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

(C) The energy required to remove an electron from an orbit $n$ is given by the ionization energy formula: $E_n = \frac{13.6 \, Z^2}{n^2} \, eV$.
For a hydrogen atom,the atomic number $Z = 1$.
Given the state $n = 10$,we substitute these values into the formula:
$E_{10} = \frac{13.6 \times (1)^2}{(10)^2} \, eV$.
$E_{10} = \frac{13.6}{100} \, eV$.
$E_{10} = 0.136 \, eV$.

(D) The binding energy of an electron in a $H$-atom is given by the formula:
Binding Energy $= \frac{13.6}{n^2} \ eV$
Given $n = 4$,we substitute this value into the formula:
Binding Energy $= \frac{13.6}{4^2} \ eV$
Binding Energy $= \frac{13.6}{16} \ eV$
Binding Energy $= 0.85 \ eV$

(A) The ground state of a hydrogen atom corresponds to the principal quantum number $n = 1$.
The second excited state corresponds to the principal quantum number $n = 3$.
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 second excited state $(n = 3)$,the energy is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} = -1.51 \; eV$.
The energy required to ionize the atom is the energy needed to bring the electron from the current state to the state where $E = 0$.
Therefore,the ionization energy $E_{ion} = 0 - E_3 = 0 - (-1.51) = 1.51 \; eV$.

(C) The energy difference between two levels is related to the wavelength of the emitted photon by the relation $\Delta E = \frac{hc}{\lambda}$.
For the given transitions:
$1$. Transition from $C$ to $B$: $E_C - E_B = \frac{hc}{\lambda_1}$
$2$. Transition from $B$ to $A$: $E_B - E_A = \frac{hc}{\lambda_2}$
$3$. Transition from $C$ to $A$: $E_C - E_A = \frac{hc}{\lambda_3}$
From the energy level diagram,we can see that:
$(E_C - E_A) = (E_C - E_B) + (E_B - E_A)$
Substituting the energy expressions in terms of wavelength:
$\frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$
Dividing both sides by $hc$:
$\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$
$\frac{1}{\lambda_3} = \frac{\lambda_2 + \lambda_1}{\lambda_1 \lambda_2}$
Therefore,$\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$.

(A) The ionization potential $I.P.$ is given by $E_{\infty} - E_1 = 122.4 \, V$. Since $E_n = -\frac{I.P.}{n^2}$,the energy of the ground state is $E_1 = -122.4 \, eV$.
The energy of the $n^{th}$ state is $E_n = -\frac{122.4}{n^2} \, eV$.
The first excited state corresponds to $n = 2$. The excitation potential is $E_2 - E_1 = -\frac{122.4}{4} - (-122.4) = 122.4(1 - \frac{1}{4}) = 122.4 \times \frac{3}{4} = 91.8 \, V$.
The second excited state corresponds to $n = 3$. The excitation potential is $E_3 - E_1 = -\frac{122.4}{9} - (-122.4) = 122.4(1 - \frac{1}{9}) = 122.4 \times \frac{8}{9} = 108.8 \, V$.

(D) The wavelength of the absorbed radiation is $\lambda = 975 \, \text{\AA} = 975 \times 10^{-10} \, m$.
Using the Rydberg formula for the transition from the ground state $(n_1 = 1)$ to an excited state $(n_2 = n)$:
$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Substituting the values $(R \approx 1.097 \times 10^7 \, m^{-1})$:
$\frac{1}{975 \times 10^{-10}} = 1.097 \times 10^7 \left( 1 - \frac{1}{n^2} \right)$
$1.0256 \times 10^7 \approx 1.097 \times 10^7 \left( 1 - \frac{1}{n^2} \right)$
$0.935 \approx 1 - \frac{1}{n^2}$
$\frac{1}{n^2} \approx 1 - 0.935 = 0.065$
$n^2 \approx \frac{1}{0.065} \approx 15.38$
Since $n$ must be an integer, we round to the nearest perfect square, $n^2 = 16$, which gives $n = 4$.
Thus, the atom transitions to the $n = 4$ energy level.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.
The ionization potential of the $n^{th}$ orbit is the energy required to remove an electron from that orbit to infinity $(n = \infty)$,which is given by $I.P._n = E_{\infty} - E_n = 0 - E_n = -E_n$.
For the third orbit $(n = 3)$:
$I.P._3 = -E_3 = -\left( -\frac{13.6}{3^2} \right) \ eV$.
$I.P._3 = \frac{13.6}{9} \ eV \approx 1.51 \ eV$.

