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

(A) The energy of an electron in the $n^{th}$ orbit is $E_n = -\frac{13.6}{n^2} \text{ eV}$.
For the first orbit $(n=1)$: $E_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV}$.
For the second orbit $(n=2)$: $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV}$.
The energy required to excite the electron from the first orbit to the second orbit is given by the difference in energy levels:
$\Delta E = E_2 - E_1$
$\Delta E = -3.4 \text{ eV} - (-13.6 \text{ eV})$
$\Delta E = -3.4 + 13.6 = 10.2 \text{ eV}$.
Therefore,the energy required is $10.2 \text{ eV}$.

(A) represents the excitation of an electron from the $n=2$ orbit to the $n=4$ orbit,which corresponds to an absorption line in the Balmer series.
$E$ represents the transition of an electron from the $\infty$ orbit to the $n=1$ orbit,which corresponds to the energy released equal to the ionization potential of hydrogen $(13.6 \ eV)$.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.
To knock out the electron (i.e.,to ionize it),we need to provide energy equal to the magnitude of the binding energy of that orbit.
The energy required to remove the electron from the $n^{th}$ orbit is $E = +\frac{13.6}{n^2} \ eV$.
For the third orbit,$n = 3$.
Therefore,the energy required is $E = +\frac{13.6}{3^2} \ eV = +\frac{13.6}{9} \ eV$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \text{ eV}$.
To remove an electron from the $n = 10$ state to infinity $(E_{\infty} = 0)$, the energy required is the ionization energy for that state.
Energy required $= E_{\infty} - E_{10} = 0 - \left(-\frac{13.6}{10^2}\right) \text{ eV}$.
Energy required $= \frac{13.6}{100} \text{ eV} = 0.136 \text{ eV}$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \text{ eV}}{n^2}$.
For the first excited state, the principal quantum number is $n_1 = 2$.
For the second excited state, the principal quantum number is $n_2 = 3$.
The energy in the first excited state is $E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} \text{ eV}$.
The energy in the second excited state is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \text{ eV}$.
The ratio of the energies is $\frac{E_2}{E_3} = \frac{-13.6/4}{-13.6/9} = \frac{9}{4}$.

(C) The Rydberg formula for the wavelength $\lambda$ is given by: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Given $n_1 = 2$,$n_2 = 4$,and $R = 10^5 \text{ cm}^{-1} = 10^7 \text{ m}^{-1}$.
$\frac{1}{\lambda} = 10^5 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 10^5 \left( \frac{1}{4} - \frac{1}{16} \right) = 10^5 \left( \frac{3}{16} \right) \text{ cm}^{-1}$.
Thus,$\lambda = \frac{16}{3} \times 10^{-5} \text{ cm} = \frac{16}{3} \times 10^{-7} \text{ m}$.
The frequency $\nu$ is given by $\nu = \frac{c}{\lambda}$,where $c = 3 \times 10^8 \text{ m/s}$.
$\nu = \frac{3 \times 10^8}{\frac{16}{3} \times 10^{-7}} = \frac{9}{16} \times 10^{15} \text{ Hz}$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \, eV$.
To remove an electron from the $n = 2$ state,we need to provide energy equal to the magnitude of its binding energy in that state.
The energy required is $E = |E_2| = |-\frac{13.6}{2^2}| \, eV$.
$E = \frac{13.6}{4} \, eV = 3.4 \, eV$.
Therefore,the correct option is $D$.

(D) For the transition from $2E$ to $E$:
$\Delta E_1 = 2E - E = E = \frac{hc}{\lambda} \implies E = \frac{hc}{\lambda}$ (Equation $1$)
For the transition from $\frac{4E}{3}$ to $E$:
$\Delta E_2 = \frac{4E}{3} - E = \frac{E}{3} = \frac{hc}{\lambda'}$ (Equation $2$)
Substituting the value of $E$ from Equation $1$ into Equation $2$:
$\frac{1}{3} \left( \frac{hc}{\lambda} \right) = \frac{hc}{\lambda'}$
$\frac{1}{3\lambda} = \frac{1}{\lambda'} \implies \lambda' = 3\lambda$
Therefore,the correct option is $(D)$.

