Spectral Series of Hydrogen Atom Questions in English

(B) The wavelength $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant.
$1$. For the first Lyman line,the transition is from $n_2 = 2$ to $n_1 = 1$:
$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_L = \frac{4}{3R}$.
$2$. For the second Balmer line,the transition is from $n_2 = 4$ to $n_1 = 2$:
$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4-1}{16} \right) = \frac{3R}{16} \implies \lambda_B = \frac{16}{3R}$.
$3$. The ratio of the wavelengths is:
$\frac{\lambda_L}{\lambda_B} = \frac{4/3R}{16/3R} = \frac{4}{16} = \frac{1}{4}$.
Thus,the ratio is $1: 4$.

(C) The wavelength $\lambda$ for a hydrogen atom transition is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the shortest wavelength,the transition occurs from $n_2 = \infty$ to $n_1$.
Thus,$\frac{1}{\lambda} = \frac{R}{n_1^2}$,or $\lambda = \frac{n_1^2}{R}$.
For the Balmer series,$n_1 = 2$,so $\lambda_{Balmer} = \frac{2^2}{R} = \frac{4}{R}$.
For the Bracket series,$n_1 = 4$,so $\lambda_{Bracket} = \frac{4^2}{R} = \frac{16}{R}$.
The ratio of the shortest wavelength of the Bracket series to the Balmer series is $\frac{\lambda_{Bracket}}{\lambda_{Balmer}} = \frac{16/R}{4/R} = \frac{16}{4} = 4:1$.

(C) For hydrogen atom,the Rydberg formula is $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,$n_1 = 2$. The minimum wavelength corresponds to $n_2 = \infty$.
$\frac{1}{\lambda_{\text{min, Balmer}}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4} \Rightarrow \lambda_{\text{min, Balmer}} = \frac{4}{R}$.
For the Brackett series,$n_1 = 4$. The maximum wavelength corresponds to $n_2 = 5$.
$\frac{1}{\lambda_{\text{max, Brackett}}} = R \left( \frac{1}{4^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{16} - \frac{1}{25} \right) = R \left( \frac{25 - 16}{400} \right) = \frac{9R}{400}$.
$\lambda_{\text{max, Brackett}} = \frac{400}{9R}$.
The ratio is $\frac{\lambda_{\text{min, Balmer}}}{\lambda_{\text{max, Brackett}}} = \frac{4/R}{400/9R} = \frac{4}{R} \times \frac{9R}{400} = \frac{36}{400} = \frac{9}{100}$.

(D) The spectral lines of the hydrogen atom are categorized into series based on the final energy level $n_f$ of the electron transition.
For the Lyman series,the electron transitions to the ground state,$n_f = 1$.
The energy of the emitted photon is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
Transitions to $n_f = 1$ result in high-energy photons that fall within the ultraviolet region of the electromagnetic spectrum.
Among the given options,the transition $2 \rightarrow 1$ belongs to the Lyman series and thus emits a photon in the ultraviolet region.

(B) For the Lyman series,the shortest wavelength corresponds to the transition from $n_2 = \infty$ to $n_1 = 1$. Using the Rydberg formula $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,we get:
$\frac{1}{\lambda_1} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R_H$ ...$(1)$
For the Balmer series,the shortest wavelength corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$. Using the Rydberg formula,we get:
$\frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R_H}{4}$ ...$(2)$
From equation $(1)$,$R_H = \frac{1}{\lambda_1}$. Substituting this into equation $(2)$,we have $\frac{1}{\lambda_2} = \frac{1}{4\lambda_1}$. However,the question asks for the expression of $R_H$ in terms of $\lambda_1$ and $\lambda_2$.
Subtracting the equations: $\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = R_H - \frac{R_H}{4} = \frac{3R_H}{4}$.
Therefore,$R_H = \frac{4}{3} \left( \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \right) = \frac{4(\lambda_2 - \lambda_1)}{3 \lambda_1 \lambda_2}$.

(C) The wavelength $\lambda$ for the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Paschen series,the transition occurs to the $n_1 = 3$ energy level.
The shortest wavelength corresponds to the transition from the highest energy level,$n_2 = \infty$.
Substituting these values into the formula:
$\frac{1}{\lambda_S} = R_H \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{9} - 0 \right) = \frac{R_H}{9}$.
Therefore,$\lambda_S = \frac{9}{R_H}$.
Given $R_H = 1.097 \times 10^7 \ m^{-1}$,we have:
$\lambda_S = \frac{9}{1.097 \times 10^7} \ m \approx 8.204 \times 10^{-7} \ m$.
Converting to nanometers $(1 \ nm = 10^{-9} \ m)$:
$\lambda_S = 820.4 \ nm$.

