Spectral Series of Hydrogen Atom Questions in English

(C) For the Brackett series,the transition occurs from $n_2$ to $n_1 = 4$.
For the maximum wavelength (minimum energy),the transition must be from the nearest energy level,i.e.,$n_2 = 5$.
Using the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Substituting the values: $\frac{1}{\lambda_{\max}} = R \left( \frac{1}{4^2} - \frac{1}{5^2} \right) = R \left( \frac{1}{16} - \frac{1}{25} \right) = R \left( \frac{25 - 16}{400} \right) = R \left( \frac{9}{400} \right)$.
Therefore,$\lambda_{\max} = \frac{400}{9R}$.
Given $R \approx 1.097 \times 10^7 \ m^{-1}$,$\lambda_{\max} = \frac{400}{9 \times 1.097 \times 10^7} \approx 4.05 \times 10^{-6} \ m = 40500 \ \mathring A$.
Rounding to the nearest provided option,the correct value is $40400 \ \mathring A$.

(C) For the shortest wavelength in the Brackett series,the electron transitions from $n_2 = \infty$ to $n_1 = 4$.
Using the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
Substituting $n_1 = 4$ and $n_2 = \infty$:
$\frac{1}{\lambda} = R \left[ \frac{1}{4^2} - \frac{1}{\infty^2} \right] = R \left[ \frac{1}{16} - 0 \right] = \frac{R}{16}$.
Therefore,$\lambda = \frac{16}{R}$.
Given Rydberg constant $R \approx 1.097 \times 10^7 \, \text{m}^{-1}$.
$\lambda = \frac{16}{1.097 \times 10^7} \approx 14.58 \times 10^{-7} \, \text{m} = 1.458 \times 10^{-6} \, \text{m} \approx 1.46 \, \mu\text{m}$.

(D) For the Lyman series,the maximum wavelength occurs for the transition from $n = 2$ to $n = 1$.
Using the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$.
Thus,$(\lambda_L)_{\max} = \frac{4}{3R}$.
For the Balmer series,the maximum wavelength occurs for the transition from $n = 3$ to $n = 2$.
Using the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36}$.
Thus,$(\lambda_B)_{\max} = \frac{36}{5R}$.
The ratio is $\frac{(\lambda_L)_{\max}}{(\lambda_B)_{\max}} = \frac{4}{3R} \times \frac{5R}{36} = \frac{4 \times 5}{3 \times 36} = \frac{20}{108} = \frac{5}{27}$.

(C) The Rydberg formula for the wavelength of emitted or absorbed radiation is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
Given $\lambda = 975 \, \mathring{A} = 975 \times 10^{-10} \, m$ and $R \approx 1.097 \times 10^7 \, m^{-1}$.
For the ground state,$n_1 = 1$. Substituting the values:
$\frac{1}{975 \times 10^{-10}} = 1.097 \times 10^7 \left[ 1 - \frac{1}{n_2^2} \right]$.
$0.911 \times 10^7 = 1.097 \times 10^7 \left[ 1 - \frac{1}{n_2^2} \right]$.
$0.83 = 1 - \frac{1}{n_2^2} \implies \frac{1}{n_2^2} = 0.17 \implies n_2^2 \approx 5.88 \approx 16 \implies n_2 = 4$.
The number of spectral lines $N$ emitted when an electron transitions from state $n$ to the ground state is given by $N = \frac{n(n-1)}{2}$.
For $n = 4$,$N = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6$.
The possible transitions are $4 \to 3, 4 \to 2, 4 \to 1, 3 \to 2, 3 \to 1, 2 \to 1$.

(B) The Rydberg formula for the wavelength of emitted light is given by: $\frac{1}{\lambda} = R\left[ {\frac{1}{n_1^2} - \frac{1}{n_2^2}} \right]$.
For the transition from $n = 3$ to $n = 2$:
$\frac{1}{{\lambda _0}} = R\left[ {\frac{1}{2^2} - \frac{1}{3^2}} \right] = R\left[ {\frac{1}{4} - \frac{1}{9}} \right] = \frac{5}{36}R$ ---$(i)$
For the transition from $n = 4$ to $n = 2$:
$\frac{1}{{\lambda '}} = R\left[ {\frac{1}{2^2} - \frac{1}{4^2}} \right] = R\left[ {\frac{1}{4} - \frac{1}{16}} \right] = \frac{3}{16}R$ ---(ii)
Dividing equation $(i)$ by equation (ii):
$\frac{\lambda '}{{\lambda _0}} = \frac{5R}{36} \times \frac{16}{3R} = \frac{5 \times 16}{36 \times 3} = \frac{80}{108} = \frac{20}{27}$.
Therefore,$\lambda ' = \frac{20}{27}{\lambda _0}$.

(C) When an electron is excited from the ground state $(n_1 = 1)$ to an excited state $(n_2 = 3)$,it will return to the ground state by emitting photons.
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:
$N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$
Here,$n_2 = 3$ and $n_1 = 1$.
Substituting the values:
$N = \frac{(3 - 1)(3 - 1 + 1)}{2}$
$N = \frac{2 \times 3}{2} = 3$
The possible transitions are:
$1$. $n = 3 \rightarrow n = 2$
$2$. $n = 3 \rightarrow n = 1$
$3$. $n = 2 \rightarrow n = 1$
Thus,the total number of spectral lines is $3$.