(D) The energy of a photon emitted during a transition is given by $E = h\nu = \frac{hc}{\lambda}$.
Since the wavelength of blue light is shorter than that of red light $(\lambda_{\text{blue}} < \lambda_{\text{red}})$,the energy of the photon corresponding to the blue line must be greater than the energy of the photon corresponding to the red line $(E_{\text{blue}} > E_{\text{red}})$.
Given that the transition from $Q$ to $S$ corresponds to a red line,we need a transition with a larger energy difference than $E_{Q \to S}$.
Looking at the energy level diagram,the transition from $R$ to $G$ involves a larger energy gap compared to $Q$ to $S$.
Therefore,the transition $R$ to $G$ will emit a photon with higher energy,corresponding to the blue line.
Thus,option $(d)$ is correct.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.
For the first excited state,$n_1 = 2$.
For the third excited state,$n_2 = 4$.
The energy of the emitted photon is given by the difference in energies: $\Delta E = E_{n_2} - E_{n_1} = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \ eV$.
Substituting the values: $\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \left( \frac{4-1}{16} \right) = 13.6 \times \frac{3}{16} = 2.55 \ eV$.

(C) In an absorption spectrum,atoms absorb photons to transition from the ground state to higher energy levels.
In the given diagram,the ground state is represented by level $X$.
Therefore,only transitions starting from level $X$ will appear in the absorption spectrum.
Looking at the diagram,the transitions originating from level $X$ are lines $1, 2,$ and $3$.
Thus,lines $1, 2,$ and $3$ will occur in the absorption spectrum.

(A) For a hydrogen atom,the kinetic energy $(K.E.)$ and potential energy $(P.E.)$ of an electron in an orbit of radius $r$ are given by:
$K.E. = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{2r}$
$P.E. = -\frac{1}{4 \pi \epsilon_0} \frac{e^2}{r}$
As the hydrogen atom is raised from the ground state to an excited state,the principal quantum number $n$ increases,which causes the orbital radius $r$ to increase $(r \propto n^2)$.
Since $K.E. \propto \frac{1}{r}$,as $r$ increases,$K.E.$ decreases.
Since $P.E. \propto -\frac{1}{r}$,as $r$ increases,the magnitude of $P.E.$ decreases,but because it is negative,the value of $P.E.$ increases (becomes less negative).
Therefore,$P.E.$ increases and $K.E.$ decreases.

(C) According to the Bohr model of the hydrogen atom,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \, eV$.
For the lowest orbit,$n = 1$.
Substituting $n = 1$ into the formula,we get $E_1 = -\frac{13.6}{1^2} = -13.6 \, eV$.
Since $-13.6 \, eV$ is the most negative value possible for the energy levels of a hydrogen atom,it represents the minimum energy state (ground state).

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \, eV$.
For the $n = 3$ state,the energy is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \, eV$.
$E_3 = -1.51 \, eV$.
To ionize the atom,we need to provide enough energy to bring the electron to the $n = \infty$ state,where the energy is $0 \, eV$.
The required ionization energy is $\Delta E = E_{\infty} - E_3 = 0 - (-1.51) = 1.51 \, eV$.

(B) The frequency of the emitted photon is given by $\nu = \frac{\Delta E}{h}$,where $\Delta E$ is the energy difference between the two levels. Since $\nu \propto \Delta E$,the lowest frequency corresponds to the transition with the smallest energy difference.
From the energy level diagram:
For $n = 2$ to $n = 1$: $\Delta E = -3.4 - (-13.6) = 10.2 \, eV$
For $n = 4$ to $n = 3$: $\Delta E = -0.85 - (-1.51) = 0.66 \, eV$
For $n = 3$ to $n = 1$: $\Delta E = -1.51 - (-13.6) = 12.09 \, eV$
For $n = 4$ to $n = 2$: $\Delta E = -0.85 - (-3.4) = 2.55 \, eV$
The transition $n = 4$ to $n = 3$ has the smallest energy difference $(0.66 \, eV)$,and therefore emits a photon of the lowest frequency. Thus,the correct option is $B$.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
The energy change for a transition from $n_i$ to $n_f$ is $\Delta E = E_{n_i} - E_{n_f} = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
Calculating the energy change for each option:
$(A)$ $n = 2$ to $n = 1$: $\Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \times (1 - 0.25) = 10.2 \text{ eV}$.
$(B)$ $n = 3$ to $n = 1$: $\Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 13.6 \times (1 - 0.111) = 12.09 \text{ eV}$.
$(C)$ $n = 4$ to $n = 2$: $\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \times (0.25 - 0.0625) = 2.55 \text{ eV}$.
$(D)$ $n = 3$ to $n = 2$: $\Delta E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \times (0.25 - 0.111) = 1.89 \text{ eV}$.
Comparing these values,the transition from $n = 3$ to $n = 1$ involves the maximum energy change of $12.09 \text{ eV}$. Thus,option $(B)$ is correct.