(B) The Rydberg formula for the wavelength of spectral lines is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) Z^2$.
For the Lyman series,the electron transitions to the ground state,so $n_1 = 1$.
The first spectral line of the Lyman series corresponds to the transition from the first excited state to the ground state,so $n_2 = 2$.
For hydrogen,the atomic number $Z = 1$.
Substituting these values into the formula:
$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) (1)^2 = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$.
Therefore,$\lambda = \frac{4}{3R}$.
Given that $\frac{1}{R} \approx 911.6 \mathring A$ (often approximated as $912 \mathring A$),we have:
$\lambda = \frac{4}{3} \times 911.6 \mathring A \approx 1215.5 \mathring A$.
Rounding to the nearest option,the wavelength is $1215 \mathring A$.

(B) The energy of the ground state of a hydrogen atom is $E_1 = -13.6 eV$.
When energy $\Delta E = 12.1 eV$ is supplied,the electron is excited to a higher energy level $n$ such that $E_n = E_1 + \Delta E$.
$E_n = -13.6 eV + 12.1 eV = -1.5 eV$.
Using the formula $E_n = -\frac{13.6}{n^2} eV$,we have $-\frac{13.6}{n^2} = -1.5$.
$n^2 = \frac{13.6}{1.5} \approx 9.07$,which implies $n = 3$.
The electron is excited to the second excited state $(n = 3)$.
The number of spectral lines emitted when an electron transitions from state $n$ to the ground state is given by $N = \frac{n(n-1)}{2}$.
Substituting $n = 3$,we get $N = \frac{3(3-1)}{2} = \frac{3 \times 2}{2} = 3$ lines.

(C) According to the Bohr model of the hydrogen atom,the energy levels are denoted by the principal quantum number $n$.
The innermost orbit corresponds to the ground state,where $n=1$.
When an electron transitions from any higher energy level $(n_2 > 1)$ to the ground state $(n_1 = 1)$,the emitted electromagnetic radiation falls in the ultraviolet region of the spectrum.
This specific set of spectral lines is known as the Lyman series.

(B) The wavelength of the spectral lines in the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \dots$
For the second line of the Balmer series,$n = 4$. Given $\lambda_2 = 4861 Å$:
$\frac{1}{4861} = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right) \implies R = \frac{16}{3 \times 4861} \dots (i)$
For the first line of the Balmer series,$n = 3$:
$\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$
Substituting the value of $R$ from equation $(i)$:
$\frac{1}{\lambda_1} = \left( \frac{16}{3 \times 4861} \right) \times \left( \frac{5}{36} \right) = \frac{80}{108 \times 4861} = \frac{20}{27 \times 4861}$
$\lambda_1 = \frac{27 \times 4861}{20} = \frac{131247}{20} = 6562.35 Å \approx 6563 Å$.

(B) The Rydberg formula for the wavelength of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$.
For hydrogen atom $(H)$,$Z = 1$. Thus,$\frac{1}{\lambda} = R (1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$.
For doubly ionized lithium $(Li^{2+})$,$Z = 3$. Let the wavelength be $\lambda'$.
Then,$\frac{1}{\lambda'} = R (3)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 9 R \left( \frac{5}{36} \right)$.
Comparing the two equations,$\frac{1}{\lambda'} = 9 \left( \frac{1}{\lambda} \right)$,which implies $\lambda' = \frac{\lambda}{9}$.

(D) For the Lyman series,the wavelengths are given by $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$ where $n = 2, 3, 4, \dots$.
Shortest wavelength $(\lambda_{L, min})$ occurs at $n = \infty$: $\frac{1}{\lambda_{L, min}} = R \implies \lambda_{L, min} = \frac{1}{R}$.
Longest wavelength $(\lambda_{L, max})$ occurs at $n = 2$: $\frac{1}{\lambda_{L, max}} = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_{L, max} = \frac{4}{3R}$.
$\Delta \lambda_L = \lambda_{L, max} - \lambda_{L, min} = \frac{4}{3R} - \frac{1}{R} = \frac{1}{3R}$.
For the Balmer series,the wavelengths are given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$ where $n = 3, 4, 5, \dots$.
Shortest wavelength $(\lambda_{B, min})$ occurs at $n = \infty$: $\frac{1}{\lambda_{B, min}} = \frac{R}{4} \implies \lambda_{B, min} = \frac{4}{R}$.
Longest wavelength $(\lambda_{B, max})$ occurs at $n = 3$: $\frac{1}{\lambda_{B, max}} = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36} \implies \lambda_{B, max} = \frac{36}{5R}$.
$\Delta \lambda_B = \lambda_{B, max} - \lambda_{B, min} = \frac{36}{5R} - \frac{4}{R} = \frac{36 - 20}{5R} = \frac{16}{5R}$.
Therefore,$\frac{\Delta \lambda_B}{\Delta \lambda_L} = \frac{16/5R}{1/3R} = \frac{16}{5} \times 3 = \frac{48}{5} = 9.6$.

