Spectral Series of Hydrogen Atom Questions in English

(C) The wavelength $\lambda$ for a transition in the hydrogen spectrum 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 last line of the Lyman series,$n_1 = 1$ and $n_2 = \infty$. Thus,$\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$,which gives $\lambda_L = \frac{1}{R}$.
For the last line of the Balmer series,$n_1 = 2$ and $n_2 = \infty$. Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4}$,which gives $\lambda_B = \frac{4}{R}$.
The ratio of the wavelengths is $\frac{\lambda_L}{\lambda_B} = \frac{1/R}{4/R} = \frac{1}{4} = 0.25$.

(D) The energy of a photon emitted during a transition from state $n_i$ to $n_f$ is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
Ultraviolet radiation corresponds to the Lyman series $(n_f = 1)$.
Visible radiation corresponds to the Balmer series $(n_f = 2)$.
Infrared radiation corresponds to the Paschen series $(n_f = 3)$,Brackett series $(n_f = 4)$,or Pfund series $(n_f = 5)$.
Looking at the options:
Option $A$: $n=3$ to $n=1$ is Lyman series (Ultraviolet).
Option $B$: $n=4$ to $n=2$ is Balmer series (Visible).
Option $C$: $n=6$ to $n=2$ is Balmer series (Visible).
Option $D$: $n=5$ to $n=3$ is Paschen series (Infrared).
Therefore,the correct transition for infrared radiation is $n=5$ to $n=3$.

(C) The Rydberg formula for the wavelength of a spectral line 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 Paschen series,$n_1 = 3$. The first line corresponds to $n_2 = 4$. Thus,$\frac{1}{\lambda_1} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16-9}{144} \right) = \frac{7R}{144}$. So,$\lambda_1 = \frac{144}{7R}$.
For the Brackett series,$n_1 = 4$. The first line corresponds to $n_2 = 5$. Thus,$\frac{1}{\lambda_2} = R \left( \frac{1}{4^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{16} - \frac{1}{25} \right) = R \left( \frac{25-16}{400} \right) = \frac{9R}{400}$. So,$\lambda_2 = \frac{400}{9R}$.
The ratio $\frac{\lambda_1}{\lambda_2} = \frac{144}{7R} \times \frac{9R}{400} = \frac{144 \times 9}{7 \times 400} = \frac{1296}{2800} = \frac{81}{175}$.

(C) The frequency of a spectral line in the hydrogen atom is given by $v = R c Z^2 (1/n_f^2 - 1/n_i^2)$.
For the Balmer series limit,$n_f = 2$ and $n_i = \infty$,so $v_1 = R c (1/2^2 - 1/\infty^2) = R c / 4$.
For the Paschen series limit,$n_f = 3$ and $n_i = \infty$,so $v_3 = R c (1/3^2 - 1/\infty^2) = R c / 9$.
For the first line of the Balmer series,$n_f = 2$ and $n_i = 3$,so $v_2 = R c (1/2^2 - 1/3^2) = R c (1/4 - 1/9) = R c (5/36)$.
Comparing the values: $v_1 - v_3 = R c (1/4 - 1/9) = R c (5/36) = v_2$.
Thus,the correct relation is $v_1 - v_3 = v_2$.

(D) Using Rydberg's 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 $n_2 = 2$:
$\frac{1}{\lambda_1} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R \left[ 1 - \frac{1}{4} \right] = \frac{3R}{4}$.
For the first line of the Paschen series,$n_1 = 3$ and $n_2 = 4$:
$\frac{1}{\lambda_2} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{9} - \frac{1}{16} \right] = R \left[ \frac{16 - 9}{144} \right] = \frac{7R}{144}$.
Now,calculating the ratio $\frac{\lambda_2}{\lambda_1}$:
$\frac{\lambda_2}{\lambda_1} = \frac{1}{\lambda_1} \times \lambda_2 = \left( \frac{3R}{4} \right) \times \left( \frac{144}{7R} \right) = \frac{3 \times 36}{7} = \frac{108}{7}$.
Thus,$\lambda_2 : \lambda_1 = 108 : 7$.