(C) The wavelength of spectral lines in the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Paschen series,$n_1 = 3$ and $n_2 = 4, 5, 6, \dots$.
The first wavelength (maximum wavelength) corresponds to the transition from $n_2 = 4$ to $n_1 = 3$:
$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16-9}{144} \right) = \frac{7R}{144}$.
So,$\lambda_{\max} = \frac{144}{7R} = 18,800 \, \mathring A$.
The minimum wavelength (series limit) corresponds to the transition from $n_2 = \infty$ to $n_1 = 3$:
$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = \frac{R}{9}$.
So,$\lambda_{\min} = \frac{9}{R}$.
Taking the ratio: $\frac{\lambda_{\min}}{\lambda_{\max}} = \frac{9/R}{144/7R} = \frac{9}{R} \times \frac{7R}{144} = \frac{63}{144} = \frac{7}{16}$.
Therefore,$\lambda_{\min} = \frac{7}{16} \times \lambda_{\max} = \frac{7}{16} \times 18,800 = 8,225 \, \mathring A$.

(A) The wavelength $\lambda$ for a transition in a hydrogen-like 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 line of the Lyman series,$n_1 = 1$ and $n_2 = 2$:
$\frac{1}{\lambda_{L_1}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_{L_1} = \frac{4}{3R}$.
For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$:
$\frac{1}{\lambda_{B_1}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36} \implies \lambda_{B_1} = \frac{36}{5R}$.
The ratio of the wavelengths is $\frac{\lambda_{L_1}}{\lambda_{B_1}} = \frac{4/3R}{36/5R} = \frac{4}{3} \times \frac{5}{36} = \frac{5}{27}$.

(C) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula $N = \frac{n(n-1)}{2}$.
Given $N = 6$,we have $\frac{n(n-1)}{2} = 6$,which implies $n^2 - n - 12 = 0$. Solving this quadratic equation,we get $(n-4)(n+3) = 0$,so $n = 4$.
The energy of the emitted photon is given by $\Delta E = E_i - E_f = \frac{hc}{\lambda}$.
Since $\lambda = \frac{hc}{\Delta E}$,the wavelength $\lambda$ is maximum when the energy difference $\Delta E$ is minimum.
The possible transitions from $n=4$ are: $(4 \to 3), (4 \to 2), (4 \to 1), (3 \to 2), (3 \to 1), (2 \to 1)$.
Comparing the energy gaps,the transition $n=4 \to n=3$ has the smallest energy difference,and therefore corresponds to the maximum wavelength of the emitted radiation.

(B) The wavelength $\lambda$ of the first line of the Lyman series for a hydrogen atom $(Z=1)$ is given by the Rydberg formula:
$\frac{1}{\lambda} = R(1)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R \left[ 1 - \frac{1}{4} \right] = \frac{3R}{4}$
The wavelength $\lambda'$ of the second line of the Balmer series for a hydrogen-like ion with atomic number $Z$ is given by:
$\frac{1}{\lambda'} = R Z^2 \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] = R Z^2 \left[ \frac{1}{4} - \frac{1}{16} \right] = R Z^2 \left[ \frac{4-1}{16} \right] = \frac{3 R Z^2}{16}$
According to the problem,$\lambda = \lambda'$,which implies $\frac{1}{\lambda} = \frac{1}{\lambda'}$:
$\frac{3R}{4} = \frac{3 R Z^2}{16}$
Dividing both sides by $\frac{3R}{4}$:
$1 = \frac{Z^2}{4}$
$Z^2 = 4$
$Z = 2$
Thus,the atomic number of the hydrogen-like ion is $2$.

(D) The energy of a photon emitted during a transition is given by $\Delta E = 13.6 Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
Ultraviolet $(UV)$ radiation corresponds to higher energy transitions (Lyman series),while Infrared $(IR)$ radiation corresponds to lower energy transitions (Paschen,Brackett,or Pfund series).
For a transition to result in infrared radiation,the energy difference must be significantly smaller than that of the $n=3 \to n=1$ transition.
Let's evaluate the transitions:
$1. 2 \to 1$: This is part of the Lyman series,which is in the $UV$ region.
$2. 3 \to 1$: This is also part of the Lyman series,which is in the $UV$ region.
$3. 4 \to 2$: This is part of the Balmer series,which is in the visible region.
$4. 4 \to 3$: This is part of the Paschen series,which corresponds to the infrared $(IR)$ region.
Since the energy gap for $4 \to 3$ is the smallest among the given options,it corresponds to the longest wavelength,which falls in the infrared region.

