$ and $c = 3 \times 10^8 \, m/s$,we have $hc = 12 \times 10^{-7} \, eV \cdot m = 1200 \, eV \cdot nm$.
Thus,$\lambda = \frac{hc}{\Delta E} = \frac{1200 \, eV \cdot nm}{12.75 \, eV} \approx 94.11 \, nm$.
Therefore,the wavelength is approximately $94.1 \, nm$.

(D) For the hydrogen atom,the energy of a transition is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
Case $1$: Second excited state $(n_i = 3)$ to first excited state $(n_f = 2)$.
$\frac{hc}{\lambda_0} = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{5}{36} \right) \ldots (i)$
Case $2$: Third excited state $(n_i = 4)$ to second orbit $(n_f = 2)$.
$\frac{hc}{\lambda'} = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \left( \frac{3}{16} \right) \ldots (ii)$
Given $\lambda' = \frac{20}{x} \lambda_0$,we divide equation $(i)$ by $(ii)$:
$\frac{\lambda'}{\lambda_0} = \frac{13.6 \left( \frac{5}{36} \right)}{13.6 \left( \frac{3}{16} \right)} = \frac{5}{36} \times \frac{16}{3} = \frac{5 \times 4}{9 \times 3} = \frac{20}{27}$
Comparing $\frac{\lambda'}{\lambda_0} = \frac{20}{27}$ with $\frac{20}{x}$,we get $x = 27$.

(D) The energy of a photon emitted during a transition is given by $\Delta E = \frac{hc}{\lambda}$.
Given $\lambda = 124.1 \, nm = 124.1 \times 10^{-9} \, m$.
Using $hc \approx 1241 \, eV \cdot nm$,the required energy difference is $\Delta E = \frac{1241 \, eV \cdot nm}{124.1 \, nm} = 10 \, eV$.
From the figure:
Transition $A$: $\Delta E = 0 - (-2.2) = 2.2 \, eV$.
Transition $B$: $\Delta E = 0 - (-5.2) = 5.2 \, eV$.
Transition $C$: $\Delta E = -2.2 - (-5.2) = 3.0 \, eV$.
Transition $D$: $\Delta E = 0 - (-10.0) = 10.0 \, eV$.
Since transition $D$ corresponds to an energy difference of $10 \, eV$,it results in the emission of a photon with wavelength $124.1 \, nm$.

(C) 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 $Li^{2+}$,the atomic number $Z = 3$.
The second excited state corresponds to $n = 3$.
Therefore,the energy of the electron in the second excited state is $E_3 = -13.6 \times \frac{3^2}{3^2} = -13.6\,eV$.
The energy required to remove the electron (binding energy) is the magnitude of this energy,which is $13.6\,eV$.
We are given that this energy is $x \times 10^{-1}\,eV$.
So,$13.6 = x \times 10^{-1} \implies x = 136$.

(D) The energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
The energy released when an electron jumps from an initial state $n_i$ to a final state $n_f$ is $\Delta E = E_{n_i} - E_{n_f} = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
Given $Z = 4$,$n_i = 4$,and $n_f = 2$:
$\Delta E = 13.6 \times (4)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \text{ eV}$.
$\Delta E = 13.6 \times 16 \left( \frac{1}{4} - \frac{1}{16} \right) \text{ eV}$.
$\Delta E = 13.6 \times 16 \left( \frac{4-1}{16} \right) \text{ eV}$.
$\Delta E = 13.6 \times 3 \text{ eV} = 40.8 \text{ eV}$.

(D) The energy of a photon emitted during a transition is given by $\Delta E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength.
From this relation,we have $\lambda = \frac{hc}{\Delta E}$,which implies $\lambda \propto \frac{1}{\Delta E}$.
For the shortest wavelength,the energy difference $\Delta E$ must be the maximum.
Looking at the provided energy level diagram:
Transition $A$ is from $n=4$ to $n=3$.
Transition $B$ is from $n=4$ to $n=2$.
Transition $C$ is from $n=3$ to $n=2$.
Transition $D$ is from $n=3$ to $n=1$.
The energy gap is largest for the transition from $n=3$ to $n=1$,which corresponds to transition $D$.
Therefore,the correct option is $D$.

