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આ MCQ મોડ્યુલ આના પર આધારિત છે: Electromagnetic Waves Spectra
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Electromagnetic Waves Spectra
🎓 Class 11
Chemistry
CBSE
Theory
Ch 2 – Structure of Atom
⏱ ~14 min
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આ મૂલ્યાંકન આના પર આધારિત હશે: Electromagnetic Waves Spectra
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
Electromagnetic Waves, Planck's Quantum Theory and Hydrogen Spectrum
Introduction: A Crisis for Classical Physics
By 1900, physicists had a problem. Rutherford's model would only arrive in 1911, but even before that, two phenomena were refusing to fit the tidy wave picture of light from Maxwell's equations: black-body radiation and the photoelectric effect. At the same time, atomic spectra showed sharp, precise lines rather than a smeared rainbow. Resolving these anomalies between 1900 and 1913 produced the quantum theory — the framework inside which Bohr would shortly build his model of hydrogen.
Part 2 develops three strands that feed into Bohr's work (Part 3): (i) the classical wave description of electromagnetic radiation, (ii) Planck's and Einstein's quantum view, and (iii) the beautifully regular line spectrum of hydrogen.
2.3 Developments Leading to the Bohr Model
2.3.1 Wave Nature of Electromagnetic Radiation
In the 1860s James Clerk Maxwell showed that electromagnetic radiation — visible light, radio waves, X-rays, microwaves — consists of oscillating electric and magnetic fields at right angles to each other and to the direction of propagation. Key wave quantities:
- Wavelength (λ): distance between two successive crests. Measured in m, nm, or Å.
- Frequency (ν, Greek "nu"): number of waves passing a fixed point per second. SI unit Hz (= s−1).
- Wave number (\(\bar{\nu}\)): reciprocal of wavelength. Common unit cm−1. \(\bar{\nu}=1/\lambda\).
- Amplitude: maximum displacement from mean; determines intensity/brightness.
c = νλ, with c = 3 × 108 m s−1 in vacuum
| Region | λ (approx) | Typical source / use |
|---|---|---|
| γ-rays | < 10−11 m | Nuclear decay; cancer therapy |
| X-rays | 10−10 – 10−8 m | Medical imaging; crystal diffraction |
| UV | 10–400 nm | Sterilisation; suntan |
| Visible | 400–750 nm | Human vision, photography |
| IR | 750 nm – 1 mm | Heat, remote controls, molecular vibrations |
| Microwave | 1 mm – 1 m | Radar, microwave ovens, telecom |
| Radio | > 1 m | Broadcasting, MRI |
2.3.2 Particle Nature — Planck's Quantum Theory (1900)
A glowing piece of iron emits mostly red light; hotter, and it turns orange, then yellow, finally bluish-white. A black body (a perfect absorber/emitter) shows a characteristic intensity-vs-wavelength curve that classical physics could not explain — classical theory predicted infinite emission at short wavelengths, the so-called ultraviolet catastrophe.
Max Planck (1900) rescued the theory with a radical postulate: radiation is emitted or absorbed by the walls of a body only in discrete packets of energy called quanta:
E = hν = hc/λ, h = 6.626 × 10−34 J·s
For light this single quantum is called a photon. A bulb of any macroscopic brightness emits billions of billions of photons per second, which is why the discreteness is invisible in daily life.
The Photoelectric Effect (Einstein, 1905)
Heinrich Hertz (1887) noticed that UV light falling on a clean metal surface ejects electrons. The experimental facts were:
- Electrons are ejected only when the frequency of the incident light exceeds a minimum value, the threshold frequency (ν0), no matter how intense the light.
- Above ν0, the number of electrons ejected per second is proportional to light intensity, but their maximum kinetic energy is not.
- The maximum kinetic energy of ejected electrons grows linearly with frequency of the incident light.
- There is no detectable time lag between light arriving and electrons coming out.
Classical wave theory predicts instead that with enough time or intensity, any frequency should eject electrons. Einstein explained the experiment in one sentence: one photon transfers all its energy to one electron. Energy conservation gives:
hν = W0 + ½mev2
KEmax = h(ν − ν0)
KEmax = h(ν − ν0)
where W0 = hν0 is the work function of the metal (binding energy of the least-tightly-held electron). Einstein received the 1921 Nobel Prize for this idea.
