Atomic models and Planck's quantum theory Questions in English

(C) According to the photoelectric effect equation:
$h\nu = KE_{max} + h\nu_0$
Where $h$ is Planck's constant $(6.63 \times 10^{-34} \ J \cdot s)$,$\nu$ is the incident frequency,$KE_{max}$ is the maximum kinetic energy,and $\nu_0$ is the threshold frequency.
Rearranging for $\nu_0$:
$h\nu_0 = h\nu - KE_{max}$
$\nu_0 = \nu - \frac{KE_{max}}{h}$
Substituting the given values:
$\nu_0 = (4 \times 10^{14} \ s^{-1}) - \frac{6.63 \times 10^{-20} \ J}{6.63 \times 10^{-34} \ J \cdot s}$
$\nu_0 = 4 \times 10^{14} \ s^{-1} - 1 \times 10^{14} \ s^{-1}$
$\nu_0 = 3 \times 10^{14} \ s^{-1}$

(A) The binding energy (work function) for $1 \ mol$ of electrons is $250 \ kJ \ mol^{-1}$.
To find the energy for a single electron,divide by Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$:
$E = \frac{250 \times 10^3 \ J \ mol^{-1}}{6.022 \times 10^{23} \ mol^{-1}} = 4.151 \times 10^{-19} \ J$.
The threshold energy is given by $E = h \nu_0$,where $h = 6.626 \times 10^{-34} \ J \ s$.
$\nu_0 = \frac{E}{h} = \frac{4.151 \times 10^{-19} \ J}{6.626 \times 10^{-34} \ J \ s} \approx 6.26 \times 10^{14} \ s^{-1}$.
Thus,the correct option is $A$.

(B) The energy of a single photon is given by $E = h \nu$,where $h = 6.626 \times 10^{-34} \ J \ s$ and $\nu = 3 \times 10^{12} \ Hz$.
Energy of one photon $= 6.626 \times 10^{-34} \times 3 \times 10^{12} = 1.9878 \times 10^{-21} \ J$.
One mole of photons contains $N_A = 6.022 \times 10^{23}$ photons.
Therefore,half a mole of photons contains $\frac{1}{2} \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$ photons.
Total energy for half a mole $= (1.9878 \times 10^{-21} \ J) \times (3.011 \times 10^{23}) = 598.5 \ J \approx 0.598 \ kJ$.
Thus,the energy is $0.598 \ kJ \ mol^{-1}$.
Hence,the correct option is $(B)$.

(D) According to Einstein's photoelectric equation: $KE = h v - h v_0$.
For the first case: $KE_1 = h(4.0 \times 10^{16}) - h v_0$.
For the second case: $KE_2 = h(2.0 \times 10^{16}) - h v_0$.
Given that $KE_1 = 4 KE_2$,we have:
$h(4.0 \times 10^{16}) - h v_0 = 4(h(2.0 \times 10^{16}) - h v_0)$.
Dividing by $h$: $4.0 \times 10^{16} - v_0 = 8.0 \times 10^{16} - 4 v_0$.
Rearranging the terms: $3 v_0 = 4.0 \times 10^{16}$.
$v_0 = \frac{4.0 \times 10^{16}}{3} = 1.33 \times 10^{16} \ s^{-1}$.

(B) According to the photoelectric effect,the energy of the incident photon is equal to the sum of the work function (threshold energy) and the kinetic energy of the ejected electron.
$h\nu = h\nu_0 + \frac{1}{2}m_ev^2$
Rearranging for kinetic energy:
$\frac{1}{2}m_ev^2 = h\nu - h\nu_0 = h(\nu - \nu_0)$
Solving for velocity $(v)$:
$v^2 = \frac{2h(\nu - \nu_0)}{m_e}$
$v = \sqrt{\frac{2h(\nu - \nu_0)}{m_e}}$

