Atomic models and Planck's quantum theory Questions in English

(C) The energy of a photon emitted during an electronic transition is given by $\Delta E = E_{n_2} - E_{n_1} = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
Since $\Delta E = \frac{hc}{\lambda}$,the wavelength $\lambda$ is inversely proportional to the energy difference $\Delta E$ $(\lambda = \frac{hc}{\Delta E})$.
To obtain the lowest wavelength,we need the transition with the highest energy difference $\Delta E$.
Comparing the transitions:
$(A)$ $n_2=\infty$ to $n_1=2$: $\Delta E \propto (1/4 - 0) = 0.25$
$(B)$ $n_2=4$ to $n_1=3$: $\Delta E \propto (1/9 - 1/16) = 0.0486$
$(C)$ $n_2=2$ to $n_1=1$: $\Delta E \propto (1/1 - 1/4) = 0.75$
$(D)$ $n_2=5$ to $n_1=3$: $\Delta E \propto (1/9 - 1/25) = 0.0711$
The transition $n_2=2$ to $n_1=1$ has the largest energy difference,therefore it emits radiation of the lowest wavelength.

(C) For the Lyman series,the Rydberg formula is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$,where $n_1 = 1$.
Given $\lambda = \frac{16}{15 R}$,so $\frac{1}{\lambda} = \frac{15 R}{16}$.
Substituting the values: $\frac{15 R}{16} = R \left[ \frac{1}{1^2} - \frac{1}{n_2^2} \right]$.
Dividing both sides by $R$: $\frac{15}{16} = 1 - \frac{1}{n_2^2}$.
Rearranging the terms: $\frac{1}{n_2^2} = 1 - \frac{15}{16} = \frac{1}{16}$.
Therefore,$n_2^2 = 16$,which gives $n_2 = 4$.

(D) . Rydberg's constant $(R_H)$ has units of $cm^{-1}$ or $m^{-1}$,and wave number $(\bar{\nu})$ also has units of $cm^{-1}$ or $m^{-1}$. Thus,this statement is correct.
$B$. Lyman series corresponds to transitions to $n=1$,which falls in the ultraviolet region. This statement is correct.
$C$. According to Bohr's postulate,angular momentum $mvr = \frac{nh}{2\pi}$. For the ground state $(n=1)$,angular momentum is $\frac{h}{2\pi}$. This statement is correct.
$D$. The radius of the first Bohr orbit of the hydrogen atom is $0.529 \times 10^{-8} \ cm$ (or $0.529 \ \mathring{A}$). The value given in the option $(2116 \times 10^{-8} \ cm)$ is incorrect.

(B) Energy of incident photon $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{310 \times 10^{-9}} \approx 6.416 \times 10^{-19} \ J$.
Converting to $eV$: $E = \frac{6.416 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 4.01 \ eV$.
Kinetic energy $K.E. = E - \Phi = 4.01 \ eV - 3.55 \ eV = 0.46 \ eV$.
$K.E. = 0.46 \times 1.6 \times 10^{-19} \ J = 0.736 \times 10^{-19} \ J$.
Using $K.E. = \frac{1}{2} m_e v^2$:
$0.736 \times 10^{-19} = \frac{1}{2} \times 9 \times 10^{-31} \times v^2$.
$v^2 = \frac{1.472 \times 10^{-19}}{9 \times 10^{-31}} \approx 0.1635 \times 10^{12} = 16.35 \times 10^{10}$.
$v \approx 4.04 \times 10^5 \ ms^{-1}$.
Thus,$x \approx 4$.

(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Converting wavelength to meters: $\lambda = 331.5 \times 10^{-9} \ m$.
Energy per photon: $E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{331.5 \times 10^{-9}} = 6 \times 10^{-19} \ J$.
Energy per mole: $E_{mol} = E \times N_A = 6 \times 10^{-19} \times 6 \times 10^{23} = 3.6 \times 10^5 \ J \ mol^{-1}$.
Using the photoelectric equation: $E_{mol} = W + KE_{mol}$,where $W$ is the work function.
$W = 3.6 \times 10^5 - 1.2 \times 10^5 = 2.4 \times 10^5 \ J \ mol^{-1}$.
To convert to $eV$ per atom: $W_{eV} = \frac{2.4 \times 10^5}{6 \times 10^{23} \times 1.6 \times 10^{-19}} = \frac{2.4 \times 10^5}{9.6 \times 10^4} = 2.5 \ eV$.