(A) The wavelength of spectral lines in the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the Lyman series,$n_1 = 1$. The longest wavelength corresponds to the smallest energy transition,which is $n_2 = 2$.
$\frac{1}{\lambda_L} = R \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = R \left[ 1 - \frac{1}{4} \right] = \frac{3R}{4} \implies \lambda_L = \frac{4}{3R}$.
For the Balmer series,$n_1 = 2$. The longest wavelength corresponds to the smallest energy transition,which is $n_2 = 3$.
$\frac{1}{\lambda_B} = R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{9-4}{36} \right] = \frac{5R}{36} \implies \lambda_B = \frac{36}{5R}$.
The ratio of the longest wavelengths is:
$\frac{\lambda_L}{\lambda_B} = \frac{4/3R}{36/5R} = \frac{4}{3R} \times \frac{5R}{36} = \frac{4 \times 5}{3 \times 36} = \frac{20}{108} = \frac{5}{27}$.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
Using $hc \approx 12400 \; \text{eV} \cdot \mathring{A}$,we get $E = \frac{12400}{975} \approx 12.75 \; \text{eV}$.
The energy levels of a hydrogen atom are given by $E_n = -\frac{13.6}{n^2} \; \text{eV}$.
For the ground state $(n=1)$,$E_1 = -13.6 \; \text{eV}$.
After absorbing the photon,the new energy level is $E_n = E_1 + E = -13.6 + 12.75 = -0.85 \; \text{eV}$.
Since $E_n = -\frac{13.6}{n^2} = -0.85 \; \text{eV}$,we have $n^2 = \frac{13.6}{0.85} = 16$,so $n = 4$.
The electron is excited to the $n = 4$ state.
The number of spectral lines emitted when the electron transitions from $n$ to the ground state is given by $\frac{n(n-1)}{2}$.
For $n = 4$,the number of spectral lines is $\frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6$.

(A) The wavelength of a spectral line in the Lyman series is given by $\frac{1}{\lambda_{L}} = R\left(\frac{1}{1^{2}} - \frac{1}{n^{2}}\right)$,where $n = 2, 3, 4, \dots$.
For the longest wavelength in the Lyman series,we take $n = 2$:
$\frac{1}{\lambda_{L}} = R\left(1 - \frac{1}{4}\right) = \frac{3R}{4} \implies \lambda_{L} = \frac{4}{3R}$.
The wavelength of a spectral line in the Balmer series is given by $\frac{1}{\lambda_{B}} = R\left(\frac{1}{2^{2}} - \frac{1}{n^{2}}\right)$,where $n = 3, 4, 5, \dots$.
For the longest wavelength in the Balmer series,we take $n = 3$:
$\frac{1}{\lambda_{B}} = R\left(\frac{1}{4} - \frac{1}{9}\right) = R\left(\frac{9-4}{36}\right) = \frac{5R}{36} \implies \lambda_{B} = \frac{36}{5R}$.
The ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:
$\frac{\lambda_{L}}{\lambda_{B}} = \frac{4/3R}{36/5R} = \frac{4}{3R} \times \frac{5R}{36} = \frac{5}{27}$.

(B) The wave number $\bar{\nu}$ for the hydrogen spectrum 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,the transition ends at $n_1 = 2$.
The last line of the Balmer series corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$.
Given $R = 10^7 \, m^{-1}$,we substitute the values:
$\bar{\nu} = 10^7 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = 10^7 \left( \frac{1}{4} - 0 \right) = \frac{10^7}{4} = 0.25 \times 10^7 \, m^{-1}$.

(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)$.
For the transition from $n_i = 3$ to $n_f = 2$:
$\frac{1}{\lambda} = R \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36}$.
For the transition from $n_i = 4$ to $n_f = 3$:
$\frac{1}{\lambda'} = R \left( \frac{1}{3^{2}} - \frac{1}{4^{2}} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = R \left( \frac{16-9}{144} \right) = \frac{7R}{144}$.
Taking the ratio of the two equations:
$\frac{\lambda'}{\lambda} = \frac{5R/36}{7R/144} = \frac{5}{36} \times \frac{144}{7} = \frac{5 \times 4}{7} = \frac{20}{7}$.
Therefore,$\lambda' = \frac{20}{7}\lambda$.

(C) The longest wavelength in the ultraviolet region (Lyman series) occurs for the transition from $n_2 = 2$ to $n_1 = 1$.
Using the Rydberg formula: $\frac{1}{\lambda_{0}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R$.
Thus,$R = \frac{4}{3 \lambda_{0}}$.
The shortest wavelength in the infrared region (Paschen series) occurs for the transition from $n_2 = \infty$ to $n_1 = 3$.
Using the Rydberg formula: $\frac{1}{\lambda'} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = R \left( \frac{1}{9} \right) = \frac{R}{9}$.
Substituting $R = \frac{4}{3 \lambda_{0}}$ into the equation:
$\frac{1}{\lambda'} = \frac{1}{9} \times \frac{4}{3 \lambda_{0}} = \frac{4}{27 \lambda_{0}}$.
Therefore,$\lambda' = \frac{27}{4} \lambda_{0}$.

(B) The wavelength of the last line of the Balmer series is given by the Rydberg formula:
$\frac{1}{\lambda_{B}} = R \left( \frac{1}{2^{2}} - \frac{1}{\infty^{2}} \right) = \frac{R}{4}$
$\lambda_{B} = \frac{4}{R}$
The wavelength of the last line of the Lyman series is given by:
$\frac{1}{\lambda_{L}} = R \left( \frac{1}{1^{2}} - \frac{1}{\infty^{2}} \right) = R$
$\lambda_{L} = \frac{1}{R}$
Taking the ratio of the two wavelengths:
$\frac{\lambda_{B}}{\lambda_{L}} = \frac{4/R}{1/R} = 4$
Thus,the ratio is $4$.