(C) For the photoelectric effect to occur,the energy of the incident photon $(E)$ must be greater than or equal to the work function $(W_0)$ of the metal,i.e.,$E \ge W_0$.
Energy of incident photon $(E) = \frac{hc}{\lambda}$.
Given $\lambda = 540 \ nm = 540 \times 10^{-9} \ m$,$h = 6.626 \times 10^{-34} \ J \cdot s$,and $c = 3 \times 10^8 \ m/s$.
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{540 \times 10^{-9}} \ J = \frac{19.878 \times 10^{-26}}{540 \times 10^{-9}} \ J = 0.0368 \times 10^{-17} \ J = 3.68 \times 10^{-19} \ J$.
Converting $E$ to $eV$: $E = \frac{3.68 \times 10^{-19}}{1.602 \times 10^{-19}} \ eV \approx 2.297 \ eV$.
Photoelectric effect occurs if $E \ge W_0$.
Comparing $E = 2.297 \ eV$ with the work functions:
$Li (2.42 \ eV) > 2.297 \ eV$ (No)
$K (2.25 \ eV) < 2.297 \ eV$ (Yes)
$Mg (3.70 \ eV) > 2.297 \ eV$ (No)
$Ag (4.30 \ eV) > 2.297 \ eV$ (No)
$Cu (4.80 \ eV) > 2.297 \ eV$ (No)
Only $K$ undergoes the photoelectric effect. Thus,the number of metals is $1$.

(B) According to Einstein's photoelectric equation,$K.E. = h(v - v_0)$.
For frequency $v_1 = 1.6 \times 10^{16} \ Hz$,$K.E._1 = h(1.6 \times 10^{16} - v_0) \longrightarrow \text{Eq. } (i)$.
For frequency $v_2 = 1.0 \times 10^{16} \ Hz$,$K.E._2 = h(1.0 \times 10^{16} - v_0) \longrightarrow \text{Eq. } (ii)$.
Given that $K.E._1 = 2 \times K.E._2$,we have:
$h(1.6 \times 10^{16} - v_0) = 2 \times h(1.0 \times 10^{16} - v_0)$.
$1.6 \times 10^{16} - v_0 = 2.0 \times 10^{16} - 2v_0$.
$2v_0 - v_0 = 2.0 \times 10^{16} - 1.6 \times 10^{16}$.
$v_0 = 0.4 \times 10^{16} \ Hz = 4 \times 10^{15} \ Hz$.

(C) Given: Work function $(W_0) = 4.80 \ eV$ and Kinetic energy $(KE) = 0.20 \ eV$.
According to the photoelectric effect equation:
Total energy $(E_T) = W_0 + KE = 4.80 \ eV + 0.20 \ eV = 5.0 \ eV$.
Convert energy to Joules: $E_T = 5.0 \times 1.602 \times 10^{-19} \ J = 8.01 \times 10^{-19} \ J$.
Using the relation $E_T = h\nu$,where $h = 6.626 \times 10^{-34} \ J \cdot s$:
$\nu = \frac{E_T}{h} = \frac{8.01 \times 10^{-19} \ J}{6.626 \times 10^{-34} \ J \cdot s} \approx 1.21 \times 10^{15} \ Hz$.

(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: $L = \frac{n h}{2 \pi}$,where $n$ is an integer $(n = 1, 2, 3, \dots)$.
This can be rewritten as $L = n \times (0.5 \frac{h}{\pi})$.
For the given options:
$A) 1.25 \frac{h}{\pi} = 2.5 \times \frac{h}{2 \pi}$ (Not an integer multiple of $\frac{h}{2 \pi}$)
$B) 1 \frac{h}{\pi} = 2 \times \frac{h}{2 \pi}$ ($n = 2$,permitted)
$C) 1.5 \frac{h}{\pi} = 3 \times \frac{h}{2 \pi}$ ($n = 3$,permitted)
$D) 0.5 \frac{h}{\pi} = 1 \times \frac{h}{2 \pi}$ ($n = 1$,permitted)
Therefore,$1.25 \frac{h}{\pi}$ is not a permitted value.

(C) Statement-$I$ is correct because according to classical electromagnetic theory,an accelerating charged particle must emit energy. Thus,an electron revolving around the nucleus should lose energy and eventually fall into the nucleus,which Rutherford's model could not explain.
Statement-$II$ is incorrect because,in the electromagnetic spectrum,the wavelength of $X$-rays ($10^{-10} \ m$ to $10^{-8} \ m$) is significantly smaller than the wavelength of microwaves ($10^{-3} \ m$ to $10^{-1} \ m$).
Therefore,Statement-$I$ is correct,but Statement-$II$ is not correct.