(B) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the longest wavelength line,the transition occurs between adjacent energy levels,i.e.,$n_2 = n_1 + 1$.
For the Balmer series,$n_1 = 2$,so $n_2 = 3$. The longest wavelength $\lambda_{BL}$ is:
$\frac{1}{\lambda_{BL}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \implies \lambda_{BL} = \frac{36}{5R} \quad ... (A)$
For the Paschen series,$n_1 = 3$,so $n_2 = 4$. The longest wavelength $\lambda_{PL}$ is:
$\frac{1}{\lambda_{PL}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{7}{144} \right) \implies \lambda_{PL} = \frac{144}{7R} \quad ... (B)$
Taking the ratio of the longest wavelengths:
$\frac{\lambda_{BL}}{\lambda_{PL}} = \frac{36}{5R} \times \frac{7R}{144} = \frac{36 \times 7}{5 \times 144} = \frac{7}{5 \times 4} = \frac{7}{20}$.

(D) The frequency $\nu$ of a spectral line in the hydrogen atom is given by $\nu = c \cdot \bar{\nu} = Rc \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series,$n_1 = 1$.
The first Lyman line corresponds to the transition from $n_2 = 2$ to $n_1 = 1$:
$\nu_1 = Rc \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = Rc \left( 1 - \frac{1}{4} \right) = \frac{3Rc}{4}$.
The second Lyman line corresponds to the transition from $n_2 = 3$ to $n_1 = 1$:
$\nu_2 = Rc \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = Rc \left( 1 - \frac{1}{9} \right) = \frac{8Rc}{9}$.
The difference between the frequencies is $\Delta \nu = \nu_2 - \nu_1 = \frac{8Rc}{9} - \frac{3Rc}{4}$.
Taking the common denominator as $36$:
$\Delta \nu = \frac{32Rc - 27Rc}{36} = \frac{5Rc}{36}$.

(C) The frequency of a spectral line in the hydrogen atom is given by the formula: $\nu = Rc \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Paschen series,the lower energy level is $n_1 = 3$.
The first Paschen line corresponds to the transition from $n_2 = 4$ to $n_1 = 3$. Its frequency is $\nu_1 = Rc \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = Rc \left( \frac{1}{9} - \frac{1}{16} \right) = Rc \left( \frac{16 - 9}{144} \right) = \frac{7Rc}{144}$.
The second Paschen line corresponds to the transition from $n_2 = 5$ to $n_1 = 3$. Its frequency is $\nu_2 = Rc \left( \frac{1}{3^2} - \frac{1}{5^2} \right) = Rc \left( \frac{1}{9} - \frac{1}{25} \right) = Rc \left( \frac{25 - 9}{225} \right) = \frac{16Rc}{225}$.
The difference between the frequencies is $\Delta \nu = \nu_2 - \nu_1 = Rc \left( \frac{16}{225} - \frac{7}{144} \right)$.
Calculating the common denominator: $225 = 3^2 \times 5^2$ and $144 = 3^2 \times 4^2$. The $LCM$ is $3^2 \times 5^2 \times 4^2 = 9 \times 25 \times 16 = 3600$.
$\Delta \nu = Rc \left( \frac{16 \times 16 - 7 \times 25}{3600} \right) = Rc \left( \frac{256 - 175}{3600} \right) = Rc \left( \frac{81}{3600} \right) = \frac{9Rc}{400}$.

(C) The wavelength $\lambda$ of a spectral line emitted during a transition from orbit $n_i$ to $n_f$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$,where $R$ is the Rydberg constant.
For the transition $3 \rightarrow 2$: $\frac{1}{\lambda_1} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36}$. Thus,$\lambda_1 = \frac{36}{5R}$.
For the transition $2 \rightarrow 1$: $\frac{1}{\lambda_2} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$. Thus,$\lambda_2 = \frac{4}{3R}$.
The ratio of the wavelengths is $\frac{\lambda_1}{\lambda_2} = \frac{36/5R}{4/3R} = \frac{36}{5R} \times \frac{3R}{4} = \frac{9 \times 3}{5} = \frac{27}{5}$.