(B) Given threshold frequency,$\nu_0 = 1.0 \times 10^{15} \ s^{-1}$.
According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{max})$ is given by $K.E._{max} = h(\nu - \nu_0)$,where $\nu$ is the frequency of incident radiation.
For $\nu_1 = 1.5 \times 10^{15} \ s^{-1}$:
$K.E._1 = h(1.5 \times 10^{15} - 1.0 \times 10^{15}) = h(0.5 \times 10^{15})$.
For $\nu_2 = 2.0 \times 10^{15} \ s^{-1}$:
$K.E._2 = h(2.0 \times 10^{15} - 1.0 \times 10^{15}) = h(1.0 \times 10^{15})$.
The ratio of maximum kinetic energies is:
$\frac{K.E._1}{K.E._2} = \frac{h(0.5 \times 10^{15})}{h(1.0 \times 10^{15})} = \frac{0.5}{1.0} = 1:2$.

(C) The angular momentum $(L)$ of an electron in a hydrogen atom is given by Bohr's postulate:
$L = \frac{nh}{2\pi}$
For the ground state,the principal quantum number $n = 1$.
Substituting the values:
$L = \frac{1 \times 6.625 \times 10^{-34}}{2 \times 3.14159} \ J \ s$
$L \approx \frac{6.625 \times 10^{-34}}{6.283} \ J \ s$
$L \approx 1.0545 \times 10^{-34} \ J \ s$
To match the options,we express this as:
$1.0545 \times 10^{-34} = 1055 \times 10^{-37} \ J \ s$
Therefore,the correct option is $C$.

(C) Power $(P) = 100 \ W = 100 \ J \cdot s^{-1}$.
Number of photons emitted per second $(n) = 2.0 \times 10^{20} \ s^{-1}$.
Energy of one photon $(E) = \frac{hc}{\lambda}$.
Total power emitted is given by $P = n \times E = n \times \frac{hc}{\lambda}$.
Substituting the values: $100 = \frac{2.0 \times 10^{20} \times 6.63 \times 10^{-34} \times 3 \times 10^8}{\lambda}$.
$\lambda = \frac{2.0 \times 6.63 \times 3 \times 10^{20-34+8}}{100} \ m$.
$\lambda = \frac{39.78 \times 10^{-6}}{100} \ m = 39.78 \times 10^{-8} \ m$.
Converting to $\mathring{A}$: $\lambda = 39.78 \times 10^{-8} \times 10^{10} \ \mathring{A} = 3978 \ \mathring{A}$.
Thus,$x = 3978$.

(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(K.E._{\max})$ of an emitted photoelectron is given by:
$K.E._{\max} = hv - \phi$
where $h$ is Planck's constant,$v$ is the frequency of incident light,and $\phi$ is the work function of the metal.
Since the stopping potential $(V_0)$ is related to the maximum kinetic energy by the equation $K.E._{\max} = eV_0$,we can substitute this into the photoelectric equation:
$eV_0 = hv - \phi$
Dividing both sides by the charge of an electron $(e)$,we get:
$V_0 = \frac{hv}{e} - \frac{\phi}{e}$
This equation represents a linear relationship between the stopping potential $(V_0)$ and the frequency $(v)$.

(A) For the Lyman series,the transition is from $n_2$ to $n_1 = 1$. The Rydberg formula is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the longest wavelength (lowest energy),the transition is from $n_2 = 2$ to $n_1 = 1$: $\frac{1}{\lambda_{max}} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R_H$,so $\lambda_{max} = \frac{4}{3 R_H}$.
For the shortest wavelength (highest energy),the transition is from $n_2 = \infty$ to $n_1 = 1$: $\frac{1}{\lambda_{min}} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H (1 - 0) = R_H$,so $\lambda_{min} = \frac{1}{R_H}$.
The ratio of the highest wavelength to the lowest wavelength is $\frac{\lambda_{max}}{\lambda_{min}} = \frac{4 / (3 R_H)}{1 / R_H} = \frac{4}{3}$.