(D) Thomson's model proposed that an atom consists of a positively charged sphere with electrons embedded in it,similar to a watermelon or plum pudding.
It correctly accounted for the overall electrical neutrality of the atom.
However,the model failed to explain the results of later experiments,such as Rutherford's alpha-particle scattering experiment.
Therefore,the statement that 'this model could not explain the overall neutrality of the atom' is false,as the model was specifically designed to explain it.

(A) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $r_n = a_0 \times \frac{n^2}{Z}$,where $a_0$ is the Bohr radius,$n$ is the principal quantum number,and $Z$ is the atomic number.
For the hydrogen atom $(Z=1)$ at $n=2$: $x = a_0 \times \frac{2^2}{1} = 4a_0$,which implies $a_0 = \frac{x}{4}$.
For the $He^{+}$ ion $(Z=2)$ at $n=3$: $r_3 = a_0 \times \frac{3^2}{2} = a_0 \times \frac{9}{2}$.
Substituting $a_0 = \frac{x}{4}$ into the equation for $r_3$: $r_3 = (\frac{x}{4}) \times \frac{9}{2} = \frac{9}{8} x$.

(B) The radius of the $n^{th}$ orbit of a hydrogen atom is given by the formula $r_n = 0.529 \times n^2 \mathring{A} = 52.9 \times n^2 \text{ pm}$.
Given $r_n = 476.1 \text{ pm}$, we have $52.9 \times n^2 = 476.1$.
$n^2 = \frac{476.1}{52.9} = 9$.
Therefore, $n = 3$.
A transition to the $n = 3$ orbit corresponds to the Paschen series.

(D) The energy of an electron in the $n^{th}$ orbit is given by $E_n = \frac{E_1}{n^2}$,where $E_1 = -2.18 \times 10^{-18} \ J$.
For the fifth orbit $(n = 5)$,$E_5 = \frac{-2.18 \times 10^{-18}}{5^2} = \frac{-2.18 \times 10^{-18}}{25} = -0.0872 \times 10^{-18} \ J$.
The energy difference required for excitation is $\Delta E = E_5 - E_1 = (-0.0872 \times 10^{-18}) - (-2.18 \times 10^{-18}) = 2.0928 \times 10^{-18} \ J$.
Using the relation $\Delta E = h \nu$,the frequency $\nu$ is $\nu = \frac{\Delta E}{h} = \frac{2.0928 \times 10^{-18}}{6.6 \times 10^{-34}} \approx 3.17 \times 10^{15} \ Hz$.

(A) The radius of an orbit in a hydrogen-like species is given by the formula $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For the second orbit of a hydrogen atom $(H)$,$n_1 = 2$ and $Z_1 = 1$. So,$r_1 = 0.529 \times \frac{2^2}{1} = 0.529 \times 4$.
For the orbit $n$ of an ion $x$ with atomic number $Z_2$,$r_2 = 0.529 \times \frac{n^2}{Z_2}$.
Given $r_1 = r_2$,we have $4 = \frac{n^2}{Z_2}$,which implies $n^2 = 4Z_2$.
Checking the options:
For $A$: $Be^{3+}$ has $Z=4$. $n^2 = 4 \times 4 = 16$,so $n=4$. This matches.
For $B$: $Li^{2+}$ has $Z=3$. $n^2 = 4 \times 3 = 12$ (not a perfect square).
For $C$: $Be^{2+}$ has $Z=4$. $n^2 = 16$,$n=4$. However,$Be^{2+}$ is not a hydrogen-like species (it has $2$ electrons).
For $D$: $He^{+}$ has $Z=2$. $n^2 = 4 \times 2 = 8$ (not a perfect square).
Thus,the correct option is $A$.