(C) The frequency of a spectral line in the hydrogen atom is given by $f = R c \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant and $c$ is the speed of light.
For the Lyman series,$n_1 = 1$. The first line is $n_2 = 2$ and the second line is $n_2 = 3$.
$f_1 = R c \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R c \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R c$
$f_2 = R c \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R c \left( 1 - \frac{1}{9} \right) = \frac{8}{9} R c$
The difference is $f = f_2 - f_1 = R c \left( \frac{8}{9} - \frac{3}{4} \right) = R c \left( \frac{32 - 27}{36} \right) = \frac{5}{36} R c$. Thus,$R c = \frac{36 f}{5}$.
For the Balmer series,$n_1 = 2$. The first line is $n_2 = 3$ and the second line is $n_2 = 4$.
$f'_1 = R c \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R c \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5}{36} R c$
$f'_2 = R c \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R c \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3}{16} R c$
The difference is $f' = f'_2 - f'_1 = R c \left( \frac{3}{16} - \frac{5}{36} \right) = R c \left( \frac{27 - 20}{144} \right) = \frac{7}{144} R c$.
Substituting $R c = \frac{36 f}{5}$,we get $f' = \frac{7}{144} \times \frac{36 f}{5} = \frac{7 f}{4 \times 5} = \frac{7 f}{20}$.

(C) The wavelength $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant.
For the second line of the Balmer series,$n_1 = 2$ and $n_2 = 4$. Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right)$. So,$\lambda_B = \frac{16}{3R}$.
For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$. Thus,$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$. So,$\lambda_L = \frac{4}{3R}$.
The ratio of the wavelengths is $\frac{\lambda_B}{\lambda_L} = \frac{16/3R}{4/3R} = \frac{16}{4} = 4:1$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.
At room temperature, hydrogen atoms are in the ground state $(n=1)$, where the energy is $E_1 = -13.6 \ eV$.
When the atom is bombarded with electrons of $13.6 \ eV$ energy, the atom can absorb this energy and be excited to the ionization limit $(n = \infty)$ or any intermediate state.
However, the question asks for the series of the emitted spectral line. When an excited electron returns to the ground state $(n=1)$ from any higher energy level $(n > 1)$, the emitted radiation falls in the Lyman series.
Since the electrons can excite the hydrogen atoms to higher states and they eventually decay to the ground state, the emitted spectral lines belong to the Lyman series.

(B) The wavelength of radiation emitted from a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the first case ($n_2 = 4$ to $n_1 = 2$):
$\frac{1}{\lambda_1} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{4} - \frac{1}{16} \right] = R \left[ \frac{4-1}{16} \right] = \frac{3R}{16}$.
For the second case ($n_2 = 3$ to $n_1 = 2$):
$\frac{1}{\lambda_2} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{9-4}{36} \right] = \frac{5R}{36}$.
Now,taking the ratio of the wavelengths:
$\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{5R/36}{3R/16} = \frac{5}{36} \times \frac{16}{3} = \frac{5 \times 4}{9 \times 3} = \frac{20}{27}$.
Thus,the ratio is $20:27$.

(C) The Rydberg formula for the Balmer series is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \dots$.
For the shortest wavelength $\lambda_1$,$n = \infty$:
$\frac{1}{\lambda_1} = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \implies R = \frac{4}{\lambda_1}$.
For the longest wavelength $\lambda_2$,$n = 3$:
$\frac{1}{\lambda_2} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \implies R = \frac{36}{5\lambda_2}$.
To express $R$ in terms of $\lambda_1$ and $\lambda_2$,we use the relation:
$\frac{1}{\lambda_1} = \frac{R}{4}$ and $\frac{1}{\lambda_2} = \frac{5R}{36}$.
Multiplying the first by $9$: $\frac{9}{\lambda_1} = \frac{9R}{4}$.
Subtracting the second: $\frac{9}{\lambda_1} - \frac{1}{\lambda_2} = \frac{9R}{4} - \frac{5R}{36} = \frac{81R - 5R}{36} = \frac{76R}{36} = \frac{19R}{9}$.
Alternatively,using the derived values:
From $\frac{1}{\lambda_1} = \frac{R}{4}$,$R = \frac{4}{\lambda_1}$.
From $\frac{1}{\lambda_2} = \frac{5R}{36}$,$R = \frac{36}{5\lambda_2}$.
Equating or manipulating the options,we find that for $R = \frac{9}{\lambda_1} - \frac{9}{\lambda_2}$,we get $R = 9(\frac{R}{4} - \frac{5R}{36}) = 9(\frac{9R-5R}{36}) = 9(\frac{4R}{36}) = R$.
Thus,the correct expression is $R = \frac{9}{\lambda_1} - \frac{9}{\lambda_2}$.