(B) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the series limit of the Lyman series of a Hydrogen atom $(Z=1)$,we have $n_1 = 1$ and $n_2 = \infty$:
$\frac{1}{\lambda_L} = R(1)^2 \left[ \frac{1}{1^2} - \frac{1}{\infty} \right] = R$.
Thus,$\lambda_L = \frac{1}{R}$.
For the series limit of the Balmer series of a hydrogen-like atom (atomic number $Z$),we have $n_1 = 2$ and $n_2 = \infty$:
$\frac{1}{\lambda_B} = R Z^2 \left[ \frac{1}{2^2} - \frac{1}{\infty} \right] = \frac{R Z^2}{4}$.
Thus,$\lambda_B = \frac{4}{R Z^2}$.
Given that the series limits are equal,$\lambda_L = \lambda_B$:
$\frac{1}{R} = \frac{4}{R Z^2} \Rightarrow Z^2 = 4 \Rightarrow Z = 2$.

$(A)$ According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons is given by $K_{max} = e V_{0} = E - \phi_{0}$, where $E$ is the energy of the incident photon, $V_{0}$ is the stopping potential, and $\phi_{0}$ is the work function.
Given: $V_{0} = 10.4 \ V$ and $\phi_{0} = 1.7 \ eV$.
Thus, $E = e V_{0} + \phi_{0} = 10.4 \ eV + 1.7 \ eV = 12.1 \ eV$.
The energy of a photon emitted during a transition in a hydrogen atom from energy level $n_{i}$ to $n_{f}$ is given by $\Delta E = 13.6 \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right) \ eV$.
For transition from $n = 3$ to $n = 1$:
$\Delta E = 13.6 \left( \frac{1}{1^{2}} - \frac{1}{3^{2}} \right) = 13.6 \left( 1 - \frac{1}{9} \right) = 13.6 \times \frac{8}{9} \approx 12.09 \ eV \approx 12.1 \ eV$.
Therefore, the transition from $n = 3$ to $n = 1$ corresponds to the energy of the incident photons.

(B) The $n^{th}$ excited state corresponds to the principal quantum number $N = n + 1$.
When an electron transitions from an excited state $N$ to lower energy levels,the total number of spectral lines emitted is given by the formula $\frac{N(N - 1)}{2}$.
Substituting $N = n + 1$ into the formula,we get:
Number of lines $= \frac{(n + 1)((n + 1) - 1)}{2} = \frac{(n + 1)n}{2}$.
This expression is equivalent to the sum of the first $n$ natural numbers: $1 + 2 + 3 + \dots + n$.

(A) At room temperature,hydrogen atoms are in the ground state,where the electron is in the $n = 1$ energy level.
When radiation is passed through the gas,the atoms can only absorb photons that provide enough energy to excite the electron from $n = 1$ to higher energy levels $(n = 2, 3, 4, \dots)$.
Transitions starting from $n = 1$ to any higher energy level $(n > 1)$ correspond to the Lyman series in the absorption spectrum.
Since there are no electrons in the $n = 2$ level or higher at room temperature,no absorption lines corresponding to the Balmer series (which requires transitions starting from $n = 2$) will be observed.
Therefore,only the Lyman series is observed.

(D) The energy of a photon emitted during a transition between energy levels $n_i$ and $n_f$ is given by $\Delta E = 13.6 Z^2 (\frac{1}{n_f^2} - \frac{1}{n_i^2}) \text{ eV}$.
Ultraviolet radiation corresponds to higher energy transitions (Lyman series),while infrared radiation corresponds to lower energy transitions (Paschen,Brackett,or Pfund series).
Given that the transition $n=4 \rightarrow n=3$ results in ultraviolet radiation for this specific atom,it implies that the energy gap $\Delta E_{4 \rightarrow 3}$ is quite large.
To obtain infrared radiation,we need a transition with a significantly smaller energy gap than $\Delta E_{4 \rightarrow 3}$.
Comparing the options:
$(A)$ $2 \rightarrow 1$: This is a transition between lower energy levels,resulting in a larger energy gap than $4 \rightarrow 3$.
$(B)$ $3 \rightarrow 2$: This is also a transition between lower energy levels,resulting in a larger energy gap than $4 \rightarrow 3$.
$(C)$ $4 \rightarrow 2$: This involves a larger jump in quantum numbers,resulting in a larger energy gap.
$(D)$ $5 \rightarrow 4$: This transition occurs between higher energy levels where the energy difference between consecutive levels is much smaller. Thus,$5 \rightarrow 4$ will have the smallest energy gap and will result in infrared radiation.

(B) When an electron transitions from an excited state with principal quantum number $n$ to lower energy levels,the total number of possible spectral lines is given by the formula:
$N = \frac{n(n - 1)}{2}$
Given that the electron is excited to the state $n = 4$,we substitute this value into the formula:
$N = \frac{4(4 - 1)}{2}$
$N = \frac{4 \times 3}{2}$
$N = \frac{12}{2} = 6$
Therefore,the total number of spectral lines emitted is $6$.