(C) Regarding the photoelectric effect,the statement in option-$(C)$ is not correct.
Because,the electrons are ejected from the metal surface instantaneously as soon as the beam of light strikes the surface,meaning there is no time lag.

(A) Given,kinetic energy of an emitted electron $(KE) = 1.986 \times 10^{-19} \ J$.
Frequency of radiation $(\nu) = 1.0 \times 10^{15} \ s^{-1}$.
Planck's constant $(h) = 6.62 \times 10^{-34} \ J \ s$.
According to the photoelectric effect equation: $h\nu = h\nu_0 + KE$,where $\nu_0$ is the threshold frequency.
Rearranging for $\nu_0$: $h\nu_0 = h\nu - KE$.
$\nu_0 = \nu - \frac{KE}{h}$.
Substituting the values: $\nu_0 = 1.0 \times 10^{15} \ s^{-1} - \frac{1.986 \times 10^{-19} \ J}{6.62 \times 10^{-34} \ J \ s}$.
$\nu_0 = 1.0 \times 10^{15} \ s^{-1} - 0.3 \times 10^{15} \ s^{-1}$.
$\nu_0 = 0.7 \times 10^{15} \ s^{-1} = 7.0 \times 10^{14} \ s^{-1}$.

(A) According to Bohr's postulate,the angular momentum $(L)$ of an electron in a hydrogen atom is quantized and is given by the formula: $L = n \frac{h}{2 \pi}$,where $n$ is an integer $(n = 1, 2, 3, \dots)$.
This can be rewritten as $L = \frac{n}{2} \cdot \frac{h}{\pi}$.
For $n = 1$,$L = 0.5 \frac{h}{\pi}$.
For $n = 2$,$L = 1.0 \frac{h}{\pi}$.
For $n = 3$,$L = 1.5 \frac{h}{\pi}$.
For $n = 4$,$L = 2.0 \frac{h}{\pi}$.
Comparing these values with the given options,$1.25 \frac{h}{\pi}$ does not correspond to any integer value of $n$ (as $n = 2.5$ is not an integer). Therefore,it is not a permitted value.

(A) The kinetic energy $(KE)$ of $1 \ mol$ of electrons is $6.023 \times 10^4 \ J$.
Since $1 \ mol = 6.023 \times 10^{23} \ \text{atoms}$,the $KE$ of $1 \ \text{electron}$ is:
$KE = \frac{6.023 \times 10^4 \ J}{6.023 \times 10^{23}} = 1.0 \times 10^{-19} \ J$.
The energy of the incident photon $(E)$ is given by $E = \frac{hc}{\lambda}$:
$E = \frac{6.626 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m/s}{600 \times 10^{-9} \ m} = 3.313 \times 10^{-19} \ J$.
The threshold energy $(Phi)$ is the difference between the incident photon energy and the kinetic energy:
$\Phi = E - KE = 3.313 \times 10^{-19} \ J - 1.0 \times 10^{-19} \ J = 2.313 \times 10^{-19} \ J$.

(A) Given: $E = 3 \times 10^{-12} \ erg$,$h = 6.62 \times 10^{-27} \ erg \cdot s$,$c = 3 \times 10^{10} \ cm/s$.
Using the formula $E = \frac{hc}{\lambda}$,we get $\lambda = \frac{hc}{E}$.
$\lambda = \frac{6.62 \times 10^{-27} \times 3 \times 10^{10}}{3 \times 10^{-12}} \ cm$.
$\lambda = 6.62 \times 10^{-5} \ cm$.
Since $1 \ cm = 10^7 \ nm$,$\lambda = 6.62 \times 10^{-5} \times 10^7 \ nm = 6.62 \times 10^2 \ nm = 662 \ nm$.

(A) The energy $E$ of a photon is given by $E = h \nu = \frac{hc}{\lambda}$.
Given: $\lambda = 2000 \ \text{Å} = 2000 \times 10^{-8} \ \text{cm} = 2 \times 10^{-5} \ \text{cm}$.
Speed of light $c = 3 \times 10^{10} \ \text{cm/s}$.
Planck's constant $h = 6.626 \times 10^{-27} \ \text{erg} \cdot \text{s}$.
Frequency $\nu = \frac{c}{\lambda} = \frac{3 \times 10^{10}}{2 \times 10^{-5}} = 1.5 \times 10^{15} \ \text{s}^{-1}$.
Energy $E = h \nu = (6.626 \times 10^{-27} \ \text{erg} \cdot \text{s}) \times (1.5 \times 10^{15} \ \text{s}^{-1}) \approx 9.94 \times 10^{-12} \ \text{erg}$.