(A) The longest wavelength corresponds to the transition between the closest energy levels.
For the Lyman series,the transition is from $n_2 = 2$ to $n_1 = 1$.
Using the Rydberg formula: $\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$.
Thus,$\lambda_L = \frac{4}{3R}$.
For the Balmer series,the transition is from $n_2 = 3$ to $n_1 = 2$.
Using the Rydberg formula: $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36}$.
Thus,$\lambda_B = \frac{36}{5R}$.
The ratio of the longest wavelengths is $\frac{\lambda_L}{\lambda_B} = \frac{4/3R}{36/5R} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27}$.

(C) The energy of the emitted photon during a transition from $n_i = 3$ to $n_f = 2$ is given by the Rydberg formula:
$E = 13.6 \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right] \text{ eV}$
Substituting the values $n_f = 2$ and $n_i = 3$:
$E = 13.6 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] \text{ eV}$
$E = 13.6 \left[ \frac{1}{4} - \frac{1}{9} \right] \text{ eV}$
$E = 13.6 \left[ \frac{9 - 4}{36} \right] \text{ eV}$
$E = 13.6 \times \frac{5}{36} \text{ eV} \approx 1.89 \text{ eV}$
For the photoelectric effect to occur, the energy of the incident photon must be greater than or equal to the work function $(\Phi)$ of the metal.
Therefore, the maximum work function is $\Phi_{\text{max}} = 1.89 \text{ eV}$.

(C) The frequency $\nu$ of emitted radiation for a transition in a hydrogen atom is given by $\nu = cR \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the series limit,the transition occurs from $n_2 = \infty$ to $n_1$.
For the Balmer series,$n_1 = 2$,so $\nu_{B} = cR \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right] = \frac{cR}{4}$.
For the Brackett series,$n_1 = 4$,so $\nu' = cR \left[ \frac{1}{4^2} - \frac{1}{\infty^2} \right] = \frac{cR}{16}$.
Dividing the two expressions: $\frac{\nu'}{\nu_{B}} = \frac{cR/16}{cR/4} = \frac{4}{16} = \frac{1}{4}$.
Therefore,$\nu' = \frac{\nu_{B}}{4}$.

(B) The Rydberg formula for the Balmer series is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \dots$
For the maximum wavelength (minimum energy),we take $n = 3$:
$\frac{1}{\lambda_{max}} = 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) = 10^7 \left( \frac{5}{36} \right) \implies \lambda_{max} = \frac{36}{5} \times 10^{-7} \ m = 7.2 \times 10^{-7} \ m = 7200 \ Å$.
For the minimum wavelength (maximum energy),we take $n = \infty$:
$\frac{1}{\lambda_{min}} = 10^7 \left( \frac{1}{4} - 0 \right) = 10^7 \left( \frac{1}{4} \right) \implies \lambda_{min} = 4 \times 10^{-7} \ m = 4000 \ Å$.
The difference in wavelength is $\Delta \lambda = \lambda_{max} - \lambda_{min} = 7200 \ Å - 4000 \ Å = 3200 \ Å$.

(D) According to Bohr's model,the wavelength $\lambda$ of the emitted radiation when an electron jumps from orbit $n_2$ to $n_1$ is given by the Rydberg formula:
$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$:
$\frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_1 = \frac{4}{3R}$
For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$:
$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36} \implies \lambda_2 = \frac{36}{5R}$
The ratio of the wavelengths is given as $\frac{\lambda_1}{\lambda_2} = 9 \alpha$:
$\frac{\lambda_1}{\lambda_2} = \left( \frac{4}{3R} \right) \times \left( \frac{5R}{36} \right) = \frac{4 \times 5}{3 \times 36} = \frac{20}{108} = \frac{5}{27}$
Given $\frac{\lambda_1}{\lambda_2} = 9 \alpha$,we have:
$9 \alpha = \frac{5}{27} \implies \alpha = \frac{5}{27 \times 9} = \frac{5}{243} \approx 0.02057 \approx 0.021$