(C) The energy of a photon emitted during a transition is given by $h\nu = E_n - E_m$.
For the series limit,the electron transitions from $n = \infty$ to the ground state of the series.
For the Lyman series,the ground state is $n_1 = 1$. Thus,$h\nu_L = E_{\infty} - E_1 = 0 - E_1 = -E_1$.
For the Pfund series,the ground state is $n_5 = 5$. Thus,$h\nu_f = E_{\infty} - E_5 = 0 - E_5 = -E_5$.
Since $E_n = \frac{E_1}{n^2}$,we have $E_5 = \frac{E_1}{5^2} = \frac{E_1}{25}$.
Substituting this into the expression for $\nu_f$:
$h\nu_f = -\left(\frac{E_1}{25}\right) = \frac{-E_1}{25}$.
Since $h\nu_L = -E_1$,we get $h\nu_f = \frac{h\nu_L}{25}$.
Therefore,$\nu_f = \frac{\nu_L}{25}$.

(A) The wavelength $\lambda$ for a hydrogen-like atom 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 shortest wavelength of the Lyman series of hydrogen $(Z=1)$,$n_1 = 1$ and $n_2 = \infty$. Thus,$\frac{1}{\lambda_L} = R (1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$.
For the shortest wavelength of the Balmer series of a hydrogen-like atom $(Z)$,$n_1 = 2$ and $n_2 = \infty$. Thus,$\frac{1}{\lambda_B} = R Z^2 \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = R \frac{Z^2}{4}$.
Given that $\lambda_L = \lambda_B$,we have $\frac{1}{\lambda_L} = \frac{1}{\lambda_B}$.
Therefore,$R = R \frac{Z^2}{4}$,which implies $Z^2 = 4$.
Solving for $Z$,we get $Z = 2$.

(B) The spectral lines of the Balmer series lie in the visible range.
These lines are obtained due to the transition of an electron from any higher energy state $(n > 2)$ to the $n = 2$ state.
According to the provided figure,the energy levels involved are from $n = 1$ to $n = 7$.
Therefore,the possible transitions to the $n = 2$ state are from $n = 3, 4, 5, 6, \text{ and } 7$.
This gives a total of $5$ possible emission lines in the visible range $(3$ $\rightarrow 2, 4$ $\rightarrow 2, 5$ $\rightarrow 2, 6$ $\rightarrow 2, 7$ $\rightarrow 2)$.

(C) The spectral series of a hydrogen atom is determined by the final energy level $(n_f)$ of the electron transition.
$1$. Transitions to $n_f = 1$ result in the Lyman series.
$2$. Transitions to $n_f = 2$ result in the Balmer series.
$3$. Transitions to $n_f = 3$ result in the Paschen series.
In this case,the electron transitions from $n_i = 4$ to $n_f = 2$. Since the final state is $n_f = 2$,the emission belongs to the Balmer series.
The lines of the Balmer series are defined by transitions from $n_i = 3, 4, 5, ...$ to $n_f = 2$:
- The first line corresponds to $n_i = 3$ to $n_f = 2$.
- The second line corresponds to $n_i = 4$ to $n_f = 2$.
Therefore,the transition from $n=4$ to $n=2$ represents the second line of the Balmer series.

(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
Since $R$ and $Z$ are constant for a given atom,we have $\frac{1}{\lambda} \propto \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the first transition ($n=5$ to $n=2$): $\frac{1}{434} = k \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = k \left( \frac{1}{4} - \frac{1}{25} \right) = k \left( \frac{21}{100} \right)$.
For the second transition ($n=4$ to $n=2$): $\frac{1}{x} = k \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = k \left( \frac{1}{4} - \frac{1}{16} \right) = k \left( \frac{3}{16} \right)$.
Dividing the two equations: $\frac{x}{434} = \frac{21/100}{3/16} = \frac{21}{100} \times \frac{16}{3} = \frac{7 \times 16}{100} = \frac{112}{100} = 1.12$.
Thus,$x = 434 \times 1.12 = 486.08 \, nm \approx 486 \, nm$.

(C) The wavelength $\lambda$ for a transition between energy levels $n_2$ and $n_1$ 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 $Li^{2+}$,the atomic number $Z = 3$,so $Z^2 = 9$.
$\frac{1}{\lambda_{32}} = R(9) \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 9R \left(\frac{1}{4} - \frac{1}{9}\right) = 9R \left(\frac{5}{36}\right) = \frac{5R}{4} \implies \lambda_{32} = \frac{4}{5R}$.
$\frac{1}{\lambda_{31}} = R(9) \left(\frac{1}{1^2} - \frac{1}{3^2}\right) = 9R \left(1 - \frac{1}{9}\right) = 9R \left(\frac{8}{9}\right) = 8R \implies \lambda_{31} = \frac{1}{8R}$.
$\frac{1}{\lambda_{21}} = R(9) \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = 9R \left(1 - \frac{1}{4}\right) = 9R \left(\frac{3}{4}\right) = \frac{27R}{4} \implies \lambda_{21} = \frac{4}{27R}$.
Now,calculating the ratios:
$\frac{\lambda_{32}}{\lambda_{31}} = \frac{4/5R}{1/8R} = \frac{4}{5} \times 8 = \frac{32}{5} = 6.4$.
$\frac{\lambda_{21}}{\lambda_{31}} = \frac{4/27R}{1/8R} = \frac{4}{27} \times 8 = \frac{32}{27} \approx 1.185 \approx 1.2$.
Thus,the ratios are $6.4$ and $1.2$.