(C) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$.
For transition $1$ ($n_i = 3$ to $n_f = 1$): $\frac{1}{\lambda_1} = R \left[ \frac{1}{1^2} - \frac{1}{3^2} \right] = R \left[ 1 - \frac{1}{9} \right] = \frac{8}{9} R$.
For transition $2$ ($n_i = 2$ to $n_f = 1$): $\frac{1}{\lambda_2} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R \left[ 1 - \frac{1}{4} \right] = \frac{3}{4} R$.
Taking the ratio of the two wavelengths:
$\frac{\lambda_1}{\lambda_2} = \frac{1/\lambda_2}{1/\lambda_1} = \frac{\frac{3}{4} R}{\frac{8}{9} R} = \frac{3}{4} \times \frac{9}{8} = \frac{27}{32}$.
Given that $\frac{\lambda_1}{\lambda_2} = \frac{x}{32}$,we have $\frac{x}{32} = \frac{27}{32}$,which implies $x = 27$.

(C) The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n^2} - \frac{1}{m^2} \right)$.
For the minimum wavelength (shortest wavelength),the transition occurs from $m = \infty$ to the ground state $n$ of the series.
For the Lyman series,$n = 1$ and $m = \infty$:
$\frac{1}{\lambda_{L}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \implies \lambda_{L} = \frac{1}{R}$.
For the Balmer series,$n = 2$ and $m = \infty$:
$\frac{1}{\lambda_{B}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4} \implies \lambda_{B} = \frac{4}{R}$.
The ratio of the minimum wavelengths is $\frac{\lambda_{L}}{\lambda_{B}} = \frac{1/R}{4/R} = \frac{1}{4} = 0.25$.

(B) The Rydberg formula for the wavelength $\lambda$ is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series,the transition occurs to the ground state,so $n_1 = 1$.
For the maximum wavelength $(\lambda_{\max})$,the transition is from the nearest energy level,$n_2 = 2$:
$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R \implies \lambda_{\max} = \frac{4}{3R}$.
For the minimum wavelength $(\lambda_{\min})$,the transition is from the highest energy level,$n_2 = \infty$:
$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R(1 - 0) = R \implies \lambda_{\min} = \frac{1}{R}$.
Therefore,the ratio of the maximum to the minimum wavelength is:
$\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{4/3R}{1/R} = \frac{4}{3}$.

(C) The shortest wavelength in the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{\infty} \right) = \frac{R}{4}$.
Thus,$\lambda_B = \frac{4}{R}$.
The shortest wavelength in the Paschen series is given by: $\frac{1}{\lambda_P} = R \left( \frac{1}{3^2} - \frac{1}{\infty} \right) = \frac{R}{9}$.
Thus,$\lambda_P = \frac{9}{R}$.
The ratio of the shortest wavelength in the Balmer series to that in the Paschen series is:
$\frac{\lambda_B}{\lambda_P} = \frac{4/R}{9/R} = \frac{4}{9}$.

(B) In the hydrogen atom spectrum,the Lyman series falls in the ultraviolet region.
The Balmer series falls in the visible region.
The Paschen,Brackett,and Pfund series fall in the infrared region.
Therefore,the correct answer is the Balmer series.

(A) The wavelength $\lambda$ of the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = 2$ for the Balmer series.
For the series limit,the transition occurs from $n_2 = \infty$.
Substituting these values,we get: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4}$.
Therefore,the wavelength is $\lambda = \frac{4}{R}$.
The frequency $v$ is related to the velocity of light $C$ and wavelength $\lambda$ by the formula $v = \frac{C}{\lambda}$.
Substituting the value of $\lambda$,we get: $v = \frac{C}{4/R} = \frac{RC}{4}$.

(B) The Lyman series corresponds to transitions of electrons from higher energy levels to the ground state $(n_1 = 1)$.
The least energetic photon in the Lyman series corresponds to the transition from the first excited state $(n_2 = 2)$ to the ground state $(n_1 = 1)$.
The energy difference for this transition is given as $\Delta E = 10.2 \text{ eV}$.
The relationship between energy and wavelength is given by $\lambda = \frac{hc}{\Delta E}$.
Substituting the given values: $\lambda = \frac{1240 \text{ eV-nm}}{10.2 \text{ eV}}$.
$\lambda \approx 121.57 \text{ nm}$.
Rounding to the nearest whole number, we get $\lambda \approx 122 \text{ nm}$.