(D) The radius of an orbit in a hydrogen atom is given by $r_n = 0.0529 \times n^2 \ nm$.
For $r_1 = 1.3225 \ nm$,$n_1^2 = 1.3225 / 0.0529 = 25$,so $n_1 = 5$.
For $r_2 = 0.2116 \ nm$,$n_2^2 = 0.2116 / 0.0529 = 4$,so $n_2 = 2$.
The energy of an electron in the $n$-th orbit is $E_n = -2.18 \times 10^{-18} / n^2 \ J$.
The energy change is $\Delta E = E_{n_2} - E_{n_1} = -2.18 \times 10^{-18} \times (1/n_2^2 - 1/n_1^2)$.
$\Delta E = -2.18 \times 10^{-18} \times (1/4 - 1/25) = -2.18 \times 10^{-18} \times (0.25 - 0.04) = -2.18 \times 10^{-18} \times 0.21 = -0.4578 \times 10^{-18} \ J$.
The energy of the emitted radiation is $|\Delta E| = 0.4578 \times 10^{-18} \ J$.

(B) For a single-electron species,the energy is given by the formula: $E_{n} = -13.6 \times \frac{Z^2}{n^2} \ \text{eV}$.
For $Li^{2+}$,the atomic number $Z = 3$. At $n = 3$,the energy is $E_3 = -13.6 \times \frac{3^2}{3^2} = -13.6 \ \text{eV}$.
For $Be^{3+}$,the atomic number $Z = 4$. At $n = 4$,the energy is $E_4 = -13.6 \times \frac{4^2}{4^2} = -13.6 \ \text{eV}$.
Since both species have the same energy of $-13.6 \ \text{eV}$,the correct pair is $Li^{2+}(n=3)$ and $Be^{3+}(n=4)$.

(B) The radius of a hydrogen-like ion is given by $r = \frac{0.0529 \times n^2}{Z} \ nm$.
Given $r = 1.763 \times 10^{-2} \ nm$ and $n = 1$.
Substituting the values: $1.763 \times 10^{-2} = \frac{0.0529 \times (1)^2}{Z}$.
$Z = \frac{0.0529}{0.01763} \approx 3$.
The energy of an orbit is given by $E = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \ J$.
Substituting $Z = 3$ and $n = 1$: $E = -2.18 \times 10^{-18} \times \frac{3^2}{1^2} \ J$.
$E = -2.18 \times 10^{-18} \times 9 \ J = -19.62 \times 10^{-18} \ J$.
$E = -1.962 \times 10^{-17} \ J$.

(D) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = a_0 \times \frac{n^2}{Z}$, where $a_0$ is the Bohr radius, $n$ is the principal quantum number, and $Z$ is the atomic number.
For the $3^{rd}$ orbit of $H$ atom $(n=3, Z=1)$: $R = a_0 \times \frac{3^2}{1} = 9 a_0$.
For the $2^{nd}$ orbit of $He^{+}$ ion $(n=2, Z=2)$: $r_2 = a_0 \times \frac{2^2}{2} = 2 a_0$.
From the first equation, $a_0 = \frac{R}{9}$.
Substituting this into the second equation: $r_2 = 2 \times (\frac{R}{9}) = \frac{2}{9} R$.

(C) The radius of the $n^{\text{th}}$ orbit of a hydrogen-like species is given by the formula: $r_n = 0.529 \times \frac{n^2}{Z} \mathring{A}$.
For the first orbit of the hydrogen atom $(n=1, Z=1)$,the radius is $r_1 = 0.529 \mathring{A}$.
For the radius of the $n^{\text{th}}$ orbit of species $X$ to be equal to the first Bohr orbit of hydrogen,we set $r_n = r_1$,which implies $\frac{n^2}{Z} = 1$,or $n^2 = Z$.
Testing the options:
For $Be^{3+}$,the atomic number $Z = 4$. Substituting $Z=4$ into $n^2 = Z$,we get $n^2 = 4$,which gives $n = 2$.
Thus,for $n = 2$ and $X = Be^{3+}$,the condition is satisfied.