(D) 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 $(n=1)$,the energy is $E_1 = -13.6 \; eV$.
To excite the atom to the first excited state $(n=2)$,the energy is $E_2 = -\frac{13.6}{2^2} = -3.4 \; eV$.
The minimum energy required for excitation is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \; eV$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{13.6}{n^2} \, eV$.
For the $3^{rd}$ orbit $(n=3)$: $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \, eV$.
For the $4^{th}$ orbit $(n=4)$: $E_4 = -\frac{13.6}{4^2} = -\frac{13.6}{16} = -0.85 \, eV$.
The energy difference for a transition between the $4^{th}$ and $3^{rd}$ orbit is: $\Delta E = E_4 - E_3 = -0.85 - (-1.51) = 0.66 \, eV$.

(D) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
For the transition $n = 4 \to n = 2$,the energy emitted corresponds to blue light.
For the transition $n = 5 \to n = 2$,the energy difference is $\Delta E_{5 \to 2} = 13.6 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{25} \right) = 13.6 \left( \frac{21}{100} \right) = 2.856 \text{ eV}$.
For the transition $n = 4 \to n = 2$,the energy difference is $\Delta E_{4 \to 2} = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \left( \frac{3}{16} \right) = 2.55 \text{ eV}$.
Since $\Delta E_{5 \to 2} > \Delta E_{4 \to 2}$,the frequency of the emitted light will be higher,and the wavelength will be shorter.
In the visible spectrum,violet light has a higher frequency and shorter wavelength than blue light. Therefore,the atom emits violet light.

(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 first excited state,the quantum number is $n = 2$.
Substituting $n = 2$ into the formula:
$E_2 = \frac{-13.6 \ eV}{2^2} = \frac{-13.6 \ eV}{4} = -3.4 \ eV$.

(B) For an electron in a hydrogen atom,the total energy $(E)$,kinetic energy $(K)$,and potential energy $(U)$ are related as follows:
$E = -K$
$U = 2E = -2K$
Given that the total energy of the electron in the first excited state is $E = -3.4 \ eV$.
Therefore,the kinetic energy $K = -E = -(-3.4 \ eV) = +3.4 \ eV$.
Thus,the correct option is $B$.

(C) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \, eV$.
For the ground state $(n = 1)$,$E_1 = -13.6 \, eV$.
For the first excited state $(n = 2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \, eV$.
The energy required to excite the electron from $n = 1$ to $n = 2$ is $\Delta E = E_2 - E_1 = -3.4 \, eV - (-13.6 \, eV) = 10.2 \, eV$.
To convert this energy into Joules,we multiply by the conversion factor $1.6 \times 10^{-19} \, J/eV$:
$\Delta E = 10.2 \times 1.6 \times 10^{-19} \, J = 16.32 \times 10^{-19} \, J = 1.632 \times 10^{-18} \, J$.
Thus,the correct option is $C$.

(C) The existence of discrete spectral lines in the emission or absorption spectra of atoms provides direct evidence for the quantization of electron energy levels. According to the Bohr model,when an electron transitions between two energy levels $E_2$ and $E_1$,it emits or absorbs a photon with energy $h\nu = E_2 - E_1$. Since only specific energy levels are allowed,only specific frequencies of light are observed,resulting in distinct spectral lines.

(B) The total energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{13.6}{n^2} \text{ eV}$.
Since the value of $n$ (the principal quantum number) is always a positive integer $(n = 1, 2, 3, ...)$,the term $\frac{13.6}{n^2}$ is always positive.
Therefore,the total energy $E_n$ is always negative,which indicates that the electron is bound to the nucleus.