(C) The transition $I$ represents the absorption of a photon because the electron moves from a lower energy level to a higher energy level.
Among the remaining transitions $(II, III, IV)$,all represent the emission of a photon as the electron moves from a higher energy level to a lower energy level.
The energy of the emitted photon is given by $\Delta E = E_{initial} - E_{final} = 13.6 \text{ eV} \times Z^2 \left( \frac{1}{n_{final}^2} - \frac{1}{n_{initial}^2} \right)$.
Comparing the transitions:
Transition $II$: $n=4$ to $n=3$,$\Delta E \propto (1/9 - 1/16) = 7/144 \approx 0.048$
Transition $III$: $n=2$ to $n=1$,$\Delta E \propto (1/1 - 1/4) = 3/4 = 0.75$
Transition $IV$: $n=4$ to $n=2$,$\Delta E \propto (1/4 - 1/16) = 3/16 = 0.1875$
Comparing the values,transition $III$ corresponds to the largest energy difference,thus representing the emission of a photon with the highest energy.

(B) The binding 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 ground state $(n = 1)$:
$E_1 = \frac{13.6}{1^2} = 13.6 \ eV$.
For the first excited state $(n = 2)$:
$E_2 = \frac{13.6}{2^2} = \frac{13.6}{4} = 3.4 \ eV$.
For the second excited state $(n = 3)$:
$E_3 = \frac{13.6}{3^2} = \frac{13.6}{9} \approx 1.51 \ eV$.
Thus,the energies required to remove the electron from these levels are $13.6 \ eV, 3.4 \ eV,$ and $1.5 \ eV$ respectively.

(D) The energy of an emitted photon is given by $\Delta E = \frac{hc}{\lambda}$.
For the first transition from $2E$ to $E$:
$\Delta E_1 = 2E - E = E = \frac{hc}{\lambda} \implies E = \frac{hc}{\lambda}$.
For the second transition from $\frac{4E}{3}$ to $E$:
$\Delta E_2 = \frac{4E}{3} - E = \frac{E}{3} = \frac{hc}{\lambda'}$.
Substituting $E = \frac{hc}{\lambda}$ into the second equation:
$\frac{1}{3} \left( \frac{hc}{\lambda} \right) = \frac{hc}{\lambda'}$.
Therefore,$\lambda' = 3\lambda$.

(A) The frequency $f$ of the emitted radiation is given by $f = c / \lambda = cR [1/(n-1)^2 - 1/n^2]$.
Simplifying the expression: $f = Rc [n^2 - (n-1)^2] / [n^2(n-1)^2]$.
$f = Rc [n^2 - (n^2 - 2n + 1)] / [n^2(n-1)^2] = Rc (2n - 1) / [n^2(n-1)^2]$.
Since $n >> 1$,we can approximate $(n-1) \approx n$ and $(2n - 1) \approx 2n$.
Substituting these approximations: $f \approx Rc (2n) / [n^2(n^2)] = 2Rc / n^3$.
Therefore,$f \propto 1/n^3$.

(A) For hydrogen-like atoms,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
The binding energy is the magnitude of this energy,$E = 13.6 \times \frac{Z^2}{n^2} \ eV$.
For a $Li^{++}$ ion,the atomic number $Z = 3$.
The first excited state corresponds to $n = 2$.
Substituting these values into the formula:
$E = 13.6 \times \frac{3^2}{2^2} \ eV$
$E = 13.6 \times \frac{9}{4} \ eV$
$E = 13.6 \times 2.25 \ eV$
$E = 30.6 \ eV$.

(C) The energy of an electron in the $n^{th}$ orbit is given by $E_n = - \frac{13.6}{n^2} \ eV$.
For the first orbit $(n = 1)$: $E_1 = - \frac{13.6}{1^2} = - 13.6 \ eV$.
For the third orbit $(n = 3)$: $E_3 = - \frac{13.6}{3^2} = - \frac{13.6}{9} \approx - 1.51 \ eV$.
The energy required to move the electron from the first to the third orbit is $\Delta E = E_3 - E_1$.
$\Delta E = - 1.51 - (- 13.6) = 12.09 \ eV$.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like ion is given by the formula: $E_n = -\frac{13.6 \times Z^2}{n^2} \ eV$.
For $Li^{++}$ (Lithium ion),the atomic number $Z = 3$.
The first excited state corresponds to $n = 2$.
Substituting these values into the formula:
$E_2 = -\frac{13.6 \times 3^2}{2^2} \ eV$
$E_2 = -\frac{13.6 \times 9}{4} \ eV$
$E_2 = -30.6 \ eV$.
The energy required to remove the electron (ionization energy from this state) is the energy needed to bring the electron to $n = \infty$ (where $E = 0$).
Therefore,the required energy is $E_{\infty} - E_2 = 0 - (-30.6 \ eV) = 30.6 \ eV$.