(C) The Rydberg formula for the wavelength $\lambda$ of spectral lines in the hydrogen atom is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant.
For the Balmer series,$n_1 = 2$. The shortest wavelength occurs when $n_2 = \infty$.
$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4} \implies \lambda = \frac{4}{R} \quad \dots (i)$
For the Brackett series,$n_1 = 4$. The shortest wavelength occurs when $n_2 = \infty$.
$\frac{1}{\lambda_{\text{Brackett}}} = R \left( \frac{1}{4^2} - \frac{1}{\infty^2} \right) = \frac{R}{16}$
$\lambda_{\text{Brackett}} = \frac{16}{R}$
Substituting $\frac{1}{R} = \frac{\lambda}{4}$ from equation $(i)$:
$\lambda_{\text{Brackett}} = 16 \times \left( \frac{\lambda}{4} \right) = 4 \lambda$.

(A) The wavelength of spectral lines for a hydrogen atom is given by the Rydberg formula:
$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
For the Balmer series,$n_1 = 2$ and $n_2 = 3, 4, 5, \ldots$.
For the minimum wavelength (series limit),$n_2 = \infty$:
$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4}$
$\lambda_{\min} = \frac{4}{R}$
For the maximum wavelength,$n_2 = 3$:
$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = \frac{5R}{36}$
$\lambda_{\max} = \frac{36}{5R}$
Now,the ratio of maximum to minimum wavelength is:
$\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{36 / 5R}{4 / R} = \frac{36}{5R} \times \frac{R}{4} = \frac{9}{5}$

(C) The wavelength of a spectral line in the hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the Balmer series,the final state is $n_f = 2$.
Substituting the values,we have: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n_i^2} \right) = R \left( \frac{1}{4} - \frac{1}{n_i^2} \right)$.
Given that $\lambda = \frac{16}{3 R}$,we substitute this into the equation:
$\frac{1}{(16 / 3 R)} = R \left( \frac{1}{4} - \frac{1}{n_i^2} \right) \Rightarrow \frac{3 R}{16} = R \left( \frac{1}{4} - \frac{1}{n_i^2} \right)$.
Dividing both sides by $R$,we get: $\frac{3}{16} = \frac{1}{4} - \frac{1}{n_i^2}$.
Rearranging the terms to solve for $n_i^2$: $\frac{1}{n_i^2} = \frac{1}{4} - \frac{3}{16} = \frac{4 - 3}{16} = \frac{1}{16}$.
Therefore,$n_i^2 = 16$,which gives $n_i = 4$.

(C) For the $H$-atom, the Rydberg formula for the Lyman series is given by:
$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n_2^2} \right)$
For the shortest wavelength, $n_2 = \infty$:
$\frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R(1 - 0) = R$
Given $\lambda_{\text{min}} = 912 \ \text{Å}$, so $R = \frac{1}{912} \ \text{Å}^{-1}$.
For the longest wavelength, the transition occurs from the nearest energy level, i.e., $n_2 = 2$:
$\frac{1}{\lambda_{\text{max}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$
Substituting the value of $R$:
$\frac{1}{\lambda_{\text{max}}} = \frac{1}{912} \times \frac{3}{4} = \frac{3}{3648} = \frac{1}{1216}$
Therefore, $\lambda_{\text{max}} = 1216 \ \text{Å}$.

(A) The wavelength for the Lyman series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$.
For the first line of the Lyman series,the transition is from $n = 2$ to $n = 1$.
Substituting these values: $\frac{1}{\lambda} = R \left( \frac{1}{1} - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$.
Thus,$R = \frac{4}{3\lambda}$ (Equation $i$).
The wavelength for the Balmer series is given by: $\frac{1}{\lambda^{\prime}} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$.
For the first line of the Balmer series,the transition is from $n = 3$ to $n = 2$.
Substituting these values: $\frac{1}{\lambda^{\prime}} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = R \left( \frac{5}{36} \right)$.
Substituting the value of $R$ from Equation $i$: $\frac{1}{\lambda^{\prime}} = \left( \frac{4}{3\lambda} \right) \left( \frac{5}{36} \right) = \frac{20}{108\lambda} = \frac{5}{27\lambda}$.
Therefore,$\lambda^{\prime} = \frac{27}{5} \lambda$.