(A) For the Paschen series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = 3$ and $n_2 = 4, 5, 6, \dots$
The longest wavelength $(\lambda_{\max})$ occurs for the transition from $n_2 = 4$ to $n_1 = 3$:
$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16-9}{144} \right) = \frac{7R}{144} \implies \lambda_{\max} = \frac{144}{7R}$
The shortest wavelength $(\lambda_{\min})$ occurs for the transition from $n_2 = \infty$ to $n_1 = 3$:
$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{9} - 0 \right) = \frac{R}{9} \implies \lambda_{\min} = \frac{9}{R}$
The ratio of the longest to the shortest wavelength is:
$\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{144/7R}{9/R} = \frac{144}{7 \times 9} = \frac{144}{63}$

(A) Using Rydberg's formula,$\frac{1}{\lambda} = R_H Z^2 \left[ \frac{1}{n^2} - \frac{1}{m^2} \right]$,where $R_H$ is the Rydberg constant.
For the shortest wavelength in the Balmer series of a hydrogen atom $(Z=1)$: $n=2, m=\infty$. Thus,$\frac{1}{\lambda_1} = R_H (1)^2 \left[ \frac{1}{2^2} - 0 \right] = \frac{R_H}{4}$,which gives $\lambda_1 = \frac{4}{R_H}$.
For the shortest wavelength in the Brackett series of a hydrogen-like atom: $n=4, m=\infty$. Thus,$\frac{1}{\lambda_2} = R_H Z^2 \left[ \frac{1}{4^2} - 0 \right] = \frac{R_H Z^2}{16}$,which gives $\lambda_2 = \frac{16}{R_H Z^2}$.
Given $\lambda_1 = \lambda_2$,we have $\frac{4}{R_H} = \frac{16}{R_H Z^2}$.
Simplifying,$Z^2 = \frac{16}{4} = 4$,so $Z = 2$.

(D) According to Rydberg's formula,$\frac{1}{\lambda} = R \left( \frac{1}{n^2} - \frac{1}{m^2} \right)$.
For the series limit of the Lyman series,$n=1, m=\infty$,so $\frac{1}{\lambda_1} = R(1 - 0) = R$.
For the first line of the Lyman series,$n=1, m=2$,so $\frac{1}{\lambda_2} = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$.
For the series limit of the Balmer series,$n=2, m=\infty$,so $\frac{1}{\lambda_3} = R \left( \frac{1}{2^2} - 0 \right) = \frac{R}{4}$.
Comparing these values,we observe that $\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = R - \frac{3R}{4} = \frac{R}{4}$.
Since $\frac{R}{4} = \frac{1}{\lambda_3}$,we get the relation $\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}$.

(C) The Rydberg formula for spectral series is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,$n_1 = 2$. The $2^{\text{nd}}$ line corresponds to the transition from $n_2 = 4$ to $n_1 = 2$.
$\frac{1}{\lambda_1} = R Z^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R Z^2 \left( \frac{1}{4} - \frac{1}{16} \right) = R Z^2 \left( \frac{3}{16} \right)$.
Thus,$\lambda_1 = \frac{16}{3 R Z^2}$.
For the Paschen series,$n_1 = 3$. The $1^{\text{st}}$ line corresponds to the transition from $n_2 = 4$ to $n_1 = 3$.
$\frac{1}{\lambda_2} = R Z^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R Z^2 \left( \frac{1}{9} - \frac{1}{16} \right) = R Z^2 \left( \frac{7}{144} \right)$.
Thus,$\lambda_2 = \frac{144}{7 R Z^2}$.
Taking the ratio $\frac{\lambda_1}{\lambda_2} = \left( \frac{16}{3 R Z^2} \right) \times \left( \frac{7 R Z^2}{144} \right) = \frac{16 \times 7}{3 \times 144} = \frac{112}{432} = \frac{7}{27}$.

(D) The wavelength of the Balmer series is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = 2$ for the Balmer series.
For the series limit,the transition occurs from $n_2 = \infty$.
Substituting these values,we get: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4}$.
The frequency $v$ is related to the wavelength $\lambda$ and the speed of light $c$ by the equation $v = \frac{c}{\lambda}$.
Substituting $\frac{1}{\lambda} = \frac{R}{4}$ into the frequency formula,we get: $v = c \times \frac{R}{4} = \frac{Rc}{4}$.

