The Line Spectra of the Hydrogen Atom

Atoms · CBSE

Theory

What this subtopic is examined on

Deriving the Rydberg formula from Bohr's third postulate, and quoting $R=\dfrac{me^4}{8\varepsilon_0^2h^3c}$ .
Naming the series, their $n$ values, their regions, and their longest and shortest wavelengths — set almost every year.
Ratios of wavelengths in which $R$ cancels.
Electron-beam bombardment: how far the atom is excited and which lines then appear.
Reading an energy-level diagram: which transition gives the maximum or minimum photon energy or wavelength.
The correspondence-principle question ( $n$ to $n-1$ for large $n$ ).

From the third postulate to the spectrum

When an atom makes a transition from a higher state $n_i$ to a lower state $n_f$ ( $n_f<n_i$ ), the energy difference is carried away by a single photon of frequency $\nu_{if}$ :

$h\nu_{if}=E_{n_i}-E_{n_f}$

Because $n_i$ and $n_f$ are integers, only certain discrete frequencies can be radiated. That is the whole explanation of the line spectrum.
Emission lines: electrons jumping down. Absorption lines: an atom absorbs a photon whose energy is exactly the gap needed to lift the electron to a higher level, so when light with a continuous range of frequencies passes through a rarefied gas, a series of dark absorption lines appears in the continuous spectrum at exactly those frequencies.

Deriving Rydberg's formula

Using $E_n=-\dfrac{me^4}{8\varepsilon_0^2n^2h^2}$ in the frequency condition:

$h\nu=E_{n_2}-E_{n_1}=\dfrac{me^4}{8\varepsilon_0^2h^2}\left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right]$

$\nu=\dfrac{me^4}{8\varepsilon_0^2h^3}\left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right]\qquad\Longrightarrow\qquad \dfrac{1}{\lambda}=\dfrac{\nu}{c}=\dfrac{me^4}{8\varepsilon_0^2h^3c}\left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right]$

$\dfrac{1}{\lambda}=R\left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right],\qquad R=\dfrac{me^4}{8\varepsilon_0^2h^3c}=1.097\times10^{7}\ \text{m}^{-1}$

Bohr's model thus derives, from three postulates, the constant Rydberg had obtained purely by fitting data. This agreement is the reason Bohr's model was accepted, and he was awarded the Nobel Prize in Physics in 1922.
Convenient energy form: $\Delta E=13.6\left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right]$ eV, then $\lambda\,(\text{nm})=\dfrac{1240}{\Delta E\,(\text{eV})}$ .

The spectral series

The lower level $n_1$ names the series; the upper level $n_2$ names the line within it.

Series	$n_1$	$n_2$	Region	First (longest $\lambda$ ) line	Series limit (shortest $\lambda$ )
Lyman	$1$	$2,3,4,\dots,\infty$	Ultraviolet	$\dfrac{4}{3R}=1215$ Å $=121.5$ nm	$\dfrac{1}{R}=912$ Å $=91.2$ nm
Balmer	$2$	$3,4,5,\dots,\infty$	Visible	$\dfrac{36}{5R}=6563$ Å $=656.3$ nm ( $H_\alpha$ )	$\dfrac{4}{R}=3646$ Å $=364.6$ nm
Paschen	$3$	$4,5,6,\dots,\infty$	Infrared	$\dfrac{144}{7R}=18751$ Å	$\dfrac{9}{R}=8204$ Å
Brackett	$4$	$5,6,7,\dots,\infty$	Infrared	$\dfrac{400}{9R}=40520$ Å	$\dfrac{16}{R}=14585$ Å $=1.4585\ \mu$ m
Pfund	$5$	$6,7,8,\dots,\infty$	Far infrared	$\dfrac{900}{11R}=74580$ Å	$\dfrac{25}{R}=22790$ Å

Longest wavelength of a series: smallest energy gap, so $n_2=n_1+1$ . Shortest wavelength (the series limit): largest gap, so $n_2=\infty$ , giving $\lambda_{\min}=\dfrac{n_1^2}{R}$ .
The Balmer lines are named $H_\alpha$ ( $3\to2$ , $656.3$ nm, red), $H_\beta$ ( $4\to2$ ), $H_\gamma$ ( $5\to2$ ), $H_\delta$ ( $6\to2$ ). $H_\alpha$ is the standard 1-mark answer.
Note on the Balmer limit: $3646$ Å is right at the violet edge. CBSE's marking scheme calls it "visible" (the Balmer series is the visible series), so answer "visible" for it in a board paper.
Series-limit shortcut: since $\lambda_{\min}=n_1^2/R$ , if the Lyman limit is $913.4$ Å then $1/R=913.4$ Å and the Balmer limit is $4/R=4\times913.4=3653.6$ Å.