(C) The energy of an electron in a hydrogen atom is given by the formula: $E_n = \frac{-13.6 \ eV}{n^2}$.
For $n=1$,$E_1 = \frac{-13.6}{1^2} = -13.6 \ eV$.
For $n=2$,$E_2 = \frac{-13.6}{2^2} = -3.4 \ eV$.
For $n=3$,$E_3 = \frac{-13.6}{3^2} = -1.51 \ eV$.
The ratio $E_1 : E_2 : E_3$ is $\frac{-13.6}{1} : \frac{-13.6}{4} : \frac{-13.6}{9}$.
Dividing by $-13.6$,we get $1 : \frac{1}{4} : \frac{1}{9}$.
Multiplying by $36$ to clear the denominators,we get $36 : 9 : 4$.

(B) The energy of an orbit in a hydrogen-like species is given by the formula $E_n = -E_0 \times \frac{Z^2}{n^2}$,where $E_0$ is the ground state energy of hydrogen $(2.18 \times 10^{-18} \ J)$.
For the second orbit of hydrogen $(n=2, Z=1)$: $E_2 = -2.18 \times 10^{-18} \times \frac{1^2}{2^2} = -5.45 \times 10^{-19} \ J$.
For the first orbit of $Li^{2+}$ ion $(n=1, Z=3)$:
$E_1 = -2.18 \times 10^{-18} \times \frac{3^2}{1^2} \ J$
$E_1 = -2.18 \times 10^{-18} \times 9 \ J$
$E_1 = -1.962 \times 10^{-17} \ J$.

(B) The energy of the $n^{th}$ Bohr orbit is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For the hydrogen atom $(Z=1)$,the energy difference between orbits $n_1$ and $n_2$ is $\Delta E = 13.6 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$.
Difference in energy between the first $(n=1)$ and second $(n=2)$ orbits:
$\Delta E_{1-2} = 13.6 \times (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 \times (1 - \frac{1}{4}) = 13.6 \times \frac{3}{4} \text{ eV}$.
Difference in energy between the second $(n=2)$ and third $(n=3)$ orbits:
$\Delta E_{2-3} = 13.6 \times (\frac{1}{2^2} - \frac{1}{3^2}) = 13.6 \times (\frac{1}{4} - \frac{1}{9}) = 13.6 \times \frac{5}{36} \text{ eV}$.
The required ratio is $\frac{\Delta E_{1-2}}{\Delta E_{2-3}} = \frac{13.6 \times (3/4)}{13.6 \times (5/36)} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5}$.

(C) For a hydrogen atom,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$ and the energy is given by $E_n \propto \frac{1}{n^2}$.
For the second orbit $(n=2)$: $r_2 \propto (2)^2 = 4$ and $E_2 \propto \frac{1}{(2)^2} = \frac{1}{4}$.
For the third orbit $(n=3)$: $r_3 \propto (3)^2 = 9$ and $E_3 \propto \frac{1}{(3)^2} = \frac{1}{9}$.
Calculating the ratios:
$\frac{r_3}{r_2} = \frac{9}{4} \implies r_3 = \frac{9}{4} r_2$
$\frac{E_3}{E_2} = \frac{1/9}{1/4} = \frac{4}{9} \implies E_3 = \frac{4}{9} E_2$
Thus,the radius and energy of the third Bohr orbit are $\frac{9}{4} r_2$ and $\frac{4}{9} E_2$,respectively.

(B) According to Bohr's model,the velocity of an electron in the $n^{th}$ orbit is given by $v_n = \frac{2.18 \times 10^6 \ Z}{n} \ m/s$.
For the first Bohr orbit of a hydrogen atom,$n = 1$ and $Z = 1$,so $v = 2.18 \times 10^6 \ m/s$.
The speed of light in vacuum is $c = 3 \times 10^8 \ m/s$.
The ratio of the speed of light to the speed of the electron is $\frac{c}{v} = \frac{3 \times 10^8 \ m/s}{2.18 \times 10^6 \ m/s} \approx 137.6$.
Thus,the approximate ratio is $137 : 1$.

(C) According to Bohr's theory of hydrogen atom:
$(i)$ The speed of electron in the $n$th orbit is $v_n \propto \frac{1}{n}$.
$(ii)$ The energy of electron in the $n$th orbit is $E_n \propto -\frac{1}{n^2}$,so the magnitude varies as $\frac{1}{n^2}$.
$(iii)$ The radius of the electron in the $n$th orbit is $r_n \propto n^2$.
Therefore,the variation of speed,energy,and radius with $n$ is $\frac{1}{n} : \frac{1}{n^2} : n^2$.