(C) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the Lyman series,$n_1 = 1$. The longest wavelength corresponds to the smallest energy transition,which occurs at $n_2 = 2$.
$\frac{1}{\lambda_{\max(L)}} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R \left[ 1 - \frac{1}{4} \right] = \frac{3R}{4} \implies \lambda_{\max(L)} = \frac{4}{3R}$.
For the Balmer series,$n_1 = 2$. The longest wavelength corresponds to the smallest energy transition,which occurs at $n_2 = 3$.
$\frac{1}{\lambda_{\max(B)}} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{9-4}{36} \right] = \frac{5R}{36} \implies \lambda_{\max(B)} = \frac{36}{5R}$.
The ratio of the longest wavelengths is $\frac{\lambda_{\max(L)}}{\lambda_{\max(B)}} = \frac{4}{3R} \times \frac{5R}{36} = \frac{4 \times 5}{3 \times 36} = \frac{20}{108} = \frac{5}{27}$.

(B) The wavelength $\lambda$ of emitted radiation is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Ultraviolet radiation corresponds to the Lyman series $(n_1 = 1)$.
Infrared radiation corresponds to the Paschen series $(n_1 = 3)$,Brackett series $(n_1 = 4)$,or Pfund series $(n_1 = 5)$.
Comparing the options:
$(A)$ $n=6$ to $n=2$ is the Balmer series (Visible).
$(B)$ $n=5$ to $n=3$ is the Paschen series (Infrared).
$(C)$ $n=3$ to $n=5$ is an absorption process,not emission.
$(D)$ $n=4$ to $n=2$ is the Balmer series (Visible).
Therefore,the transition $n=5$ to $n=3$ results in infrared radiation.

(B) The wave number $\bar{\nu}$ for a hydrogen atom is given by the Rydberg formula: $\bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series,the transition occurs to the ground state,so $n_1 = 1$.
The first line of the Lyman series corresponds to the transition from the first excited state to the ground state,so $n_2 = 2$.
Substituting these values into the formula:
$\bar{\nu} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$
$\bar{\nu} = R \left( 1 - \frac{1}{4} \right)$
$\bar{\nu} = R \left( \frac{3}{4} \right) = \frac{3 R}{4}$.

(B) The wavelength of visible radiation in the Balmer series is given by the Rydberg formula:
$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{m^2} \right)$ for $m = 3, 4, 5, ...$
$\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{m^2} \right)$
$\frac{1}{\lambda} = R \left( \frac{m^2 - 4}{4 m^2} \right)$
Taking the reciprocal to find $\lambda$:
$\lambda = \frac{4 m^2}{R(m^2 - 4)}$
Comparing this with the given equation $\lambda = \frac{k m^2}{m^2 - 4}$,we get:
$k = \frac{4}{R}$

(C) The wavelength of a spectral line 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 $n = 3, 4, 5, \dots$ and $R$ is the Rydberg constant.
For the Balmer series,the transition to the $n=2$ orbit determines the line:
- For $n=3$ to $n=2$,we get the first line of the Balmer series.
- For $n=4$ to $n=2$,we get the second line of the Balmer series.
Therefore,the transition from the fourth orbit to the second orbit corresponds to the second line of the Balmer series.

(A) The wave number $\bar{\nu}$ is given by the Rydberg formula: $\bar{\nu} = \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,the transition occurs to the $n_1 = 2$ energy level.
The last line of a spectral series corresponds to the transition from an infinite energy level,so $n_2 = \infty$.
Substituting the values: $\bar{\nu} = 10^7 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$.
Since $\frac{1}{\infty} = 0$,we get: $\bar{\nu} = 10^7 \left( \frac{1}{4} - 0 \right) = 0.25 \times 10^7 \, m^{-1}$.
This can be written as: $\bar{\nu} = 2.5 \times 10^6 \, m^{-1}$.

(D) The electron is initially in the third excited state,which corresponds to the principal quantum number $n = 4$ (since the ground state is $n = 1$,first excited state is $n = 2$,second excited state is $n = 3$,and third excited state is $n = 4$).
When the electron transitions from an excited state $n$ to the ground state,the maximum number of spectral lines emitted is given by the formula:
$N = \frac{n(n - 1)}{2}$
Substituting $n = 4$ into the formula:
$N = \frac{4(4 - 1)}{2} = \frac{4 \times 3}{2} = 6$
Therefore,the maximum number of spectral lines emitted is $6$.