Ratios in which $R$ cancels

$\dfrac{\lambda_{\max}}{\lambda_{\min}}\bigg|_{\text{Balmer}}=\dfrac{36/5R}{4/R}=\dfrac{9}{5}\qquad\qquad \dfrac{\lambda_{\max}(\text{Lyman})}{\lambda_{\min}(\text{Balmer})}=\dfrac{4/3R}{4/R}=\dfrac{1}{3}$

Write out $1/\lambda$ for each, invert, and divide. Do not substitute a value for $R$ — it always cancels, and substituting is where arithmetic errors creep in.

The energy-level diagram

Vertical energy axis in eV, $0$ eV at the top, horizontal lines at $-13.60$ ( $n=1$ ), $-3.40$ ( $n=2$ ), $-1.51$ ( $n=3$ ), $-0.85$ ( $n=4$ ), $-0.54$ ( $n=5$ ), $-0.38$ ( $n=6$ ), converging on $0$ at $n=\infty$ ; the region above $0$ eV is a continuum of free states.
Downward arrows are emission. Bundle them by the level they land on: all arrows landing on $n=1$ form the Lyman series, on $n=2$ the Balmer series, on $n=3$ the Paschen series, on $n=4$ the Brackett series, on $n=5$ the Pfund series.
Reading such a diagram in an MCQ: (i) discard any upward arrow, which is absorption not emission; (ii) the biggest vertical drop is the highest photon energy and the shortest wavelength; (iii) the smallest drop is the longest wavelength.
Worked examples of that reading: among $2\to1$ ( $10.2$ eV), $4\to2$ ( $2.55$ eV) and $3\to2$ ( $1.89$ eV), the highest-energy photon is from $2\to1$ . Among transitions including $4\to1$ ( $12.75$ eV), $4\to1$ wins. The smallest gap among $4\to3$ ( $0.65$ eV), $3\to2$ and $4\to2$ is $4\to3$ , so that gives the longest wavelength.
Three-level relation: if $C\to B$ gives $\lambda_1$ , $B\to A$ gives $\lambda_2$ and $C\to A$ gives $\lambda_3$ , then since $E_C-E_A=(E_C-E_B)+(E_B-E_A)$ ,

$\dfrac{hc}{\lambda_3}=\dfrac{hc}{\lambda_1}+\dfrac{hc}{\lambda_2}\qquad\Longrightarrow\qquad \dfrac{1}{\lambda_3}=\dfrac{1}{\lambda_1}+\dfrac{1}{\lambda_2}\qquad\Longrightarrow\qquad \lambda_3=\dfrac{\lambda_1\lambda_2}{\lambda_1+\lambda_2}$

Excitation by an electron beam or a photon

Method: an atom in the ground state absorbing energy $E$ reaches the level $n$ with $E_n=-13.6+E$ , so $n^2=\dfrac{13.6}{\lvert E_n\rvert}$ .
A photon must match a gap exactly; an electron in a beam can transfer any amount up to its kinetic energy, so a beam excites the atom to the highest level it can reach.
$12.75$ eV beam: $E_n=-13.6+12.75=-0.85$ eV, so $n^2=13.6/0.85=16$ , $n=4$ . Number of lines $=\dfrac{4\times3}{2}=6$ : Lyman $4\to1,\,3\to1,\,2\to1$ ; Balmer $4\to2,\,3\to2$ ; Paschen $4\to3$ .
$12.5$ eV beam: $E_n=-13.6+12.5=-1.1$ eV, $n^2=13.6/1.1=12.36$ , $n=3.5$ . A quantum number cannot be fractional, so the atom is excited only to $n=3$ , and only Lyman and Balmer lines appear.
$12.1$ eV photon: $12.1=13.6-\dfrac{13.6}{n^2}$ gives $\dfrac{13.6}{n^2}=1.5$ , $n^2=9$ , $n=3$ — the second excited state.
Number of possible spectral lines when the atom de-excites from level $n$ to the ground state: $N=\dfrac{n(n-1)}{2}$ . From $n=4$ , $N=6$ .
Standard wavelengths from an $n=4$ excitation: $2\to1$ gives $1213$ Å, $3\to1$ gives $1024$ Å (Lyman); $3\to2$ gives $6548$ Å (Balmer).