(C) Quantisation refers to the restriction of physical quantities to discrete,specific values rather than a continuous range.
In the case of a staircase,a person can only stand on specific steps ($1^{st}, 2^{nd}, 3^{rd}$,etc.),which corresponds to discrete energy levels in an atom.
Conversely,a car on a road,an apple falling,or a thrown disc represent continuous motion where any position or velocity is possible.
Therefore,standing on the steps of a staircase is the best analogy for the quantised energy levels of electrons.

(C) The quantity $h \nu / K_{B}$ is related to the characteristic temperature of a system.
According to the kinetic theory of gases,the average kinetic energy of a gas molecule is given by $\frac{3}{2} K_{B} T$.
Equating the energy of a photon or quantum of energy $h \nu$ to the thermal energy $K_{B} T$,we get $h \nu = K_{B} T$.
Thus,$\frac{h \nu}{K_{B}} = T$.
Therefore,the quantity $\frac{h \nu}{K_{B}}$ has the dimensions of temperature.
Hence,option $C$ is correct.

(C) The velocity of an electron in the $n^{th}$ Bohr orbit is given by the formula $v = v_0 \frac{Z}{n}$,where $Z$ is the atomic number and $n$ is the orbit number.
For a $He^{+}$ ion,the atomic number $Z = 2$.
For the $1^{st}$ orbit $(n=1)$,the velocity is $v_1 = v_0 \frac{2}{1} = 2v_0$.
For the $3^{rd}$ orbit $(n=3)$,the velocity is $v_3 = v_0 \frac{2}{3} = \frac{2}{3}v_0$.
Therefore,the ratio $v_3 : v_1 = \frac{2}{3}v_0 : 2v_0 = \frac{1}{3} : 1 = 1 : 3$.

(B) The radius of the $n^{th}$ orbit for a hydrogen-like species is given by the formula: $r_n = 0.529 \times \frac{n^2}{Z} \mathring{A}$.
For the $1^{st}$ Bohr orbit of a hydrogen atom $(n=1, Z=1)$: $r = 0.529 \times \frac{1^2}{1} = 0.529 \mathring{A}$.
Now,checking the options:
For $Be^{3+}$ $(Z=4)$ with $n=2$: $r = 0.529 \times \frac{2^2}{4} = 0.529 \times \frac{4}{4} = 0.529 \mathring{A}$.
Thus,the $2^{nd}$ orbit of $Be^{3+}$ has the same radius as the $1^{st}$ orbit of hydrogen.

(D) The radius of an orbit in the Bohr model is given by $r_n = a_0 \times \frac{n^2}{Z}$.
For a hydrogen atom $(Z=1)$ in the ground state $(n=1)$: $r_1 = a_0 \times \frac{1^2}{1} = a_0$.
For a $He^{+}$ ion $(Z=2)$ in the ground state $(n=1)$: $r_2 = a_0 \times \frac{1^2}{2} = \frac{a_0}{2}$.
Since $r_1 = a_0$,we have $r_2 = \frac{r_1}{2}$.
Therefore,$\frac{r_2}{r_1} = \frac{1}{2}$.

(A) The velocity of an electron in the $n^{th}$ Bohr orbit of a hydrogen-like atom is given by the formula: $v_n = 2.188 \times 10^8 \times \frac{Z}{n} \ cm \ s^{-1}$.
For the first Bohr orbit of a hydrogen atom,$Z = 1$ and $n = 1$.
Substituting these values: $v_1 = 2.188 \times 10^8 \times \frac{1}{1} \ cm \ s^{-1} = 2.188 \times 10^8 \ cm \ s^{-1}$.

(C) The distance travelled in one revolution is the circumference of the orbit,which is $2 \pi r$.
Velocity $(v)$ is given by the ratio of distance to time $(T)$: $v = \frac{2 \pi r}{T}$.
From Bohr's postulate,the angular momentum is $mvr = \frac{nh}{2 \pi}$,which gives $v = \frac{nh}{2 \pi mr}$.
Equating the two expressions for velocity: $\frac{2 \pi r}{T} = \frac{nh}{2 \pi mr}$.
Solving for $T$: $T = \frac{2 \pi r \times 2 \pi mr}{nh} = \frac{4 \pi^2 m r^2}{nh}$.