(A) The energy of an electron in the $n^{th}$ orbit is $E_n = -\frac{13.6}{n^2} \, eV$.
To move an electron from $n_1 = 2$ to $n_2 = 3$,the required energy $\Delta E$ is given by:
$\Delta E = E_3 - E_2$
$\Delta E = \left( -\frac{13.6}{3^2} \right) - \left( -\frac{13.6}{2^2} \right)$
$\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$
$\Delta E = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right)$
$\Delta E = 13.6 \left( \frac{9 - 4}{36} \right) = 13.6 \times \frac{5}{36}$
$\Delta E \approx 1.888 \, eV \approx 1.9 \, eV$.

(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{K}{n^2}$,where $K = 13.6 \ eV$.
The energy difference between $n = 2$ and $n = 3$ orbits is $E = E_3 - E_2 = -\frac{K}{3^2} - (-\frac{K}{2^2}) = K(\frac{1}{4} - \frac{1}{9}) = K(\frac{5}{36})$.
Therefore,$E = \frac{5}{36}K$,which implies $K = \frac{36}{5}E$.
The energy required to remove an electron from the ground state $(n = 1)$ to infinity $(n = \infty)$ is the ionization energy: $E_{ion} = E_{\infty} - E_1 = 0 - (-\frac{K}{1^2}) = K$.
Substituting the value of $K$ from the first equation: $E_{ion} = \frac{36}{5}E = 7.2 \ E$.

(D) For the transition $3 \to 1$,the energy difference is $\Delta E_{31} = 2E - E = E = \frac{hc}{\lambda} \implies E = \frac{hc}{\lambda} \dots (i)$
For the transition $2 \to 1$,the energy difference is $\Delta E_{21} = \frac{4E}{3} - E = \frac{E}{3} = \frac{hc}{\lambda'}$
Substituting the value of $E$ from equation $(i)$: $\frac{1}{3} \left( \frac{hc}{\lambda} \right) = \frac{hc}{\lambda'}$
Therefore,$\lambda' = 3\lambda$.

(B) The energy required for a transition between orbits $n_1$ and $n_2$ is given by the formula:
$\Delta E = 13.6 \, Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \, eV$
Given $\Delta E = 47.2 \, eV$,$n_1 = 2$,and $n_2 = 3$:
$47.2 = 13.6 \times Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$
$47.2 = 13.6 \times Z^2 \left( \frac{1}{4} - \frac{1}{9} \right)$
$47.2 = 13.6 \times Z^2 \left( \frac{9 - 4}{36} \right)$
$47.2 = 13.6 \times Z^2 \left( \frac{5}{36} \right)$
$Z^2 = \frac{47.2 \times 36}{13.6 \times 5}$
$Z^2 = \frac{1699.2}{68} \approx 24.988 \approx 25$
$Z = \sqrt{25} = 5$
Thus,the atomic number $Z$ is $5$.

(B) The energy of an electron in the ground state $(n = 1)$ of a hydrogen atom is $E_1 = -13.6 \, eV$.
When energy $E_{given} = 12.4 \, eV$ is supplied,the new energy $E_n$ of the electron will be:
$E_n = E_1 + E_{given} = -13.6 \, eV + 12.4 \, eV = -1.2 \, eV$.
The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \, eV$.
Equating the two expressions:
$-\frac{13.6}{n^2} = -1.2$
$n^2 = \frac{13.6}{1.2} \approx 11.33$.
Since $n$ must be an integer,we look for the nearest orbit. The electron will transition to the $n = 3$ orbit $(E_3 = -1.51 \, eV)$ as it does not have enough energy to reach the $n = 4$ orbit $(E_4 = -0.85 \, eV)$.

(D) The total energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by $E_n = \frac{-13.6}{n^2} \text{ eV}$.
For the ground state,$n = 1$,so $E_1 = -13.6 \text{ eV}$.
For the first excited state,$n = 2$.
Therefore,the total energy in the first excited state is $E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.4 \text{ eV}$.
In any orbit,the kinetic energy $K$ is equal to the negative of the total energy $E$,i.e.,$K = -E$.
Thus,the kinetic energy of the electron in the first excited state is $K = -(-3.4 \text{ eV}) = 3.4 \text{ eV}$.

(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 ground state $(n=1)$,$E_1 = -13.6 \, eV$.
For the first excited state $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, eV$.
The excitation energy required to move the electron from the ground state to the first excited state is $\Delta E = E_2 - E_1$.
$\Delta E = -3.4 \, eV - (-13.6 \, eV) = 10.2 \, eV$.