2.3.3 Dual Behaviour of Electromagnetic Radiation
Light diffracts (wave behaviour: Young's double slit) and it kicks electrons out of metals (particle behaviour: photoelectric effect). Neither description alone is complete; both are needed depending on the experiment. A photon of frequency ν simultaneously has wave attributes λ, ν and particle attributes E = hν, momentum p = h/λ.
2.3.4 Emission and Absorption Spectra
Pass white light through a prism and a smooth rainbow results — a continuous spectrum (examples: sunlight, the filament of an incandescent bulb). But pass an electric discharge through hydrogen gas and then through a prism, and you find only a few sharp bright lines — a line spectrum (or atomic spectrum). Each element's line spectrum is unique — its "fingerprint". Sodium gives the famous yellow double line; hydrogen has a rich series of lines throughout the spectrum.
The Hydrogen Emission Spectrum
Between 1885 and 1909, Balmer, Lyman, Paschen, Brackett and Pfund catalogued five series of lines from atomic hydrogen. Their wave numbers fit a single empirical formula discovered by Rydberg:
\(\bar{\nu} = R_H \left[\dfrac{1}{n_1^{\,2}} - \dfrac{1}{n_2^{\,2}}\right]\), RH = 109 677 cm−1, n2 > n1
| Series | n1 | n2 | Region |
|---|---|---|---|
| Lyman | 1 | 2, 3, 4, … | Ultraviolet |
| Balmer | 2 | 3, 4, 5, … | Visible |
| Paschen | 3 | 4, 5, 6, … | Infrared |
| Brackett | 4 | 5, 6, 7, … | Infrared |
| Pfund | 5 | 6, 7, 8, … | Far infrared |
Worked Numericals
Problem 2.6 — Energy of one mole of photons at ν = 5 × 1014 Hz
Energy of one photon: \( E = h\nu = (6.626\times10^{-34})(5\times10^{14}) = 3.313\times10^{-19}\ \text{J} \)
Energy of one mole: \( E_{\text{mol}} = N_A E = (6.022\times10^{23})(3.313\times10^{-19}) \)
\[ E_{\text{mol}} = 1.995\times10^{5}\ \text{J mol}^{-1} = 199.5\ \text{kJ mol}^{-1} \]
Problem 2.7 — Photons per second from a 100 W bulb emitting at 400 nm
Energy of one photon:
\[ E = \dfrac{hc}{\lambda} = \dfrac{(6.626\times10^{-34})(3\times10^{8})}{400\times10^{-9}} = 4.97\times10^{-19}\ \text{J} \]
Photons per second = power ÷ energy per photon:
\[ n = \dfrac{100}{4.97\times10^{-19}} = 2.01\times10^{20}\ \text{photons s}^{-1} \]
Problem 2.8 — Work function and threshold frequency
The work function of caesium is 2.14 eV. (a) Find ν0. (b) If light of λ = 300 nm falls on it, find KEmax of the ejected electron. (1 eV = 1.602 × 10−19 J)
(a) \( W_0 = 2.14 \times 1.602\times10^{-19} = 3.428\times10^{-19}\ \text{J} \)
\( \nu_0 = W_0/h = 3.428\times10^{-19}/6.626\times10^{-34} = 5.17\times10^{14}\) Hz
(b) \( \nu = c/\lambda = (3\times10^{8})/(300\times10^{-9}) = 1.0\times10^{15}\) Hz
\[ KE_{\max}=h(\nu-\nu_0)=(6.626\times10^{-34})(1.0\times10^{15}-5.17\times10^{14}) \]
\[ KE_{\max} = 3.20\times10^{-19}\ \text{J} = 2.00\ \text{eV} \]
Example — Wavelength of the first (Hα) line of the Balmer series
For Balmer, \(n_1=2\); first line has \(n_2=3\).
\[ \bar{\nu} = R_H\left[\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\right] = 109677\left[\dfrac{1}{4}-\dfrac{1}{9}\right] \]
\[ \bar{\nu} = 109677 \times \dfrac{5}{36} = 15\,233\ \text{cm}^{-1} \]
\[ \lambda = 1/\bar{\nu} = 6.565\times10^{-5}\ \text{cm} = 656.5\ \text{nm (red)} \]
This is the red Hα line seen in every hydrogen discharge tube.