(C) The ionization energy $(IE)$ for a hydrogen-like species is given by the formula: $IE_n = 13.6 \times Z^2 / n^2 \ eV$.
For the second ionization of $He$ $(He^+ \rightarrow He^{2+} + e^-)$,the electron is removed from the $n=1$ state of a $He^+$ ion,which has an atomic number $Z=2$.
Substituting these values: $IE = 13.6 \times (2^2 / 1^2) \ eV = 13.6 \times 4 \ eV = 54.4 \ eV$.

(B) Sommerfeld introduced elliptical orbits to explain the fine structure of spectral lines in the hydrogen atom. According to his model,the electron moves in an elliptical orbit with the nucleus at one of the foci. The trajectory is a closed ellipse-like curve,which is narrower at the perihelion (closest point to the nucleus) and flatter at the aphelion (farthest point from the nucleus).

(D) In $1885$,Johann Balmer for the first time showed that the wave numbers of spectral lines present in the visible region of the hydrogen spectrum are given by the formula:
$\bar{v} (cm^{-1}) = 109677 \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$
Here,$n = 3, 4, 5, \dots$
Thus,the Balmer series of the hydrogen spectrum was discovered first,and it lies in the visible region of the electromagnetic spectrum.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For $Li^{2+}$ ion,the atomic number $Z = 3$.
The first excited state corresponds to $n = 2$.
Substituting these values into the formula: $E_2 = -13.6 \times \frac{3^2}{2^2} \ eV$.
$E_2 = -13.6 \times \frac{9}{4} \ eV$.
$E_2 = -13.6 \times 2.25 \ eV = -30.6 \ eV$.

(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = Z^2 \cdot R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$
For the transition from $n=2$ to $n=1$: $\frac{1}{\lambda} = Z^2 \cdot R_H \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = Z^2 \cdot R_H \left( \frac{3}{4} \right)$
This implies $\lambda = \frac{4}{3 Z^2 R_H}$,so $\lambda \propto \frac{1}{Z^2}$.
To obtain the shortest wavelength,the atomic number $Z$ must be the maximum.
Comparing the species: $H$ $(Z=1)$,$He^{+}$ $(Z=2)$,and $Li^{2+}$ $(Z=3)$.
Since $Li^{2+}$ has the highest $Z$,it will produce the shortest wavelength.

(C) The Rydberg formula for the hydrogen spectrum is given by $\Delta E = R h c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Paschen series,the electron transitions occur from higher energy levels to the third energy level.
Therefore,the value of $n_1$ is $3$ and $n_2$ can be any integer greater than $3$,i.e.,$n_2 = 4, 5, 6, \dots$.

(A) The total energy $E$ for $N$ moles of photons is given by $E = N_A \times h \times \nu$.
Here,$N_A = 6.022 \times 10^{23} \ mol^{-1}$,$h = 6.626 \times 10^{-34} \ J \cdot s$,and $\nu = 2500 \ s^{-1}$.
$E = (6.022 \times 10^{23}) \times (6.626 \times 10^{-34}) \times 2500 \ J$.
$E \approx 9.97 \times 10^{-7} \ J$.
Since $1 \ J = 10^7 \ erg$,$E \approx 9.97 \times 10^{-7} \times 10^7 \ erg = 9.97 \ erg \approx 10 \ erg$.

(B) According to Einstein's photoelectric equation,$KE_{max} = E - \phi$.
Given that the ratio of work functions $\phi_A : \phi_B = 1:2$,let $\phi_A = \phi$ and $\phi_B = 2\phi$.
For metal $M_A$: $(KE_{max})_A = 6 - \phi$.
For metal $M_B$: $(KE_{max})_B = 6 - 2\phi$.
The ratio of kinetic energies is given as $\frac{(KE_{max})_A}{(KE_{max})_B} = \frac{2.642}{1}$.
Substituting the expressions: $\frac{6 - \phi}{6 - 2\phi} = 2.642$.
$6 - \phi = 2.642(6 - 2\phi)$.
$6 - \phi = 15.852 - 5.284\phi$.
$4.284\phi = 9.852$.
$\phi \approx 2.3 \ eV$.
Thus,$\phi_A = 2.3 \ eV$ and $\phi_B = 4.6 \ eV$.