(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \text{ eV}}{n^2}$.
The energy difference between $n=2$ and $n=3$ is given as $E = E_3 - E_2$.
$E = -\frac{13.6}{3^2} - \left( -\frac{13.6}{2^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{5}{36} \right)$.
Thus,$13.6 = E \times \frac{36}{5} = 7.2 E$.
The ionization energy of a hydrogen atom is the energy required to move an electron from $n=1$ to $n=\infty$,which is $E_{\infty} - E_1 = 0 - (-13.6) = 13.6 \text{ eV}$.
Therefore,the ionization energy is $7.2 E$.

(A) For a hydrogen atom,the total energy $E_n$,potential energy $U_n$,and kinetic energy $K_n$ are given by:
$E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$
$U_n = 2E_n = -27.2 \frac{Z^2}{n^2} \text{ eV}$
$K_n = |E_n| = 13.6 \frac{Z^2}{n^2} \text{ eV}$
When an electron transitions from an excited state $(n > 1)$ to the ground state $(n = 1)$,the value of $n$ decreases.
As $n$ decreases,$K_n$ increases because it is proportional to $1/n^2$.
Since $E_n$ and $U_n$ are negative and their magnitudes decrease as $n$ decreases,their values become more negative,meaning they decrease.
Therefore,kinetic energy increases,while potential and total energies decrease.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula:
$E_n = \frac{-13.6}{n^2} \, eV$
Given the quantum number $n = 5$,we substitute this value into the formula:
$E_5 = \frac{-13.6}{5^2} \, eV$
$E_5 = \frac{-13.6}{25} \, eV$
$E_5 = -0.544 \, eV$
Rounding to two decimal places,we get $E_5 = -0.54 \, eV$.

(D) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
Frequency $\nu$ is related to energy by $\nu = \frac{\Delta E}{h}$.
Calculating the energy differences for the given transitions:
$(a)$ $3 \rightarrow 2$: $\Delta E \propto \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5}{36} \approx 0.138$
$(b)$ $4 \rightarrow 3$: $\Delta E \propto \left( \frac{1}{9} - \frac{1}{16} \right) = \frac{7}{144} \approx 0.048$
$(c)$ $4 \rightarrow 2$: $\Delta E \propto \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3}{16} = 0.1875$
$(d)$ $3 \rightarrow 1$: $\Delta E \propto \left( \frac{1}{1} - \frac{1}{9} \right) = \frac{8}{9} \approx 0.888$
Since the transition $3 \rightarrow 1$ has the largest energy difference,it corresponds to the highest frequency.

(C) The frequency $\nu$ of the emitted radiation is given by the formula: $\nu = RC \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$
Here,$R = 10^7 \ m^{-1}$,$C = 3 \times 10^8 \ m/s$,$n_1 = 4$,and $n_2 = 5$.
Substituting the values:
$\nu = (10^7) \times (3 \times 10^8) \left[ \frac{1}{4^2} - \frac{1}{5^2} \right]$
$\nu = 3 \times 10^{15} \left[ \frac{1}{16} - \frac{1}{25} \right]$
$\nu = 3 \times 10^{15} \left[ \frac{25 - 16}{400} \right]$
$\nu = 3 \times 10^{15} \left[ \frac{9}{400} \right]$
$\nu = \frac{27}{400} \times 10^{15} = 0.0675 \times 10^{15} \ Hz = 6.75 \times 10^{13} \ Hz$.

(D) The number of possible emission lines when an electron transitions from an excited state $n$ to lower energy states is given by the formula:
$N = \frac{n(n - 1)}{2}$
Here,the highest energy level is $n = 4$.
Substituting $n = 4$ into the formula:
$N = \frac{4(4 - 1)}{2} = \frac{4 \times 3}{2} = \frac{12}{2} = 6$
Thus,there are $6$ possible emission lines corresponding to the transitions: $4 \to 3, 4 \to 2, 4 \to 1, 3 \to 2, 3 \to 1,$ and $2 \to 1$.