(A) The series limit of the Lyman series corresponds to the transition from $n = \infty$ to $n = 1$. Using the Rydberg formula $\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2})$:
$\frac{1}{\lambda_1} = R(\frac{1}{1^2} - \frac{1}{\infty^2}) = R$
The series limit of the Balmer series corresponds to the transition from $n = \infty$ to $n = 2$:
$\frac{1}{\lambda_3} = R(\frac{1}{2^2} - \frac{1}{\infty^2}) = \frac{R}{4}$
The first line of the Lyman series corresponds to the transition from $n = 2$ to $n = 1$:
$\frac{1}{\lambda_2} = R(\frac{1}{1^2} - \frac{1}{2^2}) = R(1 - \frac{1}{4}) = R - \frac{R}{4}$
Substituting the values of $R$ and $\frac{R}{4}$ from the previous equations:
$\frac{1}{\lambda_2} = \frac{1}{\lambda_1} - \frac{1}{\lambda_3}$
Rearranging the terms,we get:
$\frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{\lambda_3}$

(D) The wave number $\bar{\nu}$ for a spectral line is given by the Rydberg formula: $\bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,$n_1 = 2$.
The last line of the Balmer series corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$.
Substituting these values: $\bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4}$.
Given $R = 10^7 \ m^{-1}$,we have $\bar{\nu} = \frac{10^7}{4} = 0.25 \times 10^7 \ m^{-1} = 25 \times 10^5 \ m^{-1}$.

(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 Balmer series,$n_1 = 2$ and $n_2 = 3, 4, 5, \dots$.
Maximum wavelength $(\lambda_{max})$ occurs for the transition $n_2 = 3$ to $n_1 = 2$:
$\frac{1}{\lambda_{max}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \implies \lambda_{max} = \frac{36}{5R}$.
Minimum wavelength $(\lambda_{min})$ occurs for the transition $n_2 = \infty$ to $n_1 = 2$:
$\frac{1}{\lambda_{min}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4} \implies \lambda_{min} = \frac{4}{R}$.
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 shortest wavelength in the Lyman series occurs for the transition from $n = \infty$ to $n = 1$:
$\frac{1}{\lambda_{L}} = R \left[ \frac{1}{1^2} - \frac{1}{\infty^2} \right] = R$
$\therefore \lambda_{L} = \frac{1}{R} = 912 \ \text{Å}$
The longest wavelength in the Paschen series occurs for the transition from $n = 4$ to $n = 3$:
$\frac{1}{\lambda_{P}} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} \right] = R \left[ \frac{1}{9} - \frac{1}{16} \right] = R \left[ \frac{16 - 9}{144} \right] = \frac{7R}{144}$
$\therefore \lambda_{P} = \frac{144}{7R}$
Substituting $\frac{1}{R} = 912 \ \text{Å}$:
$\lambda_{P} = \frac{144}{7} \times 912 \ \text{Å} \approx 18760 \ \text{Å}$

(B) The Lyman series of the hydrogen atom corresponds to transitions ending at the ground state $(n_f = 1)$,which results in photon emissions in the ultraviolet region.
The Balmer series lies in the visible region.
The Paschen,Brackett,and Pfund series lie in the infrared region.

(B) 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 series limit occurs at $n_{2} = \infty$. Thus,$\frac{1}{\lambda_{1}} = R \left( \frac{1}{2^{2}} - \frac{1}{\infty^{2}} \right) = \frac{R}{4}$,which implies $\lambda_{1} = \frac{4}{R}$.
For the Brackett series,$n_{1} = 4$. The longest wavelength occurs at $n_{2} = 5$. Thus,$\frac{1}{\lambda_{2}} = R \left( \frac{1}{4^{2}} - \frac{1}{5^{2}} \right) = R \left( \frac{1}{16} - \frac{1}{25} \right) = R \left( \frac{25 - 16}{400} \right) = \frac{9R}{400}$.
This implies $\lambda_{2} = \frac{400}{9R}$.
Now,taking the ratio: $\frac{\lambda_{2}}{\lambda_{1}} = \frac{400}{9R} \times \frac{R}{4} = \frac{100}{9} = 11.11$ (This seems to be a calculation check,let's re-evaluate).
Wait,$\frac{\lambda_{2}}{\lambda_{1}} = \frac{400}{9R} / \frac{4}{R} = \frac{400}{36} = 11.11$. Let's re-check the ratio $\frac{\lambda_{1}}{\lambda_{2}} = \frac{4}{R} \times \frac{9R}{400} = \frac{36}{400} = 0.09$.
Therefore,$\lambda_{1} = 0.09 \lambda_{2}$.