(D) Let $n_1$ be the lower energy level and $n_2$ be the higher energy level.
Given: $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$.
Adding the two equations: $2n_2 = 6 \implies n_2 = 3$.
Subtracting the two equations: $2n_1 = 2 \implies n_1 = 1$.
Using Rydberg's formula: $\frac{1}{\lambda} = R_H Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For $Li^{2+}$,$Z = 3$. Substituting the values: $\frac{1}{\lambda} = R_H (3)^2 \left[ \frac{1}{1^2} - \frac{1}{3^2} \right]$.
$\frac{1}{\lambda} = 9 R_H \left[ 1 - \frac{1}{9} \right] = 9 R_H \left( \frac{8}{9} \right) = 8 R_H$.
$\lambda = \frac{1}{8 R_H}$.
Taking $R_H \approx 1.097 \times 10^5 \ cm^{-1} \approx 1.1 \times 10^5 \ cm^{-1}$.
$\lambda = \frac{1}{8 \times 1.1 \times 10^5} \ cm = \frac{1}{8.8 \times 10^5} \ cm \approx 0.1136 \times 10^{-5} \ cm = 1.136 \times 10^{-6} \ cm$.
Thus,$\lambda \approx 1.14 \times 10^{-6} \ cm$.

(D) The energy of an electron in a stationary state is given by the formula: $E_{n} = -2.18 \times 10^{-18} \times \frac{Z^{2}}{n^{2}} \ J/atom$.
For the $3^{rd}$ orbit $(n=3)$ of $Li^{2+}$ ion $(Z=3)$:
$E_{3} = -2.18 \times 10^{-18} \times \frac{3^{2}}{3^{2}} = -2.18 \times 10^{-18} \ J$.
Thus,option $D$ is correct.

(A) For the Balmer series,the wave number $\bar{v}$ is given by the formula: $\bar{v} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right]$,where $n = 3, 4, 5, \dots$
For $n = 3$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{9-4}{36} \right] = \frac{5R}{36}$
For $n = 4$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{16} \right] = R \left[ \frac{4-1}{16} \right] = \frac{3R}{16}$
For $n = 5$: $\bar{v} = R \left[ \frac{1}{4} - \frac{1}{25} \right] = R \left[ \frac{25-4}{100} \right] = \frac{21R}{100}$
Thus,the set of spectral lines is $\frac{5R}{36}, \frac{3R}{16}, \frac{21R}{100}$.

(B) The energy of a spectral line is given by $\Delta E = 13.6 \times Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$.
For the Lyman series,the lowest energy line $(L_1)$ corresponds to the transition from $n_2 = 2$ to $n_1 = 1$: $\Delta E(L_1) = 13.6 \times 1^2 (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 \times \frac{3}{4} \text{ eV}$.
For the Balmer series,the lowest energy line $(B_1)$ corresponds to the transition from $n_2 = 3$ to $n_1 = 2$: $\Delta E(B_1) = 13.6 \times 1^2 (\frac{1}{2^2} - \frac{1}{3^2}) = 13.6 \times (\frac{1}{4} - \frac{1}{9}) = 13.6 \times \frac{5}{36} \text{ eV}$.
Given $\Delta E(L_1) = x \times \Delta E(B_1)$,we have $x = \frac{\Delta E(L_1)}{\Delta E(B_1)} = \frac{3/4}{5/36} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5} = 5.4$.
To express $x$ as $x \times 10^{-1}$,we have $5.4 = 54 \times 10^{-1}$.
Therefore,the value is $54$.

(D) The energy of a spectral line is given by $\Delta E = 13.6 \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$. For hydrogen $(Z=1)$,$\Delta E \propto (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.