(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 $(n = 3)$:
$E_3 = - \frac{13.6}{3^2} = - \frac{13.6}{9} = - 1.51 \ eV$.
The energy required to transfer the electron from the first orbit to the third orbit is $\Delta E = E_3 - E_1$.
$\Delta E = - 1.51 - (- 13.6) = - 1.51 + 13.6 = 12.09 \ eV$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.
When an electron jumps from an initial orbit $n_1 = 2$ to a final orbit $n_2 = 1$,the energy emitted is given by the difference in energy levels:
$\Delta E = E_{n_1} - E_{n_2} = -13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ (for emitted energy,we take the magnitude of the change).
$\Delta E = -13.6 \left( \frac{1}{2^2} - \frac{1}{1^2} \right) \ eV$
$\Delta E = -13.6 \left( \frac{1}{4} - 1 \right) \ eV$
$\Delta E = -13.6 \left( -\frac{3}{4} \right) \ eV$
$\Delta E = 10.2 \ eV$.
Since the question asks for the energy emitted,the value is $10.2 \ eV$. None of the provided options match this result.

(D) The energy of a photon emitted during a transition is given by $\Delta E = E_i - E_f = \frac{hc}{\lambda}$.
Since $\lambda = \frac{hc}{\Delta E}$,the wavelength $\lambda$ is inversely proportional to the energy difference $\Delta E$ of the transition.
Emission occurs when an electron moves from a higher energy level to a lower energy level.
Comparing the given transitions:
$(A)$ $n = 2$ to $n = 1$: Large energy difference.
$(B)$ $n = 1$ to $n = 2$: Absorption,not emission.
$(C)$ $n = 2$ to $n = 5$: Absorption,not emission.
$(D)$ $n = 5$ to $n = 2$: Smaller energy difference compared to $n = 2$ to $n = 1$.
Since $\Delta E$ for $n = 5 \to 2$ is smaller than $\Delta E$ for $n = 2 \to 1$,the wavelength $\lambda$ for the $n = 5 \to 2$ transition will be larger.
Therefore,the correct option is $(D)$.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
The energy difference between two successive levels $n$ 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 the principal quantum number $n$ increases,the denominator $n^2(n+1)^2$ increases much faster than the numerator $(2n+1)$.
Therefore,the energy difference $\Delta E$ decreases as $n$ increases.
From the provided image,we can see that the energy gap between levels decreases as we move to higher energy states: $E_1 > E_2 > E_3$.

(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 $Li^{++}$ (Lithium ion),the atomic number $Z = 3$.
The first excited state corresponds to $n = 2$.
Substituting these values into the formula,the energy of the electron in the first excited state is $E_2 = -\frac{13.6 \times 3^2}{2^2} = -\frac{13.6 \times 9}{4} = -30.6 \ eV$.
The binding energy (energy required to remove the electron) is the magnitude of this energy,which is $30.6 \ eV$.

(D) The energy of an electron in the $n$-th orbit is given by $E_n = \frac{-13.6}{n^2} \; eV$.
For $n = 3$,$E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} \approx -1.51 \; eV$.
For $n = 2$,$E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.4 \; eV$.
The energy of the emitted photon is equal to the difference in energy between the two states:
$\Delta E = E_3 - E_2 = -1.51 - (-3.4) = -1.51 + 3.4 = 1.89 \; eV$.
Rounding to one decimal place,the energy is approximately $1.9 \; eV$.

(B) The energy of a photon emitted during a transition in a hydrogen atom is given by the formula: $E = 13.6 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] eV$.
For the Balmer series,the transition ends at the energy level $n_1 = 2$.
The highest energy photon corresponds to the transition from the highest possible energy level,which is $n_2 = \infty$.
Substituting these values into the formula:
$E = 13.6 \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right] eV$
$E = 13.6 \left[ \frac{1}{4} - 0 \right] eV$
$E = \frac{13.6}{4} eV = 3.4 eV$.
Therefore,the energy of the highest energy photon of the Balmer series is $3.4 eV$.