Correspondence principle: the $n\to n-1$ transition

Frequency emitted in the transition from $n$ to $n-1$ :

$\nu=\dfrac{me^4}{8\varepsilon_0^2h^3}\left[\dfrac{1}{(n-1)^2}-\dfrac{1}{n^2}\right]=\dfrac{me^4}{8\varepsilon_0^2h^3}\cdot\dfrac{2n-1}{(n-1)^2n^2}$

For large $n$ , $2n-1\approx2n$ and $(n-1)^2\approx n^2$ , so

$\nu\approx\dfrac{me^4}{8\varepsilon_0^2h^3}\cdot\dfrac{2n}{n^4}=\dfrac{me^4}{4\varepsilon_0^2h^3n^3}$

The classical frequency of revolution in the $n$ th orbit is $f_n=\dfrac{v_n}{2\pi r_n}=\dfrac{me^4}{4\varepsilon_0^2h^3n^3}$ — the same expression.
Conclusion to write: for large quantum numbers the frequency of the radiation predicted by Bohr's model coincides with the classical frequency of revolution. This is Bohr's correspondence principle; for small $n$ the two are quite unrelated.

Standard results to collect

$h\nu=E_i-E_f$ ; $\dfrac{1}{\lambda}=R\left[\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right]$ ; $R=\dfrac{me^4}{8\varepsilon_0^2h^3c}=1.097\times10^7$ m $^{-1}$ .
$\lambda\,(\text{nm})=\dfrac{1240}{\Delta E\,(\text{eV})}$ ; $\Delta E=1.89$ eV gives $656$ nm, $\Delta E=10.2$ eV gives $121.5$ nm, $\Delta E=2.55$ eV gives $486$ nm.
$\lambda_{\min}$ of a series $=\dfrac{n_1^2}{R}$ ; $\lambda_{\max}=\dfrac{1}{R}\cdot\dfrac{n_1^2(n_1+1)^2}{2n_1+1}$ .
Balmer $\lambda_{\max}/\lambda_{\min}=9/5$ ; Lyman $\lambda_{\max}$ to Balmer $\lambda_{\min}$ $=1/3$ .
Number of lines from level $n$ : $\dfrac{n(n-1)}{2}$ .
Combined with the photoelectric effect: a $3\to1$ photon carries $13.6\left[1-\dfrac19\right]=12.1$ eV. If it ejects photoelectrons of maximum kinetic energy $9$ eV, the work function is $W_0=12.1-9=3.1$ eV and the threshold wavelength is $\lambda_{th}=\dfrac{1240}{3.1}=400$ nm $=4\times10^{-7}$ m.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
Name the spectral series for a hydrogen atom which lies in the visible region. Find the ratio of the maximum to the minimum wavelength of this series.
2 marksCBSE Class 12 Physics — SA · Term II 2021-22 · Section B, 2 marksAnswer 1 ↓
2
Calculate the orbital period of the electron in the first excited state of hydrogen atom.
2 marksCBSE Class 12 Physics — SA · Delhi 2019, U · Section B, 2 marksAnswer 2 ↓
3
A 12.75 eV electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelengths and the corresponding series of the lines emitted.
2 marksCBSE Class 12 Physics — SA · AI 2017, U · Section B, 2 marksAnswer 3 ↓
4
The ground state energy of hydrogen atom is $-13.6$ eV. If an electron makes a transition from an energy level $-1.51$ eV to $-3.4$ eV, calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.
2 marksCBSE Class 12 Physics — SA · AI 2017, Ap · Section B, 2 marksAnswer 4 ↓
5
Define ionization energy. How would the ionization energy change when the electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron but having the same charge?