$A$. Paschen ($1^{st}$ line)	$n_1=3, n_2=4 \rightarrow \Delta E \propto (\frac{1}{9} - \frac{1}{16}) = \frac{7}{144} \approx 0.0486$
$B$. Balmer ($2^{nd}$ line)	$n_1=2, n_2=4 \rightarrow \Delta E \propto (\frac{1}{4} - \frac{1}{16}) = \frac{3}{16} = 0.1875$
$C$. Paschen ($3^{rd}$ line)	$n_1=3, n_2=6 \rightarrow \Delta E \propto (\frac{1}{9} - \frac{1}{36}) = \frac{3}{36} = 0.0833$
$D$. Bracket ($4^{th}$ line)	$n_1=4, n_2=8 \rightarrow \Delta E \propto (\frac{1}{16} - \frac{1}{64}) = \frac{3}{64} = 0.0468$

Comparing the values: $0.0468 (D) < 0.0486 (A) < 0.0833 (C) < 0.1875 (B)$.
Therefore,the correct ascending order is $D < A < C < B$.

(D) The energy of a transition in a hydrogen-like atom is given by $\Delta E = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \ eV$.
For the first Balmer line,the transition is from $n_2 = 3$ to $n_1 = 2$:
$x = 13.6 \times (1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{5}{36} \right) \quad \dots(i)$
For the second Balmer line,the transition is from $n_2 = 4$ to $n_1 = 2$:
$\Delta E = 13.6 \times (1)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \left( \frac{3}{16} \right) \quad \dots(ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{\Delta E}{x} = \frac{13.6 \times (3/16)}{13.6 \times (5/36)} = \frac{3}{16} \times \frac{36}{5} = \frac{3 \times 9}{4 \times 5} = \frac{27}{20} = 1.35$
Therefore,$\Delta E = 1.35x$.

(B) The energy of an electron in the $n^{th}$ orbit is given by $E_{n} = -R_{H} \times \frac{Z^{2}}{n^{2}}$.
The energy difference between the first $(n=1)$ and second $(n=2)$ orbit for a hydrogen atom $(Z=1)$ is $\Delta E = E_{2} - E_{1} = R_{H} \times \left[ \frac{1}{1^{2}} - \frac{1}{2^{2}} \right]$.
Given $R_{H} = 2.18 \times 10^{-11} \ ergs = 2.18 \times 10^{-18} \ J$.
$\Delta E = 2.18 \times 10^{-18} \times \left[ 1 - \frac{1}{4} \right] = 2.18 \times 10^{-18} \times 0.75 = 1.635 \times 10^{-18} \ J \ atom^{-1}$.
To convert to $J \ mol^{-1}$,multiply by Avogadro's number $(N_{A} = 6.022 \times 10^{23} \ mol^{-1})$:
$\Delta E = 1.635 \times 10^{-18} \times 6.022 \times 10^{23} \approx 9.846 \times 10^{5} \ J \ mol^{-1}$.
Rounding to the provided option,the correct answer is $9.835 \times 10^{5} \ J \ mol^{-1}$.

(C) The ionization energy of a hydrogen-like species is given by the formula $E = E_H \times Z^2 / n^2$,where $E_H$ is the ionization energy of the hydrogen atom in the ground state $(2.18 \times 10^{-18} \text{ J atom}^{-1})$,$Z$ is the atomic number,and $n$ is the principal quantum number.
For the process $Li^{2+}(g) \to Li^{3+}(g) + e^-$,the electron is removed from the $n=1$ state of the lithium ion $(Li^{2+})$.
The atomic number of Lithium $(Li)$ is $Z = 3$.
Substituting the values into the formula:
$E = 2.18 \times 10^{-18} \times (3)^2 / (1)^2$
$E = 2.18 \times 10^{-18} \times 9$
$E = 1.962 \times 10^{-17} \text{ J atom}^{-1}$.

(B) According to Bohr's theory,the radius of an orbit is given by $r_n = a_0 \cdot n^2 / Z$,where $n$ is the principal quantum number and $Z$ is the atomic number.
We compare the ratio $n^2 / Z$ for each species:
$A$. $H$ (first orbit): $n=1, Z=1 \implies n^2/Z = 1^2/1 = 1$
$B$. $He^+$ (first orbit): $n=1, Z=2 \implies n^2/Z = 1^2/2 = 0.5$
$C$. $He^+$ (second orbit): $n=2, Z=2 \implies n^2/Z = 2^2/2 = 2$
$D$. $Li^{2+}$ (first orbit): $n=1, Z=3 \implies n^2/Z = 1^2/3 = 0.33$
$E$. $Be^{3+}$ (second orbit): $n=2, Z=4 \implies n^2/Z = 2^2/4 = 1$
Comparing the values,species $A$ and $E$ both have a ratio of $1$. Therefore,they have identical radii.