(A) According to Bohr's postulates,the centripetal force required for an electron of mass $m$ and velocity $v$ to revolve in an orbit of radius $r$ is provided by the electrostatic force of attraction between the nucleus (charge $+e$) and the electron (charge $-e$):
$\frac{mv^2}{r} = \frac{ke^2}{r^2} \implies mv^2 = \frac{ke^2}{r} \quad (1)$
From Bohr's quantization condition,the angular momentum is quantized:
$mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi mr} \quad (2)$
Substituting $(2)$ into $(1)$:
$m\left(\frac{nh}{2\pi mr}\right)^2 = \frac{ke^2}{r} \implies \frac{n^2h^2}{4\pi^2mr^2} = \frac{ke^2}{r} \implies r = \frac{n^2h^2}{4\pi^2kme^2}$
The kinetic energy $E_K$ is:
$E_K = \frac{1}{2}mv^2 = \frac{1}{2}\left(\frac{ke^2}{r}\right) = \frac{ke^2}{2} \left(\frac{4\pi^2kme^2}{n^2h^2}\right) = \frac{2\pi^2k^2me^4}{n^2h^2}$
The potential energy $E_p$ is:
$E_p = -\frac{ke^2}{r} = -ke^2 \left(\frac{4\pi^2kme^2}{n^2h^2}\right) = -\frac{4\pi^2k^2me^4}{n^2h^2}$
The total energy $E$ is the sum of kinetic and potential energy:
$E = E_K + E_p = \frac{2\pi^2k^2me^4}{n^2h^2} - \frac{4\pi^2k^2me^4}{n^2h^2} = -\frac{2\pi^2k^2me^4}{n^2h^2}$

(B) For a hydrogen atom,the total energy $(E)$,kinetic energy $(K)$,and potential energy $(U)$ are related as follows:
$E = -K$
$U = 2E$
Given the ground state energy $E = -13.6 \, eV$.
Therefore,the potential energy $U = 2 \times (-13.6 \, eV) = -27.2 \, eV$.

(D) The energy of an emitted photon is given by $\Delta E = E_{initial} - E_{final}$.
Emission occurs when an electron transitions from a higher energy level to a lower energy level.
Transitions $II$ and $IV$ represent emission.
Transition $II$ is from $n=4$ to $n=3$,and transition $IV$ is from $n=4$ to $n=2$.
The energy difference $\Delta E$ is proportional to the gap between the energy levels.
Since the gap between $n=4$ and $n=2$ is larger than the gap between $n=4$ and $n=3$,transition $IV$ releases more energy than transition $II$.
Therefore,transition $IV$ represents the emission of a photon with the most energy.

(A) According to the kinetic theory of gases,the average kinetic energy of a gas molecule at temperature $T$ is given by $E = \frac{3}{2} kT$,where $k$ is the Boltzmann constant $(k = 1.38 \times 10^{-23} \, J/K)$.
To excite a hydrogen atom to the first excited state,the kinetic energy of the colliding particles must be at least $10.2 \, eV$.
Converting the energy to Joules: $E = 10.2 \times 1.6 \times 10^{-19} \, J = 1.632 \times 10^{-18} \, J$.
Equating the energy: $1.632 \times 10^{-18} = \frac{3}{2} \times (1.38 \times 10^{-23}) \times T$.
Solving for $T$: $T = \frac{1.632 \times 10^{-18} \times 2}{3 \times 1.38 \times 10^{-23}} \approx 7.88 \times 10^4 \, K$.
Rounding to the nearest given option,$T \approx 7.9 \times 10^4 \, K$.

(C) The energy required to excite the hydrogen atom to the first excited state is $\Delta E = 10.2 \ eV$.
To find the temperature $T$,we equate the average kinetic energy of the gas atoms to the excitation energy: $\frac{3}{2} kT = \Delta E$.
Here,$k = 1.38 \times 10^{-23} \ J/K$ is the Boltzmann constant and $\Delta E = 10.2 \times 1.6 \times 10^{-19} \ J$.
Substituting the values: $T = \frac{2 \Delta E}{3k} = \frac{2 \times 10.2 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}}$.
$T = \frac{32.64 \times 10^{-19}}{4.14 \times 10^{-23}} \approx 7.88 \times 10^4 \ K$.