2 marksCBSE Class 12 Physics — SA · AI 2016 · Section B, 2 marksAnswer 5 ↓
6
An electron jumps from fourth to first orbit in an atom. How many maximum number of spectral lines can be emitted by the atom? To which series these lines correspond?
2 marksCBSE Class 12 Physics — SA · Foreign 2016 · Section B, 2 marksAnswer 6 ↓
7
The figure shows the energy level diagram of hydrogen atom, with transitions labelled A, B, C (from $n=4$ ), D, E (from $n=3$ ), and F (from $n=2$ ), all converging down to $n=2$ or $n=1$ (A, B, C originate at $n=4$ ; D, E originate at $n=3$ ; F originates at $n=2$ ; all terminate at either $n=2$ or $n=1$ ). (a) Find out the transition which results in the emission of a photon of wavelength 496 nm. (b) Which transition corresponds to the emission of radiation of maximum wavelength? Justify your answer.
2 marksCBSE Class 12 Physics — SA · AI 2015C, Ap · Section B, 2 marksAnswer 7 ↓
8
Calculate the shortest wavelength in the Balmer series of hydrogen atom. In which region (infrared, visible, ultraviolet) of hydrogen spectrum does this wavelength lie?
2 marksCBSE Class 12 Physics — SA · AI 2015 · Section B, 2 marksAnswer 8 ↓
9
An electron in a hydrogen atom makes transitions from orbits of higher energies to orbits of lower energies. (i) When will such transitions result in (a) Lyman series? (b) Balmer series? (ii) Find the ratio of the longest wavelength in Lyman series to the shortest wavelength in Balmer series.
3 marksCBSE Class 12 Physics — SA · Term II 2021-22 · Section C, 3 marksAnswer 9 ↓
10
Write Rydberg's formula for wavelengths of the spectral lines of hydrogen spectrum. Mention to which series in the emission spectrum of hydrogen, the $H_\alpha$ line belongs.
3 marksCBSE Class 12 Physics — SA · AI 2019, Ev · Section C, 3 marksAnswer 10 ↓
11
Using Rydberg formula, calculate the longest wavelengths belonging to Lyman and Balmer series of hydrogen spectrum. In which region these transitions lie? OR Using Rydberg formula, calculate the wavelengths of the spectral lines of the first member of the Lyman series and of the Balmer series.
5 marksCBSE Class 12 Physics — SA · Foreign 2015 / 2/5, Foreign 2014 · Section E, 5 marksAnswer 11 ↓
12
A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited? Calculate the wavelengths of the first member of Lyman and first member of Balmer series.
3 marksCBSE Class 12 Physics — SA · Delhi 2014, An · Section C, 3 marksAnswer 12 ↓
13
The value of ground state energy of hydrogen atom is $-13.6$ eV. (i) Find the energy required to move an electron from the ground state to the first excited state of the atom. (ii) Determine (a) the kinetic energy, and (b) orbital radius in the first excited state of the atom. (Given the value of Bohr radius $=0.53$ Å)
3 marksCBSE Class 12 Physics — SA · AI 2014C, Cr · Section C, 3 marksAnswer 13 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