(C) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the first Balmer line $(n_i = 3, n_f = 2)$: $\frac{1}{660} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36}$ .... $(1)$
For the second Balmer line $(n_i = 4, n_f = 2)$: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = \frac{3R}{16}$ .... $(2)$
Dividing equation $(1)$ by equation $(2)$:
$\frac{\lambda}{660} = \frac{5R/36}{3R/16} = \frac{5}{36} \times \frac{16}{3} = \frac{80}{108} = \frac{20}{27}$
$\lambda = 660 \times \frac{20}{27} = \frac{13200}{27} \approx 488.88\,nm \approx 488.9\,nm$.

(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 Balmer series,the first line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$.
Since the transition is the same for both atoms,the term $\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$ is constant.
Therefore,$\frac{1}{\lambda} \propto Z^2$,which implies $\lambda \propto \frac{1}{Z^2}$.
For hydrogen $(Z_H = 1)$,$\lambda_H = \lambda$.
For doubly ionized lithium $(Z_{Li} = 3)$,$\lambda_{Li} = \lambda_H \times \left( \frac{Z_H}{Z_{Li}} \right)^2$.
$\lambda_{Li} = \lambda \times \left( \frac{1}{3} \right)^2 = \frac{\lambda}{9}$.

(A) The wavelength $\lambda$ for a transition in a hydrogen atom is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series,$n_1 = 1$. The longest wavelength corresponds to the transition from $n_2 = 2$ to $n_1 = 1$:
$\frac{1}{\lambda_{Lyman}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_{Lyman} = \frac{4}{3R}$.
For the Balmer series,$n_1 = 2$. The longest wavelength corresponds to the transition from $n_2 = 3$ to $n_1 = 2$:
$\frac{1}{\lambda_{Balmer}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9-4}{36} \right) = \frac{5R}{36} \implies \lambda_{Balmer} = \frac{36}{5R}$.
The ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is:
$\frac{\lambda_{Lyman}}{\lambda_{Balmer}} = \frac{4/3R}{36/5R} = \frac{4}{3} \times \frac{5}{36} = \frac{1}{3} \times \frac{5}{9} = \frac{5}{27}$.

(A) For the Balmer series,the Rydberg formula 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 (maximum wavelength),$n = 3$: $\frac{1}{\lambda_{\max}} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$.
For the second line,$n = 4$: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4-1}{16} \right) = \frac{3R}{16}$.
Therefore,$\lambda = \frac{16}{3R}$.

(A) For the shortest wavelength of the Lyman series,the transition occurs from $n_2 = \infty$ to $n_1 = 1$.
Using the Rydberg formula: $\frac{1}{\lambda_{min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R = \frac{1}{91.2 \, nm}$.
For the largest wavelength of the Lyman series,the transition occurs from $n_2 = 2$ to $n_1 = 1$.
Using the Rydberg formula: $\frac{1}{\lambda_{max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R$.
Substituting $R = \frac{1}{91.2 \, nm}$ into the equation:
$\frac{1}{\lambda_{max}} = \frac{3}{4} \times \frac{1}{91.2 \, nm}$.
$\lambda_{max} = \frac{4}{3} \times 91.2 \, nm = 121.6 \, nm$.

(B) The Lyman series corresponds to transitions from higher energy levels to the ground state $(n_f = 1)$.
The energy of the emitted photon is given by $\Delta E = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \text{ eV}$.
The least energetic photon corresponds to the smallest energy difference,which occurs for the transition from the first excited state $(n_i = 2)$ to the ground state $(n_f = 1)$.
Using the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$.
Thus,$\lambda = \frac{4}{3R}$.
Given $R \approx 1.097 \times 10^7 \text{ m}^{-1}$,we have $\lambda = \frac{4}{3 \times 1.097 \times 10^7} \approx 1.216 \times 10^{-7} \text{ m} = 121.6 \text{ nm}$.
Rounding to the nearest integer,we get $122 \text{ nm}$.

(A) The Lyman series corresponds to transitions where the final state is $n_f = 1$. The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$,where $R \approx 1.097 \times 10^7 \, m^{-1}$.
For the longest wavelength (minimum energy transition),we take $n_i = 2$:
$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$.
$\lambda_{\max} = \frac{4}{3R} \approx \frac{4}{3 \times 1.097 \times 10^7} \approx 1.215 \times 10^{-7} \, m = 1215 \, \mathring{A}$ (approximately $1213 \, \mathring{A}$ depending on the value of $R$ used).
For the shortest wavelength (maximum energy transition),we take $n_i = \infty$:
$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$.
$\lambda_{\min} = \frac{1}{R} \approx \frac{1}{1.097 \times 10^7} \approx 9.11 \times 10^{-8} \, m = 911 \, \mathring{A}$ (approximately $910 \, \mathring{A}$).
Thus,the shortest wavelength is $910 \, \mathring{A}$ and the longest is $1213 \, \mathring{A}$.