(C) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the Paschen series,$n_1 = 3$. The longest wavelength corresponds to the transition from $n_2 = 4$.
$\frac{1}{\lambda_P} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{9} - \frac{1}{16} \right] = R \left[ \frac{16 - 9}{144} \right] = \frac{7R}{144}$.
Thus,$\lambda_P = \frac{144}{7R}$.
For the Lyman series,$n_1 = 1$. The longest wavelength corresponds to the transition from $n_2 = 2$.
$\frac{1}{\lambda_L} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R \left[ 1 - \frac{1}{4} \right] = \frac{3R}{4}$.
Thus,$\lambda_L = \frac{4}{3R}$.
Now,calculating the ratio $\frac{\lambda_P}{\lambda_L}$:
$\frac{\lambda_P}{\lambda_L} = \frac{144 / 7R}{4 / 3R} = \frac{144}{7R} \times \frac{3R}{4} = \frac{36 \times 3}{7} = \frac{108}{7} \approx 15.42$.
Since $15 < 15.42 < 16$,the correct option is $C$.

(C) When atoms are excited from the ground state to an excited state with principal quantum number $n = 20$,the number of possible spectral lines is given by the formula:
$N = \frac{n(n - 1)}{2}$
where $n$ is the principal quantum number.
Substituting the value $n = 20$ into the formula:
$N = \frac{20(20 - 1)}{2}$
$N = \frac{20 \times 19}{2}$
$N = 10 \times 19 = 190$
Therefore,the number of different wavelengths that may be observed in the spectrum is $190$.

(C) From the hydrogen spectrum, when an electron transitions from $n_2$ orbit to $n_1$ orbit, the emitted wavelength $\lambda$ 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 the Lyman series, $n_1 = 1$. The first line corresponds to $n_2 = 2$. Thus, $\frac{1}{\lambda_L} = R Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R Z^2 \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R Z^2$.
For the Balmer series, $n_1 = 2$. The first line corresponds to $n_2 = 3$. Thus, $\frac{1}{\lambda_B} = R Z^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R Z^2 \left( \frac{1}{4} - \frac{1}{9} \right) = R Z^2 \left( \frac{5}{36} \right)$.
Dividing the expression for $\frac{1}{\lambda_L}$ by $\frac{1}{\lambda_B}$, we get: $\frac{\lambda_B}{\lambda_L} = \frac{3/4}{5/36} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5} = 5.4$.
Given $\lambda_L = 1215.4 \text{ Å}$, we have $\lambda_B = 5.4 \times 1215.4 \text{ Å} = 6563.16 \text{ Å} \approx 6563 \text{ Å}$.

(A) For the shortest wavelength in the Balmer series, the transition occurs from $n_{i} = \infty$ to $n_{f} = 2$.
Using the Rydberg formula: $\frac{1}{\lambda} = R_{H} \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right)$.
Substituting the values: $\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$.
$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{1}{4} = 2742500 \ \text{m}^{-1}$.
$\lambda = \frac{1}{2742500} \ \text{m} \approx 3.646 \times 10^{-7} \ \text{m}$.
Converting to $\text{Å}$: $\lambda = 3.646 \times 10^{-7} \times 10^{10} \ \text{Å} = 3646 \ \text{Å}$.

(A) From Rydberg's equation,the wavelength $\lambda$ for an electron transition is given by:
$\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$
Here,the transition is from $n_i = 5$ to $n_f = 1$.
Substituting the values:
$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{5^2} \right)$
$\frac{1}{\lambda} = R \left( 1 - \frac{1}{25} \right)$
$\frac{1}{\lambda} = R \left( \frac{25 - 1}{25} \right)$
$\frac{1}{\lambda} = \frac{24 R}{25}$
Therefore,the wavelength is $\lambda = \frac{25}{24 R}$.

(A) For the Lyman series,the wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$,where $n = 2, 3, 4, \dots$
$1$. The minimum wavelength $(\lambda_{\min})$ occurs for the transition from $n = \infty$ to $n = 1$:
$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$
Given $\lambda_{\min} = P$,we have $P = \frac{1}{R}$.
$2$. The maximum wavelength $(\lambda_{\max})$ occurs for the transition from $n = 2$ to $n = 1$:
$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$
$3$. Substituting $R = \frac{1}{P}$ into the equation:
$\frac{1}{\lambda_{\max}} = \frac{1}{P} \cdot \frac{3}{4}$
$\lambda_{\max} = \frac{4 P}{3}$.

(B) The wavelength $\lambda$ of radiation emitted by a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$.
For each spectral series,the range of wavelengths is determined by the transition from $n_i = n_f + 1$ to $n_i = \infty$.
$1$. Lyman series $(n_f = 1)$: $\lambda$ ranges from $\frac{1}{R}$ to $\frac{4}{3R}$.
$2$. Balmer series $(n_f = 2)$: $\lambda$ ranges from $\frac{4}{R}$ to $\frac{36}{5R}$.
$3$. Paschen series $(n_f = 3)$: $\lambda$ ranges from $\frac{9}{R}$ to $\frac{144}{7R}$.
Comparing the given options,the range $\frac{7}{5R}$ to $\frac{19}{5R}$ does not correspond to any of the spectral series of the hydrogen atom.