14
The potential energy of an electron in the second excited state in hydrogen atom is
- (a)
  $-3.4$ eV
- (b)
  $-3.02$ eV
- (c)
  $-1.51$ eV
- (d)
  $-6.8$ eV
1 markCBSE Class 12 Physics — Multiple Choice · 2023, Ap · Section A, 1 markAnswer 14 ↓
15
The diagram shows four energy levels of an electron in Bohr model of hydrogen atom (transitions labelled I: $n{=}2\to n{=}1$ , II: $n{=}4\to n{=}2$ , III: $n{=}1$ to $n{=}4$ shown as an upward absorption arrow, IV: $n{=}3\to n{=}2$ ). Identify the transition in which the emitted photon will have the highest energy.
- (a)
  I
- (b)
  II
- (c)
  III
- (d)
  IV
1 markCBSE Class 12 Physics — Multiple Choice · 2023 · Section A, 1 markAnswer 15 ↓
16
The figure shows the energy level diagram of the hydrogen atom with a few transitions (transition I: $n=2\to n=1$ , absorption, going from $-13.6$ eV to $-3.4$ eV; transition II: $n=4\to n=3$ , going from $-0.85$ eV to $-1.5$ eV; transition III: $n=4\to n=1$ , from $-0.85$ eV to $-13.6$ eV; transition IV: $n=3\to n=2$ , from $-1.5$ eV to $-3.4$ eV). Which transition shows the emission of photon with maximum energy?
- (a)
  I
- (b)
  II
- (c)
  III
- (d)
  IV
1 markCBSE Class 12 Physics — Multiple Choice · 2020 · Section A, 1 markAnswer 16 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
The BALMER series for a hydrogen atom lies in the visible region. $\;\dfrac1\lambda=R\left(\dfrac1{2^2}-\dfrac1{n^2}\right)$ , where $n=3,4,\dots,\infty$ For minimum wavelength, $n=\infty$ : $\;\dfrac1{\lambda_{min}}=\dfrac R4\;\Rightarrow\;\lambda_{min}=\dfrac4R$ For maximum wavelength, $n=3$ : $\;\dfrac1{\lambda_{max}}=R\left(\dfrac14-\dfrac19\right)=\dfrac{5R}{36}\;\Rightarrow\;\lambda_{max}=\dfrac{36}{5R}$ $\;\dfrac{\lambda_{max}}{\lambda_{min}}=\dfrac{36}{5R}\times\dfrac R4=\dfrac9{5}$
2.↑ back to the question
$\;T_n=\dfrac{4\varepsilon_0^2n^3h^3}{me^4}$ . First excited state: $n=2$ . $\;T=\dfrac{2\times6.6\times10^{-34}\times0.53\times10^{-10}}{9\times10^9\times1\times(1.6\times10^{-19})^2}$ (using the equivalent form $T=2\pi r/v$ with $r$ and $v$ for $n=2$ ) $\;T\approx3.03\times10^{-16}\text{ s}$
3.↑ back to the question
$\Delta E=12.75$ eV. Energy of electron in $n^{th}$ orbit of hydrogen atom: $\;E_n=-\dfrac{13.6}{n^2}$ eV Energy of the electron in the excited state after absorbing a photon of $12.75$ eV: $\;E_n=-13.6+12.75=-0.85$ eV $\;n^2=\dfrac{-13.6}{E_n}=\dfrac{13.6}{0.85}=16\;\Rightarrow\;n=4$ So the electron gets excited to the $n=4$ state. Total number of wavelengths in the emission spectrum: $\;\dfrac{n(n-1)}2=\dfrac{4\times3}2=6$ Emitted wavelength for a jump from $E_i$ to $E_f$ : $\;\lambda_{if}=\dfrac{hc}{E_i-E_f}=\dfrac{6.6\times10^{-34}\times3\times10^8}{E_i-E_f}=\dfrac{19.8\times10^{-26}}{E_i-E_f}\text{ m}$ Wavelength for $3\to2$ : $\lambda_{32}=6547.6$ Å; for $3\to1$ : $\lambda_{31}=1023.6$ Å; for $2\to1$ : $\lambda_{21}=1213.2$ Å (and similarly for the other transitions from $n=4$ ). Lyman series: $\lambda_{21}$ ( $\approx1213$ Å) and $\lambda_{31}$ ( $\approx1024$ Å); Balmer series: $\lambda_{32}$ ( $\approx6548$ Å).
4.↑ back to the question