Example — Energy of a photon of wavelength 200 nm
\[ E = \dfrac{hc}{\lambda} = \dfrac{(6.626\times10^{-34})(3\times10^{8})}{200\times10^{-9}} \]
\[ E = 9.94\times10^{-19}\ \text{J} = 6.20\ \text{eV} \]
This UV photon is energetic enough to ionise atomic hydrogen (IP = 13.6 eV? actually not — 6.20 eV is below H ionisation, but it is above the work functions of most alkali metals, so it could trigger photoemission from caesium or sodium).
Example — de Broglie wavelength of a fast electron
Find λ for an electron moving at 106 m s−1.
\[ \lambda = \dfrac{h}{m_e v} = \dfrac{6.626\times10^{-34}}{(9.11\times10^{-31})(10^{6})} \]
\[ \lambda = 7.27\times10^{-10}\ \text{m} = 7.27\ \text{Å} \]
Comparable to interatomic distances — this is why electrons diffract from crystal lattices (Davisson–Germer, 1927).
Example — Shortest and longest wavelengths in the Lyman series
Lyman: \(n_1=1\). Longest λ ↔ smallest \(\bar{\nu}\) ↔ \(n_2=2\):
\( \bar{\nu} = 109677(1 - 1/4) = 82\,258\ \text{cm}^{-1}\), λ = 121.6 nm.
Shortest λ ↔ \(n_2\to\infty\): \( \bar{\nu} = 109\,677\) cm−1, λ = 91.2 nm (series limit).
Activity 2.2 — Why a Red Lamp Can Never Trigger Photoemission in Zinc L4 Analyse
Objective: Use the Einstein photoelectric equation to argue that intensity cannot substitute for frequency.
- Work function of zinc ≈ 4.3 eV. Convert to the threshold frequency ν0.
- A red bulb emits λ ≈ 700 nm. Compute the corresponding photon frequency and energy in eV.
- Compare with 4.3 eV — does any single photon have enough energy?
- Now imagine a 10 000-watt red searchlight shining on the zinc. Discuss, using the one-photon-one-electron rule, why the intensity does not help.
Predict first: Will increasing the intensity of red light beyond any reasonable value eject electrons from zinc?
ν0 = 4.3 × 1.602 × 10−19 / 6.626 × 10−34 ≈ 1.04 × 1015 Hz. A red photon at 700 nm has ν = 4.29 × 1014 Hz — less than a quarter of ν0. Photon energy ≈ 1.77 eV is far below the 4.3 eV binding energy, so no single photon can eject an electron. Intensity only increases the number of photons per second, not the energy of each one. Therefore zinc stays electrically silent under red light, regardless of brightness.
Photon Energy Calculator
Enter either a wavelength or a frequency and the tool returns photon energy in joules and eV, plus the energy of one mole of photons.
Enter a value and press Compute.
Competency-Based Questions
A physics lab uses a monochromatic 500-nm source and a photocell coated with a metal of work function 1.9 eV. Later, the same lab records the hydrogen emission spectrum through a diffraction grating.
Q1. The frequency of the 500-nm light is closest to:
B. ν = c/λ = 3×108/5×10−7 = 6×1014 Hz.
Q2. Determine KEmax of photoelectrons ejected from the metal by the 500-nm light.
Photon energy = hc/λ = 3.97 × 10−19 J = 2.48 eV. KEmax = 2.48 − 1.9 = 0.58 eV ≈ 9.3 × 10−20 J.
Q3. Fill in the blank: The series of hydrogen lines lying in the visible range is called the ______ series.
Balmer (n1 = 2).
Q4. State True or False: Increasing the intensity of light below the threshold frequency eventually causes photoemission. Justify.
False. Einstein's equation shows photoemission requires one photon with energy ≥ W0. Intensity only increases photon flux, not per-photon energy.
Q5. Calculate the wavelength of the spectral line emitted when an electron in a hydrogen atom falls from n = 4 to n = 2.
\(\bar{\nu} = 109677(1/4 - 1/16) = 109677(3/16) = 20 564\) cm−1. λ = 1/\(\bar{\nu}\) = 4.86 × 10−5 cm = 486 nm (blue-green, Hβ line).
Assertion–Reason Questions
Options: A. Both A and R true, R explains A. B. Both true, R does NOT explain A. C. A true, R false. D. A false, R true.