(D) According to the Rydberg formula:
$\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$
In the first case,the electron jumps from the third excited state $(n_i = 4)$ to the second excited state $(n_f = 3)$:
$\frac{1}{\lambda_1} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{9} - \frac{1}{16} \right] = R \left[ \frac{16 - 9}{144} \right] = \frac{7}{144} R$ .... $(i)$
In the second case,the electron jumps from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$:
$\frac{1}{\lambda_2} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{9 - 4}{36} \right] = \frac{5}{36} R$ .... $(ii)$
To find the ratio $\frac{\lambda_1}{\lambda_2}$,we divide the expression for $\frac{1}{\lambda_1}$ by $\frac{1}{\lambda_2}$:
$\frac{\lambda_2}{\lambda_1} = \frac{\frac{5}{36} R}{\frac{7}{144} R} = \frac{5}{36} \times \frac{144}{7} = \frac{5 \times 4}{7} = \frac{20}{7}$
Therefore,the ratio $\frac{\lambda_1}{\lambda_2} = \frac{20}{7}$.

(B) The electrostatic potential energy of the electron-proton system is defined as zero when they are at an infinite distance apart.
To pull the electron and proton apart from the state $n = 2$,we must provide energy equal to the binding energy of that state.
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 $n = 2$,the energy is $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV}$.
The work done $W$ required to move the electron to infinity (where $E_f = 0$) is $W = E_f - E_i = 0 - (-3.4 \text{ eV}) = 3.4 \text{ eV}$.
Converting this to Joules: $W = 3.4 \times 1.6 \times 10^{-19} \text{ J}$.

(C) The number of dark lines in an absorption spectrum is given by $(n - 1)$,where $n$ is the principal quantum number of the highest excited state reached by the atoms.
Given that there are $5$ dark lines,we have $n - 1 = 5$,which implies $n = 6$.
When these atoms return to the ground state,the number of bright lines in the emission spectrum is given by the formula $\frac{n(n - 1)}{2}$.
Substituting $n = 6$ into the formula,we get $\frac{6(6 - 1)}{2} = \frac{6 \times 5}{2} = 15$.
Therefore,the number of bright lines in the emission spectrum is $15$.

(C) The energy of an emitted photon is given by $\Delta E = E_2 - E_1 = \frac{hc}{\lambda}$.
Since $\lambda = \frac{hc}{\Delta E}$,the wavelength $\lambda$ is inversely proportional to the energy difference $\Delta E$.
For the shortest wavelength,the energy difference $\Delta E$ must be maximum. Looking at the diagram,transition $C$ (from $n=4$ to $n=2$) covers a larger energy gap compared to $D$ (from $n=4$ to $n=3$) or $B$ (from $n=3$ to $n=2$).
For the longest wavelength,the energy difference $\Delta E$ must be minimum. Transition $D$ (from $n=4$ to $n=3$) represents the smallest energy gap among the given transitions.
Therefore,the shortest wavelength corresponds to transition $C$ and the longest wavelength corresponds to transition $D$.

(C) The energy of an electron in the $n^{th}$ shell of a hydrogen atom is given by $E_n = -\frac{E_1}{n^2}$,where $E_1 = 2.18 \times 10^{-18} \ J$ is the ionization energy of the ground state $(n=1)$.
For $n = 3$,the energy of the electron is $E_3 = -\frac{2.18 \times 10^{-18}}{3^2} \ J$.
$E_3 = -\frac{2.18 \times 10^{-18}}{9} \ J$.
To ionize the atom,we must provide energy equal to the magnitude of the electron's energy in that state to bring it to $E = 0$.
Required Energy $= |E_3| = \frac{2.18 \times 10^{-18}}{9} \ J$.
Required Energy $= 0.2422 \times 10^{-18} \ J = 2.42 \times 10^{-19} \ J$.

(D) The energy of an electron in the $n$-th orbit of a hydrogen-like atom is given by $E_n = -\frac{Z^2 E_0}{n^2}$,where $E_0 = 13.6 \text{ eV}$.
Calculating the energy differences:
$E_{4n} - E_{2n} = -\frac{Z^2 E_0}{(4n)^2} - \left( -\frac{Z^2 E_0}{(2n)^2} \right) = Z^2 E_0 \left( \frac{1}{4n^2} - \frac{1}{16n^2} \right) = Z^2 E_0 \left( \frac{4-1}{16n^2} \right) = \frac{3 Z^2 E_0}{16n^2}$.
$E_{2n} - E_n = -\frac{Z^2 E_0}{(2n)^2} - \left( -\frac{Z^2 E_0}{n^2} \right) = Z^2 E_0 \left( \frac{1}{n^2} - \frac{1}{4n^2} \right) = Z^2 E_0 \left( \frac{4-1}{4n^2} \right) = \frac{3 Z^2 E_0}{4n^2}$.
Taking the ratio:
$\frac{E_{4n} - E_{2n}}{E_{2n} - E_n} = \frac{3 Z^2 E_0 / 16n^2}{3 Z^2 E_0 / 4n^2} = \frac{4}{16} = \frac{1}{4}$.
Since the result is a constant,it is independent of $Z$ and $n$,which can be expressed as $Z^0 n^0$.