(B) The wave number is given by the Rydberg formula: $\bar{\nu} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the $H$ atom,$Z = 1$.
For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$. Thus,$\bar{\nu}_B = R_H (\frac{1}{2^2} - \frac{1}{3^2}) = R_H (\frac{1}{4} - \frac{1}{9}) = R_H (\frac{9-4}{36}) = R_H (\frac{5}{36})$.
For the first line of the Brackett series,$n_1 = 4$ and $n_2 = 5$. Thus,$\bar{\nu}_{Br} = R_H (\frac{1}{4^2} - \frac{1}{5^2}) = R_H (\frac{1}{16} - \frac{1}{25}) = R_H (\frac{25-16}{400}) = R_H (\frac{9}{400})$.
The ratio is $\frac{\bar{\nu}_B}{\bar{\nu}_{Br}} = \frac{R_H (5/36)}{R_H (9/400)} = \frac{5}{36} \times \frac{400}{9} = \frac{5 \times 100}{9 \times 9} = \frac{500}{81}$.
Calculating the value,$\frac{500}{81} \approx 6.1728$.
Comparing with option $B$,$\frac{5}{0.81} = \frac{500}{81} \approx 6.1728$. Therefore,the correct ratio is $5 : 0.81$.

(A) For the Lyman series of the hydrogen atom $(Z=1)$,the shortest wavelength occurs for the transition from $n_2 = \infty$ to $n_1 = 1$.
Using the Rydberg formula: $1/\lambda = R Z^2 (1/n_1^2 - 1/n_2^2)$.
$1/x = R(1)^2 (1/1^2 - 1/\infty^2) = R$.
Therefore,$x = 1/R$.
For the Balmer series of $He^+$ $(Z=2)$,the longest wavelength occurs for the transition from $n_2 = 3$ to $n_1 = 2$.
$1/\lambda' = R Z^2 (1/n_1^2 - 1/n_2^2) = R(2)^2 (1/2^2 - 1/3^2)$.
$1/\lambda' = 4R (1/4 - 1/9) = 4R (5/36) = 5R/9$.
Substituting $R = 1/x$:
$1/\lambda' = 5/(9x)$.
Thus,$\lambda' = 9x/5$.

(D) The formula for the Bohr radius of a hydrogen-like species is given by $r_n = a_0 \times \frac{n^2}{Z}$,where $a_0 = 52.9 \text{ pm}$ is the Bohr radius of the hydrogen atom,$n$ is the principal quantum number,and $Z$ is the atomic number.
Given $r_n = 70.53 \text{ pm}$,we have $70.53 = 52.9 \times \frac{n^2}{Z}$.
Calculating the ratio: $\frac{n^2}{Z} = \frac{70.53}{52.9} = 1.333... = \frac{4}{3}$.
Testing the options:
For option $D$: $Li^{2+}$ has $Z = 3$. If $n = 2$,then $\frac{n^2}{Z} = \frac{2^2}{3} = \frac{4}{3}$.
This matches the calculated ratio. Therefore,the species is $Li^{2+}$ and the stationary state is $n = 2$.

(D) According to the photoelectric equation: $E = \phi + K.E.$
First,convert the work function $\phi$ from eV to Joules: $\phi = 2.3 \text{ eV} = 2.3 \times 1.602 \times 10^{-19} \text{ J} \approx 3.6846 \times 10^{-19} \text{ J}$.
Given kinetic energy $K.E. = 2.8 \times 10^{-20} \text{ J} = 0.28 \times 10^{-19} \text{ J}$.
Total energy of the incident radiation $E = 3.6846 \times 10^{-19} + 0.28 \times 10^{-19} = 3.9646 \times 10^{-19} \text{ J}$.
Using the relation $E = \frac{hc}{\lambda}$,where $h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}$ and $c = 3 \times 10^8 \text{ m/s}$:
$\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{3.9646 \times 10^{-19}} \approx 5.01 \times 10^{-7} \text{ m} = 501 \text{ nm}$.
Rounding to the nearest integer as per the format,$x = 5$,so $x = 5 \times 10^2$ nm.