Assertion: A photon of red light cannot eject electrons from a metal whose work function corresponds to blue light.
Reason: Red photons carry less energy per photon than blue photons.
A. Both true; lower per-photon energy is exactly why red light fails even if intensity is high.
Assertion: The Balmer series lies in the visible region of the spectrum.
Reason: All Balmer lines end on the n = 1 level of hydrogen.
C. Assertion is true, but Balmer lines end on n = 2 (Lyman lines end on n = 1).
Assertion: Electromagnetic radiation exhibits both wave and particle character.
Reason: Diffraction requires a wave description while the photoelectric effect requires a particle description.
A. Classic justification — different experiments reveal different facets; together they define the dual nature.
Frequently Asked Questions — Electromagnetic Waves, Planck's Quantum Theory and Hydrogen Spectrum
What is electromagnetic radiation?
Electromagnetic radiation is energy that travels through space as oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation. It includes radio waves, microwaves, infrared, visible light, ultraviolet, X-rays and gamma rays. In NCERT Class 11 Chemistry, key properties are wavelength (λ), frequency (ν), wavenumber (ν̄ = 1/λ) and speed (c = 3 × 10⁸ m/s in vacuum). The relationship c = νλ links them. Different regions of the electromagnetic spectrum interact differently with matter and underpin spectroscopy techniques used to analyse atomic structure.
What is black body radiation and how did it lead to quantum theory?
A black body is an idealised object that absorbs and emits all electromagnetic radiation falling on it. Classical physics could not explain the observed spectrum of black body radiation at different temperatures — it predicted infinite energy at short wavelengths (the ultraviolet catastrophe). In 1900, Max Planck solved this by proposing that energy is emitted in discrete packets (quanta) of size E = hν, not continuously. This radical assumption matched experiment perfectly and gave birth to quantum theory. NCERT Class 11 Chemistry presents this as the first crack in classical physics.
What is the photoelectric effect?
The photoelectric effect is the emission of electrons from a metal surface when light of sufficient frequency strikes it. Classical wave theory could not explain three observations: (1) emission only above a threshold frequency, regardless of intensity, (2) kinetic energy of emitted electrons depends on frequency, not intensity, (3) emission is instantaneous. Einstein explained the effect in 1905 using Planck's quantum idea: light consists of photons with energy E = hν. He won the 1921 Nobel Prize for this. The equation hν = W + ½mv² (work function W) is examined in NCERT Class 11 Chemistry.
What is the hydrogen line spectrum?
When hydrogen gas in a discharge tube is excited, it emits light at specific wavelengths only, producing a line spectrum (not continuous). The visible region shows four lines (Balmer series): red, blue-green, blue and violet. Other series occur in UV (Lyman), IR (Paschen, Brackett, Pfund). Each line corresponds to an electron transition between specific energy levels. The wavelengths are given by the Rydberg formula 1/λ = R(1/n₁² − 1/n₂²) where R = 1.097 × 10⁷ m⁻¹. NCERT Class 11 Chemistry uses this spectrum to motivate the Bohr model of the atom.
What is the Rydberg formula and what is its significance?
The Rydberg formula 1/λ = R_H (1/n₁² − 1/n₂²) calculates the wavelength of spectral lines in the hydrogen atom, where R_H = 1.097 × 10⁷ m⁻¹ is the Rydberg constant, n₁ is the lower energy level and n₂ is the higher level. For the Lyman series n₁ = 1, Balmer n₁ = 2, Paschen n₁ = 3, etc. The formula was originally empirical, but Bohr derived it theoretically from his model. In NCERT Class 11 Chemistry, it is used to predict wavelengths for various series and is a frequent CBSE board exam topic.
How is energy of a photon calculated?
The energy of a photon is calculated using Planck's equation E = hν, where h = 6.626 × 10⁻³⁴ J·s is Planck's constant and ν is the frequency in hertz. Alternatively, E = hc/λ where c is the speed of light and λ is wavelength. For example, a photon of green light with λ = 500 nm has E = (6.626 × 10⁻³⁴ × 3 × 10⁸) / (500 × 10⁻⁹) = 3.97 × 10⁻¹⁹ J. NCERT Class 11 Chemistry uses these calculations extensively in problems involving the photoelectric effect, atomic transitions and ionisation energy.
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