The energy levels of the H atom are given as $\;E_n=\dfrac{-13.6}{n^2}$ eV. For $E=-1.51$ eV: $\;n^2=\dfrac{13.6}{1.51}\approx9\;\Rightarrow\;n=3$ For $E=-3.4$ eV: $\;n^2=\dfrac{13.6}{3.4}=4\;\Rightarrow\;n=2$ So the electron makes a transition from $n=3$ to $n=2$ , i.e. it belongs to the BALMER series. $\;\dfrac{hc}{\lambda_{32}}=1.51-3.4$ eV in magnitude $=1.89$ eV $=1.89\times1.6\times10^{-19}=3.024\times10^{-19}$ J $\;\lambda_{32}=\dfrac{6.625\times10^{-34}\times3\times10^8}{3.024\times10^{-19}}\approx6.57\times10^{-7}\text{ m}$
5.↑ back to the question
The minimum energy required to free the electron from the ground state of the hydrogen atom is known as the ionization energy of that atom. $\;E_0=\dfrac{me^4}{8\varepsilon_0^2h^2}$ , i.e. $E_0\propto m$ So when the electron in hydrogen atom is replaced by a particle of mass 200 times that of the electron (same charge), the ionization energy INCREASES by 200 times.
6.↑ back to the question
Number of spectral lines obtained due to transition of electron from $n=4$ (3rd excited state) to $n=1$ (ground state): $\;N=\dfrac{n(n-1)}2=\dfrac{4\times3}2=6$ The transition lines ending at $n=1$ correspond to the Lyman series (3 lines: $4\to1,3\to1,2\to1$ ), lines ending at $n=2$ correspond to the Balmer series (2 lines: $4\to2,3\to2$ ), and the line ending at $n=3$ corresponds to the Paschen series (1 line: $4\to3$ ).
7.↑ back to the question
(a) $\lambda=496$ nm $=496\times10^{-9}$ m $\;E=\dfrac{hc}\lambda=\dfrac{6.62\times10^{-34}\times3\times10^8}{496\times10^{-9}\times1.6\times10^{-19}}\approx2.5\text{ eV}$ This energy corresponds to the transition from $n=4$ to $n=2$ (energy gap $=-0.85-(-3.4)=2.55\approx2.5$ eV), i.e. one of the transitions (labelled, say, B) terminating at $n=2$ from $n=4$ . (b) Since $E=\dfrac{hc}\lambda$ , i.e. $\lambda\propto\dfrac1E$ , the transition with the SMALLEST energy gap gives the maximum wavelength. Among the transitions shown, the smallest energy gap is for the transition from $n=4$ to $n=3$ (labelled, say, A): $|E_4-E_3|=|-0.85-(-1.5)|=0.65$ eV, the smallest energy difference among the transitions shown — hence it corresponds to the emission of radiation of maximum wavelength.
8.↑ back to the question
$\;\dfrac1\lambda=R_H\left[\dfrac1{2^2}-\dfrac1{n_f^2}\right]$ , where $n_f=3,4,5,\dots$ For the SHORTEST wavelength, the transition is from $n_i=\infty$ to $n_f=2$ : $\;\dfrac1{\lambda_{min}}=R_H\left[\dfrac14-0\right]=\dfrac{1.097\times10^7}4$ $\;\lambda_{min}=3.646\times10^{-7}\text{ m}=3646\text{ \text{Å}}$ This wavelength lies in the VISIBLE region of the electromagnetic spectrum.
9.↑ back to the question
(i)(a) Transitions from higher energy levels ( $n_2=2,3,\dots,\infty$ ) to the FIRST energy level ( $n_1=1$ ) constitute the Lyman series. (b) Transitions from higher energy levels ( $n_2=3,4,\dots,\infty$ ) to the SECOND energy level ( $n_1=2$ ) constitute the Balmer series. (ii) Longest wavelength of Lyman series ( $n_1=1,n_2=2$ ): $\;\dfrac1{\lambda_L}=R\left(1-\dfrac14\right)=\dfrac{3R}4\;\Rightarrow\;\lambda_L=\dfrac4{3R}$ Shortest wavelength of Balmer series ( $n_1=2,n_2=\infty$ ): $\;\dfrac1{\lambda_B}=R\left(\dfrac14-0\right)=\dfrac R4\;\Rightarrow\;\lambda_B=\dfrac4R$ $\;\dfrac{\lambda_L}{\lambda_B}=\dfrac4{3R}\times\dfrac R4=\dfrac13$
10.↑ back to the question
Rydberg's formula: $\;\dfrac1\lambda=R\left[\dfrac1{n_1^2}-\dfrac1{n_2^2}\right]$ , where $R$ is Rydberg's constant $=1.0973\times10^7$ m $^{-1}$ . The $H_\alpha$ line of the emission spectrum of hydrogen (wavelength 656.3 nm) belongs to the BALMER series ( $n_1=2,\,n_2=3$ ).