(A) The number of spectral lines emitted when an electron transitions from an energy level $n_2$ to a lower energy level $n_1$ is given by the formula $N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.
Here,the electron transitions from the $n^{th}$ excited state to the second excited state.
The $n^{th}$ excited state corresponds to the principal quantum number $n_2 = n + 1$.
The second excited state corresponds to the principal quantum number $n_1 = 3$.
Given that the number of wavelengths emitted is $10$,we have:
$10 = \frac{((n+1) - 3)((n+1) - 3 + 1)}{2}$
$10 = \frac{(n-2)(n-1)}{2}$
$20 = n^2 - 3n + 2$
$n^2 - 3n - 18 = 0$
$(n-6)(n+3) = 0$
Since $n$ must be positive,we get $n = 6$.

(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 second orbit,$n = 2$.
Substituting the value of $n$ into the formula: $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \ eV$.
The energy required to remove an electron from the second orbit is the energy needed to bring the electron from $E_2 = -3.4 \ eV$ to $E = 0 \ eV$ (infinity).
Therefore,the required energy is $0 - (-3.4) = 3.4 \ eV$.

(C) The longest wavelength of the Balmer series for a hydrogen atom corresponds to the transition of an electron from the $n=3$ state to the $n=2$ state.
This implies that when the hydrogen atom was excited by the colliding electron,the electron was promoted from the $n=1$ ground state to the $n=3$ excited state.
The energy of the $n$-th quantum state for a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
For the ground state $(n=1)$: $E_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV}$.
For the excited state $(n=3)$: $E_3 = -\frac{13.6}{3^2} = -1.51 \text{ eV}$.
The energy required for this excitation is $\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09 \text{ eV} \approx 12.1 \text{ eV}$.
Since the colliding electron loses all its kinetic energy to excite the atom,the minimum kinetic energy of the electron must be equal to the excitation energy,which is $12.1 \text{ eV}$.

(C) In a hydrogen atom,the total energy $E$,kinetic energy $K$,and potential energy $U$ are related by $U = 2E$ and $K = -E$.
For the first excited state $(n=2)$,the total energy is $E_2 = -13.6 / 2^2 = -3.4 \ eV$.
The potential energy in the first excited state is $U_2 = 2E_2 = 2(-3.4) = -6.8 \ eV$.
If we shift the potential energy scale such that the new potential energy $U'_2 = 0$,we have added $+6.8 \ eV$ to the potential energy.
Since the total energy $E = K + U$,and kinetic energy $K$ remains unchanged by a shift in the potential energy reference point,the new total energy $E'$ is $E' = E + \Delta U$.
For the ground state $(n=1)$,the original total energy is $E_1 = -13.6 \ eV$.
The shift applied is $\Delta U = +6.8 \ eV$.
Therefore,the new total energy in the ground state is $E'_1 = -13.6 + 6.8 = -6.8 \ eV$.

(B) For an electron in a hydrogen-like atom,the total energy $(TE)$ is equal to the negative of the kinetic energy $(KE)$: $TE = -KE$.
Given $TE = -3.4 \, eV$,we have $-3.4 \, eV = -KE$,which implies $KE = 3.4 \, eV$.
Thus,$E = 3.4 \, eV$.
The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p = \sqrt{2mE}$.
Substituting the values: $h = 6.63 \times 10^{-34} \, J \cdot s$,$m = 9.11 \times 10^{-31} \, kg$,and $E = 3.4 \times 1.6 \times 10^{-19} \, J$.
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.11 \times 10^{-31} \times 3.4 \times 1.6 \times 10^{-19}}}$
$\lambda = \frac{6.63 \times 10^{-34}}{\sqrt{9.92 \times 10^{-49}}} \approx \frac{6.63 \times 10^{-34}}{9.96 \times 10^{-25}} \approx 0.665 \times 10^{-9} \, m = 6.65 \times 10^{-10} \, m$.
Rounding to the given options,$\lambda \approx 6.6 \times 10^{-10} \, m$.