(A) For an electron in a hydrogen atom,the potential energy $U$ is given by $U = -\frac{ke^2}{r}$,which means $U \propto -\frac{1}{r}$.
The kinetic energy $K$ is given by $K = \frac{ke^2}{2r}$,which means $K \propto \frac{1}{r}$.
When the electron jumps from the ground state to an excited state,the principal quantum number $n$ increases,which implies the orbital radius $r$ increases.
As $r$ increases,the kinetic energy $K$ decreases.
Since $U$ is negative and proportional to $-1/r$,as $r$ increases,the magnitude of $U$ decreases,making $U$ less negative (i.e.,it increases towards zero).
Therefore,the potential energy increases and the kinetic energy decreases.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{13.6 \ eV}{n^2}$.
For the ground state,$n = 1$,so the energy is $E_1 = -13.6 \ eV$.
To remove the electron from the atom (ionization),we need to provide enough energy to bring the electron to the state where $n = \infty$,where the energy is $E_{\infty} = 0 \ eV$.
The energy required is $\Delta E = E_{\infty} - E_1 = 0 - (-13.6 \ eV) = 13.6 \ eV$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \text{ eV}}{n^2}$.
When an electron transitions from $n+1$ to $n$,the energy of the emitted photon is $\Delta E = E_{n+1} - E_n = 13.6 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) \text{ eV}$.
This simplifies to $\Delta E = 13.6 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right) = 13.6 \left( \frac{2n+1}{n^2(n+1)^2} \right)$.
For large $n$,$2n+1 \approx 2n$ and $n^2(n+1)^2 \approx n^4$.
Thus,$\Delta E \approx 13.6 \left( \frac{2n}{n^4} \right) = \frac{27.2}{n^3}$.
Since the energy of the photon is related to wavelength by $\Delta E = \frac{hc}{\lambda}$,we have $\lambda = \frac{hc}{\Delta E}$.
Substituting $\Delta E \propto \frac{1}{n^3}$,we get $\lambda \propto n^3$.

(C) The energy of an emitted photon is given by $E = h\nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
Since $\nu = E/h$,the highest frequency corresponds to the transition with the largest energy difference $\Delta E$.
For a hydrogen atom,the energy of a level $n$ is $E_n = -13.6/n^2 \ eV$.
Transitions involving absorption (like $n=1$ to $n=2$ or $n=2$ to $n=6$) do not emit photons.
Comparing emission transitions:
For $n=6$ to $n=2$,$\Delta E = 13.6 \times (1/2^2 - 1/6^2) = 13.6 \times (1/4 - 1/36) = 13.6 \times (8/36) \approx 3.02 \ eV$.
For $n=2$ to $n=1$,$\Delta E = 13.6 \times (1/1^2 - 1/2^2) = 13.6 \times (1 - 1/4) = 13.6 \times (3/4) = 10.2 \ eV$.
Since $10.2 \ eV > 3.02 \ eV$,the transition from $n=2$ to $n=1$ releases the highest energy and thus emits a photon of the highest frequency.

(A) The energy of an emitted photon is given by the difference in energy between the two levels involved in the transition: $E = E_{initial} - E_{final} = h\nu$.
For emission to occur,the electron must transition from a higher energy level to a lower energy level.
Transitions $II$ and $III$ are emission transitions (downward arrows).
Transition $I$ is an absorption transition (upward arrow).
Transition $IV$ is also an emission transition.
Comparing the energy gaps:
Transition $II$ is from $n=4$ to $n=3$.
Transition $III$ is from $n=2$ to $n=1$.
Transition $IV$ is from $n=4$ to $n=2$.
The energy difference is proportional to the gap between the levels. The gap between $n=2$ and $n=1$ is significantly larger than the gap between $n=4$ and $n=3$ or $n=4$ and $n=2$.
Therefore,transition $III$ corresponds to the largest energy change and thus emits the photon with the highest energy.

(D) The wavelength $\lambda$ of a photon emitted during a transition is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
This implies that $\lambda$ is inversely proportional to the energy difference $\Delta E$ of the transition,where $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
To have the minimum wavelength,the energy difference $\Delta E$ must be maximum.
Comparing the transitions:
$(A)$ $\Delta E \propto (1/16 - 1/25) = 0.0225$
$(B)$ $\Delta E \propto (1/9 - 1/16) = 0.0486$
$(C)$ $\Delta E \propto (1/4 - 1/9) = 0.1389$
$(D)$ $\Delta E \propto (1/1 - 1/4) = 0.7500$
The transition from $n = 2$ to $n = 1$ results in the largest energy change,and therefore,the minimum wavelength.