(B) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Comparing this with the given equation $\frac{1}{\lambda} = R \left( \frac{1}{4^2} - \frac{1}{n^2} \right)$,we find that $n_1 = 4$.
The spectral series corresponding to $n_1 = 4$ is the Brackett series.
The Brackett series transitions occur between higher energy levels $(n_2 = 5, 6, 7, \ldots)$ and the $n_1 = 4$ level.
This series lies in the infrared region of the electromagnetic spectrum.

(B) In the hydrogen emission spectrum,the wavelength $\lambda$ is given by the Rydberg formula:
$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
For the maximum wavelength of a series,the energy difference between the levels must be minimum. This occurs for the transition from $n_2 = n+1$ to $n_1 = n$.
Substituting these values into the formula:
$\frac{1}{\lambda_{\max}} = R \left[ \frac{1}{n^2} - \frac{1}{(n+1)^2} \right]$
$= R \left[ \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right]$
$= R \left[ \frac{n^2 + 2n + 1 - n^2}{n^2(n+1)^2} \right]$
$= \frac{R(2n+1)}{n^2(n+1)^2}$
Therefore,the maximum wavelength is:
$\lambda_{\max} = \frac{n^2(n+1)^2}{R(2n+1)}$

(A) 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 first line of the Balmer series,$n = 3$:
$\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36} \implies R = \frac{36}{5\lambda}$.
For the second line of the Balmer series,$n = 4$:
$\frac{1}{\lambda'} = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4-1}{16} \right) = \frac{3R}{16}$.
Substituting the value of $R$ from the first equation into the second:
$\frac{1}{\lambda'} = \frac{3}{16} \times \left( \frac{36}{5\lambda} \right) = \frac{3 \times 9}{4 \times 5 \lambda} = \frac{27}{20\lambda}$.
Therefore,$\lambda' = \frac{20}{27} \lambda$.

(A) The frequency of a spectral line is given by $f = Rc \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the series limit of the Balmer series $(n_1=2, n_2=\infty)$: $f_1 = Rc \left( \frac{1}{2^2} - 0 \right) = \frac{Rc}{4}$.
For the series limit of the Paschen series $(n_1=3, n_2=\infty)$: $f_3 = Rc \left( \frac{1}{3^2} - 0 \right) = \frac{Rc}{9}$.
For the first line of the Balmer series $(n_1=2, n_2=3)$: $f_2 = Rc \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = Rc \left( \frac{1}{4} - \frac{1}{9} \right)$.
Substituting the expressions for $f_1$ and $f_3$ into the equation for $f_2$:
$f_2 = \frac{Rc}{4} - \frac{Rc}{9} = f_1 - f_3$.
Rearranging the terms,we get $f_1 - f_2 = f_3$.

(B) The wavelength $\lambda$ emitted during an electronic transition 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 first transition: electron jumps from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$.
$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$. Thus,$\lambda = \frac{36}{5R}$.
For the second transition: electron jumps from the third excited state $(n_i = 4)$ to the second orbit $(n_f = 2)$.
$\frac{1}{\lambda'} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right)$. Thus,$\lambda' = \frac{16}{3R}$.
We are given $\lambda' = \frac{20 \lambda}{x}$. Substituting the values:
$\frac{16}{3R} = \frac{20}{x} \cdot \frac{36}{5R} \implies \frac{16}{3} = \frac{20 \cdot 36}{5x} \implies \frac{16}{3} = \frac{4 \cdot 36}{x} \implies \frac{16}{3} = \frac{144}{x}$.
$x = \frac{144 \cdot 3}{16} = 9 \cdot 3 = 27$.