(C) The energy of an electron in the $n$th orbital of a hydrogen atom is given by the formula: $E_n = -2.18 \times 10^{-18} \cdot \frac{Z^2}{n^2} \ J$.
For a hydrogen atom,$Z = 1$.
Thus,$E_n = -\frac{2.18 \times 10^{-18}}{n^2} \ J$.
For $n = \infty$,$E_{\infty} = -\frac{2.18 \times 10^{-18}}{\infty} = 0 \ J$.
For $n = 3$,$E_3 = -\frac{2.18 \times 10^{-18}}{3^2} = -\frac{2.18 \times 10^{-18}}{9} = -0.242 \times 10^{-18} \ J$.
Therefore,$E_{\infty} = 0 \ J$ and $E_3 = -0.242 \times 10^{-18} \ J$.
Hence,option $(c)$ is the correct answer.

(C) The radius of the $n^{th}$ orbit for a hydrogen-like species is given by the formula:
$r = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$
Where $n$ is the principal quantum number (orbit number) and $Z$ is the atomic number.
For the $B^{4+}$ ion:
$n = 2$
$Z = 5$ (Atomic number of Boron)
Substituting these values into the formula:
$r = 0.529 \times \frac{2^2}{5} \ \mathring{A}$
$r = 0.529 \times \frac{4}{5} \ \mathring{A}$
$r = 0.529 \times 0.8 \ \mathring{A}$
$r = 0.4232 \ \mathring{A}$

(D) The radius of an electron $(r_n)$ in any orbit is given by the formula: $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For $Li^{2+}$: $n = 2$,$Z = 3$.
For $Be^{3+}$: $n = 3$,$Z = 4$.
The ratio is given by: $\frac{r_{Li^{2+}}}{r_{Be^{3+}}} = \frac{n_1^2 / Z_1}{n_2^2 / Z_2} = \frac{n_1^2}{Z_1} \times \frac{Z_2}{n_2^2}$.
Substituting the values: $\frac{2^2}{3} \times \frac{4}{3^2} = \frac{4}{3} \times \frac{4}{9} = \frac{16}{27}$.
Thus,the correct option is $(D)$.

(C) The velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $v = 2.18 \times 10^6 \times \frac{Z}{n} \ ms^{-1}$.
For a hydrogen atom,the atomic number $Z = 1$.
For the third orbit,the principal quantum number $n = 3$.
Substituting these values into the formula:
$v = 2.18 \times 10^6 \times \frac{1}{3} \ ms^{-1}$.
$v = 0.7266 \times 10^6 \ ms^{-1}$.
$v = 7.26 \times 10^5 \ ms^{-1}$.
Therefore,option $(C)$ is the correct answer.

(B) The energy of an electron in the $n^{th}$ orbit of an $H$-like atom is given by the formula:
$E_{n} = \frac{-13.6 \times Z^2}{n^2} \ eV$.
For the $H$-atom,$Z = 1$ and for the $3^{rd}$ orbit,$n = 3$.
Substituting these values:
$E = \frac{-13.6 \times (1)^2}{(3)^2} \ eV = \frac{-13.6}{9} \ eV \approx -1.511 \ eV$.
Since $1 \ eV = 1.602 \times 10^{-19} \ J$,
$E = -1.511 \times 1.602 \times 10^{-19} \ J \approx -2.42 \times 10^{-19} \ J$.

(C) The energy levels of a hydrogen atom are given by the formula: $E_n = -13.6 / n^2 \ eV$.
For the ground state,$n = 1$,so $E_1 = -13.6 \ eV$.
An excited state occurs when $n > 1$.
For the first excited state,$n = 2$.
Substituting $n = 2$ into the formula: $E_2 = -13.6 / 2^2 = -13.6 / 4 = -3.4 \ eV$.