(B) In a standard hydrogen atom,the potential energy is $U = -\frac{27.2 \ eV}{n^2}$ and the total energy is $E = -\frac{13.6 \ eV}{n^2}$.
If we shift the reference level such that the potential energy at the ground state $(n=1)$ is zero,we add a constant $C$ to the potential energy expression.
Since $U(1) = -27.2 + C = 0$,we get $C = 27.2 \ eV$.
Thus,the new potential energy is $U' = -\frac{27.2}{n^2} + 27.2 \ eV$.
The total energy $E'$ becomes $E' = K + U' = \frac{13.6}{n^2} + (-\frac{27.2}{n^2} + 27.2) = 27.2 - \frac{13.6}{n^2} \ eV$.
For $n=1$,$E' = 27.2 - 13.6 = 13.6 \ eV$.
As $n$ increases,$\frac{13.6}{n^2}$ decreases,so $E'$ increases. Therefore,statement $B$ is incorrect.

(A) The energy of a photon emitted during an electronic transition is given by $\Delta E = E_{initial} - E_{final} = h\nu$,where $\nu$ is the frequency.
For emission to occur,the electron must transition from a higher energy level to a lower energy level. Thus,options $(b)$ and $(d)$ are excluded as they represent absorption.
Comparing the energy differences for the remaining transitions:
For $n = 2$ to $n = 1$: $\Delta E_1 = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - 0.25 \right) = 13.6 \times 0.75 = 10.2 \text{ eV}$.
For $n = 6$ to $n = 2$: $\Delta E_2 = 13.6 \left( \frac{1}{2^2} - \frac{1}{6^2} \right) = 13.6 \left( 0.25 - 0.0277 \right) = 13.6 \times 0.2223 \approx 3.02 \text{ eV}$.
Since $\nu = \frac{\Delta E}{h}$,the transition with the largest energy difference will emit the photon with the highest frequency.
Comparing $\Delta E_1$ and $\Delta E_2$,we see that $\Delta E_1 > \Delta E_2$.
Therefore,the transition from $n = 2$ to $n = 1$ emits the photon of the highest frequency.

(D) For a hydrogen-like atom,the radius of the orbit is given by $r_n \propto n^2$. As the electron transitions from an excited state to the ground state,the principal quantum number $n$ decreases,so the radius $r$ decreases.
The kinetic energy is $K.E. = \frac{kZe^2}{2r}$. Since $r$ decreases,$K.E.$ increases.
The potential energy is $P.E. = -\frac{kZe^2}{r}$. Since $r$ decreases,the magnitude of $P.E.$ increases,meaning $P.E.$ becomes more negative (decreases).
The total energy is $T.E. = -\frac{kZe^2}{2r}$. Since $r$ decreases,the magnitude of $T.E.$ increases,meaning $T.E.$ becomes more negative (decreases).
Therefore,kinetic energy increases,while potential energy and total energy decrease.

(D) From the energy level diagram,we use the relation $\Delta E = \frac{hc}{\lambda}$.
For the transition corresponding to wavelength $\lambda_1$,the energy difference is $\Delta E_1 = -E - (-2E) = E$.
Thus,$\frac{hc}{\lambda_1} = E$,which gives $\lambda_1 = \frac{hc}{E}$.
For the transition corresponding to wavelength $\lambda_2$,the energy difference is $\Delta E_2 = -E - (-\frac{4}{3}E) = \frac{1}{3}E$.
Thus,$\frac{hc}{\lambda_2} = \frac{E}{3}$,which gives $\lambda_2 = \frac{3hc}{E}$.
The ratio $r = \frac{\lambda_1}{\lambda_2} = \frac{hc/E}{3hc/E} = \frac{1}{3}$.

(C) The energy of a photon with wavelength $\lambda = 800 \ nm$ is given by $E = \frac{hc}{\lambda} = \frac{1240}{800} = 1.55 \ eV$.
Since only photons with $\lambda > 800 \ nm$ pass through the filter,the energy of the emitted photons must be $E < 1.55 \ eV$.
The energy levels of the hydrogen atom are $E_n = -\frac{13.6}{n^2} \ eV$.
For $n=1, 2, 3, 4, 5$,the energies are $-13.6, -3.4, -1.51, -0.85, -0.54 \ eV$ respectively.
The energy difference for the transition $n=4 \to n=3$ is $\Delta E = E_4 - E_3 = -0.85 - (-1.51) = 0.66 \ eV$.
Since $0.66 \ eV < 1.55 \ eV$,this photon passes through the filter.
The next possible transition $n=5 \to n=4$ has $\Delta E = -0.54 - (-0.85) = 0.31 \ eV$,which also passes. However,if the sample were in the $4^{th}$ excited state $(n=5)$,multiple transitions would result in photons passing through the filter. Given that only one type of photon passes,the sample must be in the $3^{rd}$ excited state $(n=4)$.