(B) The number of spectral lines emitted when an electron transitions from an excited state $n_2$ to the ground state $n_1=1$ is given by the formula $N = \frac{n_2(n_2-1)}{2}$.
Given $N = 6$,we have $\frac{n_2(n_2-1)}{2} = 6$,which implies $n_2^2 - n_2 - 12 = 0$.
Solving this quadratic equation,we get $(n_2-4)(n_2+3) = 0$. Since $n_2 > 0$,we find $n_2 = 4$.
The energy of the incident photon must be equal to the energy difference between the ground state $(n=1)$ and the excited state $(n=4)$:
$h\nu = E_4 - E_1 = -0.85 \ eV - (-13.6 \ eV) = 12.75 \ eV$.
Given $h = 4.25 \times 10^{-15} \ eVs$,the frequency $\nu$ is:
$\nu = \frac{12.75 \ eV}{4.25 \times 10^{-15} \ eVs} = 3 \times 10^{15} \ Hz$.
Comparing this with $x \times 10^{15} \ Hz$,we get $x = 3$.

(A) The energy levels for a hydrogen atom are given by $n=1$ (ground state),$n=2$ (first excited state),$n=3$ (second excited state),and $n=4$ (third excited state).
From the figure,$A$ corresponds to $n=2$,$B$ corresponds to $n=3$,and $C$ corresponds to $n=4$.
The wavelength $\lambda$ for a transition from $n_2$ to $n_1$ 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 transition $\lambda_1$ (from $n=3$ to $n=2$): $\frac{1}{\lambda_1} = R(1)^2 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left( \frac{5}{36} \right)$.
For transition $\lambda_2$ (from $n=4$ to $n=3$): $\frac{1}{\lambda_2} = R(1)^2 \left[ \frac{1}{3^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{9} - \frac{1}{16} \right] = R \left( \frac{7}{144} \right)$.
Now,calculate the ratio $\frac{\lambda_1}{\lambda_2}$:
$\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{R(7/144)}{R(5/36)} = \frac{7}{144} \times \frac{36}{5} = \frac{7}{4 \times 5} = \frac{7}{20}$.
Given the ratio is $\frac{7}{4n}$,we have $\frac{7}{20} = \frac{7}{4n}$,which implies $4n = 20$,so $n = 5$.

(A) The energy difference between the levels $n=2$ and $n=1$ is given by $\Delta E = E_2 - E_1 = -3.4 \,eV - (-13.6 \,eV) = 10.2 \,eV$.
By the principle of conservation of momentum, the momentum of the emitted photon must be equal in magnitude to the recoil momentum of the hydrogen atom.
$p_{\text{atom}} = p_{\text{photon}} = \frac{\Delta E}{c}$
Since $p_{\text{atom}} = m \cdot v$, where $m$ is the mass of the hydrogen atom and $v$ is the recoil speed:
$v = \frac{\Delta E}{m \cdot c}$
Substituting the given values:
$v = \frac{10.2 \,eV}{(1.6 \times 10^{-27} \,kg) \times (3 \times 10^8 \,m/s)}$
Convert $eV$ to Joules $(1 \,eV = 1.6 \times 10^{-19} \,J)$:
$v = \frac{10.2 \times 1.6 \times 10^{-19} \,J}{1.6 \times 10^{-27} \,kg \times 3 \times 10^8 \,m/s}$
$v = \frac{10.2 \times 10^{-19}}{3 \times 10^{-19}} \,m/s = 3.4 \,m/s$
Expressing $3.4 \,m/s$ as a fraction with denominator $5$:
$v = \frac{3.4 \times 5}{5} = \frac{17}{5} \,m/s$
Comparing this with $\frac{x}{5} \,m/s$, we get $x = 17$.