(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{313.52 Z^2}{n^2} \ kcal \ mol^{-1}$.
For a hydrogen atom,$Z = 1$.
The Lyman series corresponds to transitions ending at $n_1 = 1$.
The $H_\alpha$ line in the Lyman series corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.
Energy in the $n_1 = 1$ orbit: $E_1 = -\frac{313.52 \times (1)^2}{(1)^2} = -313.52 \ kcal \ mol^{-1} \approx -313.6 \ kcal \ mol^{-1}$.
Energy in the $n_2 = 2$ orbit: $E_2 = -\frac{313.52 \times (1)^2}{(2)^2} = -\frac{313.52}{4} = -78.38 \ kcal \ mol^{-1} \approx -78.4 \ kcal \ mol^{-1}$.
Thus,the energies are $-313.6 \ kcal \ mol^{-1}$ and $-78.4 \ kcal \ mol^{-1}$.

(C) The kinetic energy $(KE)$ of an electron in an orbit is given by $KE = \frac{1}{2} mv^2$.
From the electrostatic force balance,$\frac{mv^2}{r} = \frac{e^2}{r^2}$,which implies $mv^2 = \frac{e^2}{r}$.
Substituting this into the kinetic energy expression,we get $KE = \frac{1}{2} \frac{e^2}{r}$.
The potential energy $(PE)$ of the electron is $PE = \frac{-e^2}{r}$.
The total energy $(E)$ is the sum of kinetic and potential energy: $E = KE + PE = \frac{1}{2} \frac{e^2}{r} - \frac{e^2}{r} = \frac{-e^2}{2r}$.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = \frac{-1312}{n^2} \ kJ \ mol^{-1}$.
For the second orbit,$n = 2$.
Substituting the value of $n$ in the formula:
$E_2 = \frac{-1312}{2^2} \ kJ \ mol^{-1} = \frac{-1312}{4} \ kJ \ mol^{-1} = -328 \ kJ \ mol^{-1}$.

(C) Using the Rydberg formula: $\frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. For hydrogen,$Z=1$.
For transition $(i)$ $n=4$ to $n=2$: $\frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \left( \frac{3}{16} \right)$. Thus,$\lambda_1 = \frac{16}{3R_H}$.
For transition $(ii)$ $n=3$ to $n=1$: $\frac{1}{\lambda_2} = R_H \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R_H \left( 1 - \frac{1}{9} \right) = R_H \left( \frac{8}{9} \right)$. Thus,$\lambda_2 = \frac{9}{8R_H}$.
Calculating $(\lambda_1 - \lambda_2)$: $\frac{16}{3R_H} - \frac{9}{8R_H} = \frac{1}{R_H} \left( \frac{128 - 27}{24} \right) = \frac{1}{R_H} \left( \frac{101}{24} \right)$.

(B) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{n_2^2} - \frac{1}{n_1^2} \right]$.
For the Balmer series,$n_2 = 2$ and $n_1 = 3, 4, 5, \dots$.
Substituting $n_2 = 2$ and $n_1 = 5$ into the formula:
$\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{5^2} \right] = R \left[ \frac{1}{4} - \frac{1}{25} \right]$.
$\frac{1}{\lambda} = R \left[ \frac{25 - 4}{100} \right] = R \left[ \frac{21}{100} \right]$.
Therefore,$\lambda = \frac{100}{21 R}$.

(D) Given:
Kinetic energy $(K.E.) = 2 \times 10^{-19} \ J$
Frequency of incident radiation $(\nu) = 1.0 \times 10^{15} \ Hz$
Planck's constant $(h) = 6.6 \times 10^{-34} \ Js$
According to Einstein's photoelectric equation:
$K.E. = h(\nu - \nu_0)$
Where $\nu_0$ is the threshold frequency.
Substituting the values:
$2 \times 10^{-19} = 6.6 \times 10^{-34} \times (1.0 \times 10^{15} - \nu_0)$
$\frac{2 \times 10^{-19}}{6.6 \times 10^{-34}} = 1.0 \times 10^{15} - \nu_0$
$0.303 \times 10^{15} = 1.0 \times 10^{15} - \nu_0$
$\nu_0 = 1.0 \times 10^{15} - 0.303 \times 10^{15}$
$\nu_0 = 0.697 \times 10^{15} \ Hz = 6.97 \times 10^{14} \ Hz$