(B) The energy levels are $E_1 = -12\, eV$,$E_2 = -7\, eV$,$E_3 = -3\, eV$,and $E_4 = -1\, eV$.
The electron starts at $E_4 = -1\, eV$ and ends at $E_1 = -12\, eV$. The total energy released is $\Delta E = (-1\, eV) - (-12\, eV) = 11\, eV$.
Since the electron emits exactly two photons,the sum of the energies of the two photons must be $11\, eV$ $(h\nu_1 + h\nu_2 = 11\, eV)$.
Possible transitions from $E_4 = -1\, eV$ to $E_1 = -12\, eV$ in two steps:
$1$. $E_4 \to E_3 \to E_1$: Energies are $(-1 - (-3)) = 2\, eV$ and $(-3 - (-12)) = 9\, eV$.
$2$. $E_4 \to E_2 \to E_1$: Energies are $(-1 - (-7)) = 6\, eV$ and $(-7 - (-12)) = 5\, eV$.
Possible photon energies are $2\, eV, 9\, eV, 6\, eV, 5\, eV$.
Comparing these with the options,$4\, eV$ is not among the possible photon energies.

(D) The energy of an emitted photon is given by $\Delta E = E_2 - E_1 = \frac{hc}{\lambda}$.
From this relation,we see that the wavelength $\lambda = \frac{hc}{\Delta E}$.
Therefore,for the wavelength $\lambda$ to be maximum,the energy difference $\Delta E$ must be minimum.
Among the emission transitions shown $(II, III, IV)$,the transition $II$ occurs between $n=4$ and $n=3$,which has the smallest energy gap $(\Delta E = 0.66 \ eV)$.
Thus,transition $II$ corresponds to the emission of a photon with the maximum wavelength.

(B) The electron starts at $-1\,eV$ and ends at $-12\,eV$. The total energy released is $\Delta E = (-1\,eV) - (-12\,eV) = 11\,eV$.
Since the electron transitions by emitting exactly two photons, the sum of the energies of the two photons must be $11\,eV$ $(E_1 + E_2 = 11\,eV)$.
Possible intermediate states are $-3\,eV$ and $-7\,eV$.
Case $1$: Transition via $-3\,eV$ level:
Photon $1$: $(-1\,eV) - (-3\,eV) = 2\,eV$
Photon $2$: $(-3\,eV) - (-12\,eV) = 9\,eV$
Case $2$: Transition via $-7\,eV$ level:
Photon $1$: $(-1\,eV) - (-7\,eV) = 6\,eV$
Photon $2$: $(-7\,eV) - (-12\,eV) = 5\,eV$
Possible photon energies are $2\,eV, 9\,eV, 6\,eV, 5\,eV$.
Comparing with the options, $4\,eV$ is not among the possible photon energies.

(B) The energy difference for the transition from $C$ to $A$ is the sum of the energy differences for the transitions from $C$ to $B$ and $B$ to $A$.
$E_C - E_A = (E_C - E_B) + (E_B - E_A)$
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 $hc$:
$\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$
$\frac{1}{\lambda_3} = \frac{\lambda_2 + \lambda_1}{\lambda_1 \lambda_2}$
Therefore, $\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$.

(C) The wavelength of the spectral lines is given by $\lambda = \frac{1500p^2}{p^2 - 1}$.
To find the ionization energy,we consider the transition from $n = \infty$ to $n = 1$,which corresponds to the series limit (minimum wavelength).
As $p \to \infty$,$\lambda_{\min} = \lim_{p \to \infty} \frac{1500p^2}{p^2 - 1} = 1500 \mathring{A}$.
The energy of the photon corresponding to this wavelength is $\Delta E = \frac{hc}{\lambda_{\min}}$.
Substituting the given values,$\Delta E = \frac{12420 \text{ eV-} \mathring{A}}{1500 \mathring{A}} = 8.28 \text{ eV}$.
Since the ionization potential is the energy required to remove an electron from the ground state $(n=1)$ to infinity,the ionization potential is $8.28 \text{ V}$.

(A) The energy of the emitted photon 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 hydrogen atom $(Z=1)$ transitioning from $n_2 = 4$ to $n_1 = 1$:
$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{4^2} \right) = R \left( 1 - \frac{1}{16} \right) = \frac{15R}{16}$.
The momentum of the emitted photon is $p = \frac{h}{\lambda}$.
Substituting the value of $\frac{1}{\lambda}$: $p = h \left( \frac{15R}{16} \right) = \frac{15Rh}{16}$.
By the law of conservation of momentum,the momentum of the hydrogen atom must be equal and opposite to the momentum of the photon: $p_{\text{atom}} = p_{\text{photon}} = mv$.
Therefore,$mv = \frac{15Rh}{16}$,which gives $v = \frac{15}{16} \frac{Rh}{m}$.