11.↑ back to the question
For the longest wavelength of the Lyman series, $n_i=2$ : $\;\dfrac1{\lambda_{max}}=R\left(\dfrac1{1^2}-\dfrac1{2^2}\right)=\dfrac{3R}4$ $\;\lambda_{max}=\dfrac4{3R}=\dfrac4{3\times1.097\times10^7}\approx1.215\times10^{-7}\text{ m}=1215\text{ \text{Å}}$ . The lines of the Lyman series are found in the ULTRAVIOLET region. For the longest wavelength of the Balmer series, $n_i=3$ : $\;\dfrac1{\lambda_{max}}=R\left(\dfrac1{2^2}-\dfrac1{3^2}\right)=\dfrac{5R}{36}$ $\;\lambda_{max}=\dfrac{36}{5R}=\dfrac{36}{5\times1.097\times10^7}\approx6.563\times10^{-7}\text{ m}=6563\text{ \text{Å}}$ . Balmer series lie in the VISIBLE region.
12.↑ back to the question
$\Delta E=12.5$ eV. Ground state $E_1=-13.6$ eV. Energy after absorbing the photon: $\;E_n=-13.6+12.5=-1.1$ eV $\;n^2=\dfrac{-13.6}{E_n}=\dfrac{13.6}{1.1}\approx12.36\;\Rightarrow\;n\approx3.5$ Since the state of the electron cannot be a fraction, the hydrogen atom is excited (at most) to $n=3$ (2nd excited state). Wavelength of the first member of the Lyman series ( $n=2\to1$ ): $\;\dfrac1\lambda=R\left(1-\dfrac14\right)=\dfrac{3R}4\;\Rightarrow\;\lambda=\dfrac4{3R}=\dfrac4{3\times1.097\times10^7}\approx1.215\times10^{-7}\text{ m}=121\text{ nm}$ Wavelength of the first member of the Balmer series ( $n=3\to2$ ): $\;\dfrac1{\lambda'}=R\left(\dfrac14-\dfrac19\right)=\dfrac{5R}{36}\;\Rightarrow\;\lambda'=\dfrac{36}{5R}=\dfrac{36}{5\times1.097\times10^7}\approx6.56\times10^{-7}\text{ m}=656\text{ nm}$
13.↑ back to the question
(i) Energy of the photon emitted (equivalently, absorbed for excitation) during a transition of the electron from the first excited state to its ground state, taking the magnitude: $\;\Delta E=E_2-E_1=-\dfrac{13.6}4-(-13.6)=-3.40+13.6=10.2\text{ eV}$ (ii)(a) The energy levels of the H-atom are given by $\;E_n=-\dfrac{Rhc}{n^2}=-\dfrac{13.6}{n^2}$ eV. For the first excited state, $n=2$ : $\;E_2=-\dfrac{13.6}4=-3.4$ eV Kinetic energy of the electron in the $n=2$ state: $\;K_2=-E_2=+3.4$ eV (b) Radius in the first excited state: $\;r_1=n^2r_0=(2)^2\times0.53\text{ \text{Å}}=2.12\text{ \text{Å}}$
14.Answer: (B)↑ back to the question
Second excited state: $n=3$ . Total energy $E_3=-\dfrac{13.6}9\approx-1.51$ eV. Potential energy $=2\times$ total energy $=2\times(-1.51)=-3.02$ eV. Option (b).
15.Answer: (A)↑ back to the question
Energy $\propto1/\lambda$ , and photon energy $=|\Delta E|$ between the two levels. Transition III (from $n=1$ to $n=4$ ) is upward (absorption, not emission), so it is excluded from consideration for the largest EMITTED photon. Among the emission transitions I ( $n=2\to1$ ), II ( $n=4\to2$ ), and IV ( $n=3\to2$ ), the LARGEST energy gap is for transition I (from $n=2$ directly down to the ground state $n=1$ : $|E_2-E_1|=3.4-13.6=10.2$ eV, the biggest of the emission energy gaps shown). Energy is highest for the transition ending at the ground state from the lowest possible excited level shown, i.e. transition I. Option (a).
16.Answer: (C)↑ back to the question
Transition I is an ABSORPTION transition (energy increases, i.e. electron moves up), not an emission. Among the emission transitions, the maximum energy photon corresponds to the LARGEST energy gap: $\;E=E_4-E_1=-0.85-(-13.6)=12.75$ eV for transition III — the largest amongst those shown. Option (c).