(B) Concept: The wavelength $\lambda$ of electromagnetic radiation emitted is given by the Rydberg formula:
$\frac{1}{\lambda} = R_{H} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \quad \dots(1)$
The frequency $f$ of the emitted radiation is related to wavelength by:
$f = \frac{c}{\lambda} \quad \dots(2)$
Substituting equation $(1)$ into equation $(2)$,we get:
$f = c R_{H} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Given values: $c = 3 \times 10^8 \ m/s$,$R_{H} = 10^7 \ m^{-1}$,$n_1 = 2$,and $n_2 = 4$.
Substituting these values:
$f = (3 \times 10^8) \times 10^7 \times \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$
$f = 3 \times 10^{15} \times \left( \frac{1}{4} - \frac{1}{16} \right)$
$f = 3 \times 10^{15} \times \left( \frac{4-1}{16} \right)$
$f = 3 \times 10^{15} \times \frac{3}{16}$
$f = \frac{9}{16} \times 10^{15} \ Hz$

(A) The wavelength of a line in the Balmer series is given by $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$ for $n = 3, 4, 5, \dots$
For the first line,$n = 3$:
$\frac{1}{\lambda_1} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36} \Rightarrow \lambda_1 = \frac{36}{5R}$
The wavelength of a line in the Brackett series is given by $\frac{1}{\lambda} = R \left( \frac{1}{4^2} - \frac{1}{n^2} \right)$ for $n = 5, 6, 7, \dots$
For the first line,$n = 5$:
$\frac{1}{\lambda_2} = R \left( \frac{1}{16} - \frac{1}{25} \right) = R \left( \frac{25-16}{400} \right) = \frac{9R}{400} \Rightarrow \lambda_2 = \frac{400}{9R}$
Dividing $\lambda_1$ by $\lambda_2$:
$\frac{\lambda_1}{\lambda_2} = \frac{36}{5R} \times \frac{9R}{400} = \frac{36 \times 9}{5 \times 400} = \frac{324}{2000} = 0.162$

(A) The wavelength of a spectral line 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 $n = 3, 4, 5, \dots$ and $R$ is the Rydberg constant.
For the Balmer series,the transition to the $n = 2$ orbit defines the series.
The first line of the Balmer series corresponds to the transition from $n = 3$ to $n = 2$.
The second line of the Balmer series corresponds to the transition from $n = 4$ to $n = 2$.
Therefore,the transition from the fourth Bohr orbit to the second Bohr orbit represents the second line of the Balmer series.

(D) According to Rydberg's formula,$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the first case,the electron jumps from the $2^{\text{nd}}$ excited state $(n_2 = 3)$ to the $1^{\text{st}}$ excited state $(n_1 = 2)$:
$\frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$.
For the second case,the electron jumps from the $3^{\text{rd}}$ excited state $(n_2 = 4)$ to the $2^{\text{nd}}$ orbit $(n_1 = 2)$:
$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right)$.
Dividing the two equations:
$\frac{\lambda_0}{\lambda} = \frac{R(5/36)}{R(3/16)} = \frac{5}{36} \times \frac{16}{3} = \frac{5 \times 4}{9 \times 3} = \frac{20}{27}$.
Therefore,$\lambda = \frac{27}{20} \lambda_0$.
Wait,re-evaluating the ratio: $\frac{1}{\lambda} = \frac{3}{16}R$ and $\frac{1}{\lambda_0} = \frac{5}{36}R$.
$\frac{\lambda}{\lambda_0} = \frac{5/36}{3/16} = \frac{5}{36} \times \frac{16}{3} = \frac{20}{27}$.
Thus,$\lambda = \frac{20}{27} \lambda_0$.
Comparing with $\frac{20}{x} \lambda_0$,we get $x = 27$.

(D) For the Paschen series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{3^2} - \frac{1}{n^2} \right]$,where $n = 4, 5, 6, \dots$
To find the shortest wavelength,we consider the limit where $n = \infty$.
Substituting $n = \infty$ into the formula:
$\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{9} - 0 \right] = \frac{R}{9}$.
Thus,$\lambda_{\min} = \frac{9}{R}$.
Given the Rydberg constant $R \approx 1.097 \times 10^7 \ m^{-1}$:
$\lambda_{\min} = \frac{9}{1.097 \times 10^7} \approx 8.204 \times 10^{-7} \ m$.
Converting to nanometers:
$\lambda_{\min} \approx 820.4 \ nm \approx 820 \ nm$.

(B) The number of spectral lines emitted when an electron transitions from an excited state $n_2$ to a lower state $n_1$ is given by the formula:
$N = \frac{(n_2 - n_1 + 1)(n_2 - n_1)}{2}$
Here,the ground state is $n_1 = 1$ and the excited state is $n_2 = 4$.
Substituting these values into the formula:
$N = \frac{(4 - 1 + 1)(4 - 1)}{2}$
$N = \frac{(4)(3)}{2}$
$N = \frac{12}{2} = 6$
Thus,the total number of spectral lines observed is $6$.