(A) The Rydberg formula for wavenumber is $\bar{\nu} = R_H Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the first spectral line of the Lyman series of $He^{+}$ $(Z=2)$: $n_1=1, n_2=2$.
$\bar{\nu}_1 = R_H \times 2^2 \times \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R_H \times 4 \times \frac{3}{4} = 3 R_H = x$.
Thus,$R_H = \frac{x}{3}$.
For the second spectral line of the Balmer series of $Li^{2+}$ $(Z=3)$: $n_1=2, n_2=4$.
$\bar{\nu}_2 = R_H \times 3^2 \times \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R_H \times 9 \times \left[ \frac{1}{4} - \frac{1}{16} \right] = R_H \times 9 \times \frac{3}{16} = \frac{27 R_H}{16}$.
Substituting $R_H = \frac{x}{3}$:
$\bar{\nu}_2 = \frac{27}{16} \times \frac{x}{3} = \frac{9 x}{16}$.

(A) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the second line of the Balmer series of the $H$-atom $(Z=1, n_1=2, n_2=4)$:
$\frac{1}{\lambda} = R(1)^2 \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{4} - \frac{1}{16} \right] = \frac{3R}{16}$ $\Rightarrow R = \frac{16}{3\lambda}$.
For the first line of the Lyman series of the $He^{+}$ ion $(Z=2, n_1=1, n_2=2)$:
$\frac{1}{\lambda_2} = R(2)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = 4R \left[ 1 - \frac{1}{4} \right] = 4R \left( \frac{3}{4} \right) = 3R$.
Substituting $R = \frac{16}{3\lambda}$ into the expression for $\lambda_2$:
$\frac{1}{\lambda_2} = 3 \left( \frac{16}{3\lambda} \right) = \frac{16}{\lambda}$ $\Rightarrow \lambda_2 = \frac{\lambda}{16}$.

(C) According to Bohr's theory,the energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2} \ eV$.
The energy required to excite an electron from orbit $n_1$ to $n_2$ is $\Delta E = E_{n_2} - E_{n_1} = 13.6 \ Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \ eV$.
For a hydrogen atom,$Z = 1$,$n_1 = 2$,and $n_2 = 3$.
Substituting these values:
$\Delta E = 13.6 \times 1^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \ eV$
$\Delta E = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) \ eV$
$\Delta E = 13.6 \left( \frac{9 - 4}{36} \right) \ eV$
$\Delta E = 13.6 \times \frac{5}{36} \ eV \approx 1.888 \ eV \approx 1.9 \ eV$.

(B) The general formula for calculating the number of spectral lines when an electron transitions from $n_2$ to $n_1$ is given by:
$N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$
Given,$n_2 = 6$ and $n_1 = 2$.
Substituting the values:
$N = \frac{(6 - 2)(6 - 2 + 1)}{2}$
$N = \frac{4 \times 5}{2}$
$N = 10$
These spectral lines belong to the Balmer series of the hydrogen spectrum.

(A) The energy of an electron in a hydrogen-like species in the $n^{th}$ shell is given by the formula:
$E = -13.6 \times \frac{Z^2}{n^2} \ eV$
Where:
$Z = \text{atomic number of } Li = 3$
$n = \text{orbit number} = 1$
Substituting these values into the formula:
$E_1 = -13.6 \times \frac{3^2}{1^2} \ eV$
$E_1 = -13.6 \times 9 \ eV = -122.4 \ eV$

(D) The spectral series of hydrogen are:
$1$. Lyman series: Ultraviolet region.
$2$. Balmer series: Visible region.
$3$. Paschen series: Infrared region.
$4$. Brackett series: Infrared region.
$5$. Pfund series: Infrared region.
Therefore,the Lyman and Balmer series are the two series that do not belong to the infrared spectral region.

(C) When an electron in a hydrogen atom is excited to the $n$th energy level,it can return to lower energy levels by emitting photons.
The total number of possible spectral lines is given by the formula:
$\text{Number of spectral lines} = \frac{n(n-1)}{2}$
where $n$ represents the principal quantum number of the excited state.

(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.