(D) The Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the Balmer series,$n_1 = 2$. The first line corresponds to $n_2 = 3$,and the second line corresponds to $n_2 = 4$.
$\frac{1}{\lambda_B} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{4} - \frac{1}{16} \right] = R \left( \frac{3}{16} \right)$.
Given $\lambda_B = 600 \ nm$,so $\frac{1}{600} = R \left( \frac{3}{16} \right) \implies R = \frac{16}{1800} \ nm^{-1}$.
For the Lyman series,$n_1 = 1$. The third line corresponds to $n_2 = 4$ (since $n_2 = 2, 3, 4, \dots$).
$\frac{1}{\lambda_L} = R \left[ \frac{1}{1^2} - \frac{1}{4^2} \right] = R \left[ 1 - \frac{1}{16} \right] = R \left( \frac{15}{16} \right)$.
Substituting the value of $R$:
$\frac{1}{\lambda_L} = \left( \frac{16}{1800} \right) \left( \frac{15}{16} \right) = \frac{15}{1800} = \frac{1}{120}$.
Therefore,$\lambda_L = 120 \ nm$.
Since $120 \ nm$ is not among the options,the correct choice is $D$.

(C) The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series,the smallest wavelength corresponds to $n_1 = 1$ and $n_2 = \infty$.
$\frac{1}{\lambda_{L,min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R = \frac{1}{91} \ nm^{-1}$.
For the Balmer series,the largest wavelength corresponds to $n_1 = 2$ and $n_2 = 3$.
$\frac{1}{\lambda_{B,max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = \frac{1}{91} \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{1}{91} \left( \frac{5}{36} \right)$.
$\lambda_{B,max} = \frac{91 \times 36}{5} = 655.2 \ nm$.
For the Paschen series,the largest wavelength corresponds to $n_1 = 3$ and $n_2 = 4$.
$\frac{1}{\lambda_{P,max}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = \frac{1}{91} \left( \frac{1}{9} - \frac{1}{16} \right) = \frac{1}{91} \left( \frac{7}{144} \right)$.
$\lambda_{P,max} = \frac{91 \times 144}{7} = 1872 \ nm$.
The difference is $\Delta\lambda = \lambda_{P,max} - \lambda_{B,max} = 1872 - 655.2 = 1216.8 \ nm \approx 1217 \ nm$.

(B) The wavelength $\lambda$ for a transition is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the Lyman series $(n_f = 1)$,the maximum wavelength occurs at $n_i = 2$: $\frac{1}{\lambda_{max}} = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_{max} = \frac{4}{3R}$. Thus,statement $A$ is correct.
The Balmer series $(n_f = 2)$ corresponds to the visible region. Thus,statement $B$ is correct.
For the Paschen series $(n_f = 3)$,the minimum wavelength occurs at $n_i = \infty$: $\frac{1}{\lambda_{min}} = R \left( \frac{1}{3^2} - 0 \right) = \frac{R}{9} \implies \lambda_{min} = \frac{9}{R}$. Thus,statement $C$ is correct.
For the Lyman series $(n_f = 1)$,the minimum wavelength occurs at $n_i = \infty$: $\frac{1}{\lambda_{min}} = R \left( 1 - 0 \right) = R \implies \lambda_{min} = \frac{1}{R}$. Thus,statement $D$ is incorrect.
Therefore,statements $A, B,$ and $C$ are correct.

(D) The Balmer series corresponds to $n_f = 2$.
The 1st line corresponds to $n_i = 3$,and the 2nd line corresponds to $n_i = 4$.
Momentum $p = E/c = (h\nu)/c = h/\lambda$.
Since $1/\lambda = R(1/2^2 - 1/n_i^2)$,we have $p \propto (1/4 - 1/n_i^2)$.
For the 1st line,$p_1 \propto (1/4 - 1/9) = 5/36$.
For the 2nd line,$p_2 \propto (1/4 - 1/16) = 3/16$.
The ratio $p_1/p_2 = (5/36) / (3/16) = (5/36) \times (16/3) = (5 \times 4) / (9 \times 3) = 20/27$.
Thus,$\alpha = 20$ and $\beta = 27$.