(D) In a hydrogen atom, an electron can exist in any of the infinite number of energy levels $(n = 1, 2, 3, \dots, \infty)$.
When an electron transitions from a higher energy level to a lower energy level, it emits a photon corresponding to a spectral line.
Since there are an infinite number of possible energy levels, there are an infinite number of possible transitions between these levels.
Therefore, the total number of spectral lines in a hydrogen atom is infinite $(\infty)$.

(D) The energy difference between two orbits in a hydrogen atom is given by the Rydberg formula for wavenumber: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Since the frequency $f$ is related to wavelength $\lambda$ by $f = \frac{C}{\lambda}$,we can write $\frac{f}{C} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Rearranging for frequency,we get $f = RC \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Given $n_1 = 2$ and $n_2 = 3$,we substitute these values into the equation:
$f = RC \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = RC \left( \frac{1}{4} - \frac{1}{9} \right)$.
Calculating the fraction: $\frac{1}{4} - \frac{1}{9} = \frac{9-4}{36} = \frac{5}{36}$.
Thus,the frequency is $f = \frac{5RC}{36}$.

(A) The frequency of a spectral line is given by $v = RC \left[ \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right]$.
For the series limit of the Lyman series $(n_{1}=1, n_{2}=\infty)$: $v_{1} = RC \left[ 1 - \frac{1}{\infty} \right] = RC$.
For the first line of the Lyman series $(n_{1}=1, n_{2}=2)$: $v_{2} = RC \left[ 1 - \frac{1}{4} \right] = \frac{3}{4} RC$.
For the series limit of the Balmer series $(n_{1}=2, n_{2}=\infty)$: $v_{3} = RC \left[ \frac{1}{4} - \frac{1}{\infty} \right] = \frac{RC}{4}$.
Comparing these values,we see that $v_{1} - v_{2} = RC - \frac{3}{4} RC = \frac{1}{4} RC = v_{3}$.
Therefore,$v_{1} - v_{2} = v_{3}$.

(B) The spectral lines of the hydrogen atom are categorized based on the region of the electromagnetic spectrum in which they fall:
$1$. Lyman series: Ultraviolet region
$2$. Balmer series: Visible region
$3$. Paschen series: Infrared region
$4$. Brackett series: Infrared region
$5$. Pfund series: Infrared region
From this classification,it is clear that the Balmer series lies in the visible region of the electromagnetic spectrum.

(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)$.
Here,the final state is the ground state,so $n_{f} = 1$,and the initial excited state is $n_{i} = n$.
Substituting these values,we get $\frac{1}{\lambda} = R \left( \frac{1}{1^{2}} - \frac{1}{n^{2}} \right) = R \left( 1 - \frac{1}{n^{2}} \right)$.
Rearranging the equation: $\frac{1}{\lambda R} = 1 - \frac{1}{n^{2}}$.
This implies $\frac{1}{n^{2}} = 1 - \frac{1}{\lambda R} = \frac{\lambda R - 1}{\lambda R}$.
Taking the reciprocal and square root,we find $n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$.

(C) The energy of a photon emitted during a transition from $n_2$ to $n_1$ is given by $E = hc / \lambda$. The energy is least when the transition is between the closest energy levels.
For the Lyman series, the transitions occur to $n_1 = 1$. The least energetic photon corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.
The energy difference is $\Delta E = 13.6 \text{ eV} \times (1/n_1^2 - 1/n_2^2) = 13.6 \times (1/1^2 - 1/2^2) = 13.6 \times (3/4) = 10.2 \text{ eV}$.
Using the relation $\lambda = hc / \Delta E$:
$\lambda = 1240 \text{ eV nm} / 10.2 \text{ eV} \approx 121.57 \text{ nm} \approx 122 \text{ nm}$.

(D) The hydrogen spectrum consists of five main spectral series: Lyman,Balmer,Paschen,Brackett,and Pfund.
$1$. The Lyman series lies in the ultraviolet region.
$2$. The Balmer series lies in the visible region.
$3$. The Paschen,Brackett,and Pfund series lie in the infrared region.
Therefore,the Balmer series is the correct answer.