(C) The wavelength $\lambda$ of the emitted photon is given by the Rydberg formula:
$\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] = R \left[ \frac{1}{1^2} - \frac{1}{5^2} \right] = R \left[ 1 - \frac{1}{25} \right] = \frac{24}{25} R$.
According to the law of conservation of momentum,the momentum of the atom $p_{atom}$ must be equal in magnitude to the momentum of the emitted photon $p_{photon}$ because the atom was initially stationary:
$p_{atom} = p_{photon} = \frac{h}{\lambda}$.
Since $p_{atom} = mv$,we have:
$mv = \frac{h}{\lambda}$.
Substituting the value of $\frac{1}{\lambda}$:
$v = \frac{h}{m} \cdot \frac{1}{\lambda} = \frac{h}{m} \cdot \left( \frac{24}{25} R \right) = \frac{24hR}{25m}$.

(B) The energy of the photon emitted by the hydrogen atom during the transition from $n = 2$ to $n = 1$ is given by:
$E = 13.6 \text{ eV} \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
For hydrogen $(Z = 1)$,$E = 13.6 \times 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \times \frac{3}{4} = 10.2 \text{ eV}$.
This photon is absorbed by a doubly ionized lithium atom ($Li^{2+}$,$Z = 3$) in an excited state $n$. The energy required to remove an electron from the $n$-th state of a hydrogen-like ion is:
$E_n = 13.6 \text{ eV} \times \frac{Z^2}{n^2} = 13.6 \times \frac{3^2}{n^2} = 13.6 \times \frac{9}{n^2}$.
For the photon to completely remove the electron,its energy must be greater than or equal to the ionization energy of the state $n$:
$10.2 \ge 13.6 \times \frac{9}{n^2}$
$\frac{10.2}{13.6} \ge \frac{9}{n^2}$
$0.75 \ge \frac{9}{n^2}$
$n^2 \ge \frac{9}{0.75} = 12$
$n \ge \sqrt{12} \approx 3.46$.
Since $n$ must be an integer,the least quantum number is $n = 4$.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \, eV$.
For $Li^{++}$,the atomic number $Z = 3$.
The first excited state corresponds to $n = 2$.
The binding energy (energy required to remove the electron) is the magnitude of the energy in that state: $E = 13.6 \times \frac{Z^2}{n^2} \, eV$.
Substituting the values: $E = 13.6 \times \frac{3^2}{2^2} = 13.6 \times \frac{9}{4} = 13.6 \times 2.25 = 30.6 \, eV$.

(A) The energy levels of a hydrogen atom are given by $E_n = -\frac{13.6}{n^2} \ eV$.
At room temperature,hydrogen atoms are in the ground state $(n=1)$.
The energy required to excite an electron from $n=1$ to $n=2$ is $E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$.
The energy required to excite an electron from $n=1$ to $n=3$ is $E_3 - E_1 = -1.51 - (-13.6) = 12.09 \ eV$.
The energy required to excite an electron from $n=1$ to $n=4$ is $E_4 - E_1 = -0.85 - (-13.6) = 12.75 \ eV$.
Since the incident electron beam has an energy of $12.5 \ eV$,it can excite hydrogen atoms up to the $n=3$ energy level.
When these excited atoms return to the ground state,the possible transitions are:
$n=3 \rightarrow n=2$ (Balmer series)
$n=3 \rightarrow n=1$ (Lyman series)
$n=2 \rightarrow n=1$ (Lyman series)
Thus,there are $2$ lines in the Lyman series and $1$ line in the Balmer series.

(B) The possible transitions for emission are from higher energy levels to lower energy levels. The energy of the emitted photon is given by $\Delta E = E_f - E_i$. The possible transitions are:
$1$. From $E_3$ to $E_2$: $\Delta E_{32} = -2 \ eV - (-6 \ eV) = 4 \ eV$.
$2$. From $E_3$ to $E_1$: $\Delta E_{31} = -2 \ eV - (-8 \ eV) = 6 \ eV$.
$3$. From $E_2$ to $E_1$: $\Delta E_{21} = -6 \ eV - (-8 \ eV) = 2 \ eV$.
Using the formula $\lambda = \frac{hc}{\Delta E}$,where $hc \approx 1240 \ eV \cdot nm$:
- For $\Delta E = 4 \ eV$,$\lambda = \frac{1240}{4} = 310 \ nm$.
- For $\Delta E = 6 \ eV$,$\lambda = \frac{1240}{6} \approx 206.67 \ nm \approx 207 \ nm$.
- For $\Delta E = 2 \ eV$,$\lambda = \frac{1240}{2} = 620 \ nm$.
Comparing these with the given options,$465 \ nm$ is not present in the emission spectrum.