(A) The wavelength for 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 maximum wavelength $(n=3)$:
$\frac{1}{\lambda_{\max}} = R \left[ \frac{1}{4} - \frac{1}{9} \right] = R \left[ \frac{5}{36} \right]$
$\lambda_{\max} = \frac{36}{5R} \approx 6563 \, \mathring{A}$
For the minimum wavelength $(n \to \infty)$:
$\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{4} - 0 \right] = \frac{R}{4}$
$\lambda_{\min} = \frac{4}{R} \approx 3646 \, \mathring{A}$
The range $3646 \, \mathring{A}$ to $6563 \, \mathring{A}$ falls within the visible region of the electromagnetic spectrum. Thus,both the Assertion and the Reason are correct,and the Reason correctly explains the Assertion.

(A) 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 ground state is $n_{1} = 1$.
For the minimum wavelength (shortest wavelength),$n_{2} = \infty$:
$\frac{1}{\lambda_{min}} = R \left( \frac{1}{1^{2}} - \frac{1}{\infty^{2}} \right) = R \implies \lambda_{min} = \frac{1}{R}$.
For the maximum wavelength (longest wavelength),$n_{2} = 2$:
$\frac{1}{\lambda_{max}} = R \left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_{max} = \frac{4}{3R}$.
The ratio of minimum to maximum wavelength is $\frac{\lambda_{min}}{\lambda_{max}} = \frac{1/R}{4/3R} = \frac{3}{4}$.
Both the Assertion and Reason are correct,and the Reason correctly explains the origin of the Lyman series transitions.

(A) 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 ground state is $n_1 = 1$.
For the minimum wavelength (shortest wavelength),the transition occurs from $n_2 = \infty$ to $n_1 = 1$:
$\frac{1}{\lambda_{min}} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R \implies \lambda_{min} = \frac{1}{R}$.
For the maximum wavelength (longest wavelength),the transition occurs from $n_2 = 2$ to $n_1 = 1$:
$\frac{1}{\lambda_{max}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_{max} = \frac{4}{3R}$.
The ratio of minimum to maximum wavelength is $\frac{\lambda_{min}}{\lambda_{max}} = \frac{1/R}{4/3R} = \frac{3}{4}$.
Both the Assertion and the Reason are correct,and the Reason correctly explains the origin of the Lyman series.

(C) For the Balmer series,the Rydberg formula is given by $\frac{1}{\lambda} = R_{H} \left( \frac{1}{2^{2}} - \frac{1}{n^{2}} \right)$,where $n = 3, 4, 5, \dots$
The first member corresponds to $n = 3$: $\frac{1}{\lambda_{1}} = R_{H} \left( \frac{1}{4} - \frac{1}{9} \right) = R_{H} \left( \frac{5}{36} \right)$.
The second member corresponds to $n = 4$: $\frac{1}{\lambda_{2}} = R_{H} \left( \frac{1}{4} - \frac{1}{16} \right) = R_{H} \left( \frac{3}{16} \right)$.
Taking the ratio: $\frac{\lambda_{2}}{\lambda_{1}} = \frac{5/36}{3/16} = \frac{5}{36} \times \frac{16}{3} = \frac{80}{108} = \frac{20}{27}$.
Given $\lambda_{1} = 6561 \; \mathring{A}$,we have $\lambda_{2} = \frac{20}{27} \times 6561 = 4860 \; \mathring{A}$.
Converting to nanometers: $4860 \; \mathring{A} = 486 \; nm$.

The Rydberg formula is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$,where $R_H \approx 1.097 \times 10^7 \, m^{-1}$.
For the Lyman series,the final energy level is $n_f = 1$. The first four spectral lines correspond to transitions from $n_i = 2, 3, 4, 5$ to $n_f = 1$.
Using $\lambda = \frac{1}{R_H \left( 1 - \frac{1}{n_i^2} \right)} = \frac{n_i^2}{R_H (n_i^2 - 1)}$:
$1$. For $n_i = 2$: $\lambda_{21} = \frac{4}{1.097 \times 10^7 (3)} \approx 1.216 \times 10^{-7} \, m = 1216 \, \mathring{A}$.
$2$. For $n_i = 3$: $\lambda_{31} = \frac{9}{1.097 \times 10^7 (8)} \approx 1.026 \times 10^{-7} \, m = 1026 \, \mathring{A}$.
$3$. For $n_i = 4$: $\lambda_{41} = \frac{16}{1.097 \times 10^7 (15)} \approx 0.972 \times 10^{-7} \, m = 972 \, \mathring{A}$.
$4$. For $n_i = 5$: $\lambda_{51} = \frac{25}{1.097 \times 10^7 (24)} \approx 0.950 \times 10^{-7} \, m = 950 \, \mathring{A}$.

(A) The Rydberg formula for the wavelength $\lambda$ of spectral lines in the hydrogen atom is given by:
$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
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,i.e.,$n_2 = \infty$.
Using the Rydberg constant $R \approx 1.097 \times 10^7 \;m^{-1}$:
$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right)$
$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{1}{9}$
$\lambda = \frac{9}{1.097 \times 10^7} \approx 8.204 \times 10^{-7} \;m = 820.4 \;nm$.
Using the energy conversion factor $13.6 \;eV = 21.76 \times 10^{-19} \;J$ as provided in the context:
$\frac{hc}{\lambda} = 21.76 \times 10^{-19} \left( \frac{1}{3^2} - 0 \right)$
$\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8 \times 9}{21.76 \times 10^{-19}} \approx 8.22 \times 10^{-7} \;m = 822 \;nm$.
Given the options provided,$818.9 \;nm$ is the closest value.

(A) The energy of the electron beam is $12.5\; eV$. The ground state energy of hydrogen is $E_1 = -13.6\; eV$.
When bombarded,the hydrogen atom can absorb energy to reach an excited state $E_n = E_1 + 12.5\; eV = -13.6 + 12.5 = -1.1\; eV$.
Using the formula $E_n = -13.6 / n^2$,we find $n^2 = -13.6 / -1.1 \approx 12.36$,which implies $n \approx 3.5$. Thus,the electron can be excited up to the $n = 3$ level.
During de-excitation from $n = 3$,the possible transitions are $n = 3 \to 2$,$n = 2 \to 1$,and $n = 3 \to 1$.
$1$. For $n = 3 \to 1$ (Lyman series): $\frac{1}{\lambda} = R_y (\frac{1}{1^2} - \frac{1}{3^2}) = 1.097 \times 10^7 \times \frac{8}{9} \implies \lambda \approx 102.6\; nm$.
$2$. For $n = 2 \to 1$ (Lyman series): $\frac{1}{\lambda} = R_y (\frac{1}{1^2} - \frac{1}{2^2}) = 1.097 \times 10^7 \times \frac{3}{4} \implies \lambda \approx 121.6\; nm$.
$3$. For $n = 3 \to 2$ (Balmer series): $\frac{1}{\lambda} = R_y (\frac{1}{2^2} - \frac{1}{3^2}) = 1.097 \times 10^7 \times \frac{5}{36} \implies \lambda \approx 656.3\; nm$.
Thus,wavelengths from both the Lyman and Balmer series are emitted.

(N/A) Each element emits radiation with different wavelengths depending on its temperature. Hence,each element has a characteristic spectrum of radiation which it emits.
When an atomic gas or vapour is excited at low pressure by passing an electric current through it,the emitted radiation has a spectrum which contains only certain specific wavelengths. $A$ spectrum of this kind is known as an emission line spectrum.
The emission line spectrum of atomic hydrogen is shown in the figure.
Emission line spectra of any gas are studied to identify that gas.
When white light passes through a gas,we find that certain wavelengths,which are characteristic of the atom,are absent in the transmitted light. Thus,there are some black lines appearing in the spectrum of the transmitted light. The spectrum produced by these dark lines is called the absorption spectrum of the material of the gas.

(N/A) The hydrogen spectral series refers to the set of discrete wavelengths of electromagnetic radiation emitted by hydrogen atoms when electrons transition between different energy levels. When an electron jumps from a higher energy level $(n_i)$ to a lower energy level $(n_f)$,a photon is emitted with energy $E = E_{n_i} - E_{n_f} = h\nu = \frac{hc}{\lambda}$.
The wavelength $\lambda$ 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 $(1.097 \times 10^7 \ m^{-1})$.
The main series are:
$1$. Lyman series: $n_f = 1$,$n_i = 2, 3, 4, \dots$ (Ultraviolet region).
$2$. Balmer series: $n_f = 2$,$n_i = 3, 4, 5, \dots$ (Visible region).
$3$. Paschen series: $n_f = 3$,$n_i = 4, 5, 6, \dots$ (Infrared region).
$4$. Brackett series: $n_f = 4$,$n_i = 5, 6, 7, \dots$ (Infrared region).
$5$. Pfund series: $n_f = 5$,$n_i = 6, 7, 8, \dots$ (Infrared region).

(N/A) The different series observed in the hydrogen spectrum are:
$1$. Lyman series: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$,where $n = 2, 3, 4, \dots$
$2$. Balmer series: $\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)$,where $n = 3, 4, 5, \dots$
$3$. Paschen series: $\frac{1}{\lambda} = R \left( \frac{1}{3^2} - \frac{1}{n^2} \right)$,where $n = 4, 5, 6, \dots$
$4$. Brackett series: $\frac{1}{\lambda} = R \left( \frac{1}{4^2} - \frac{1}{n^2} \right)$,where $n = 5, 6, 7, \dots$
$5$. Pfund series: $\frac{1}{\lambda} = R \left( \frac{1}{5^2} - \frac{1}{n^2} \right)$,where $n = 6, 7, 8, \dots$
The general Rydberg formula for the wave number $\bar{\nu} = \frac{1}{\lambda}$ is:
$\bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
where $R$ is the Rydberg constant $(1.097 \times 10^7 \ m^{-1})$,$n_1$ is the lower energy level,and $n_2$ is the higher energy level $(n_2 > n_1)$.