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Structure of Atom - Developments Leading to Bohr's Model (Electromagnetic Radiation, Planck's Quantum Theory, Photoelectric Effect, Atomic Spectra)

Grade 11CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electromagnetic Radiation (EMR) exhibits dual nature: wave-like and particle-like properties. The wave nature is characterized by frequency (ν\nu), wavelength (λ\lambda), and velocity (cc), related by c=νλc = \nu \lambda.

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Planck's Quantum Theory: Atoms and molecules can emit or absorb energy only in discrete quantities called 'quanta'. The energy of a quantum is proportional to its frequency: E=hνE = h\nu.

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Photoelectric Effect: When light of a certain minimum frequency (threshold frequency, ν0\nu_0) strikes the surface of a metal, electrons are ejected. The kinetic energy of these electrons depends on the frequency of incident light: hν=hν0+K.E.h\nu = h\nu_0 + K.E.

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Emission Spectra: Produced when radiation from an excited sample is passed through a prism. It consists of bright lines on a dark background. The Hydrogen spectrum consists of series like Lyman (n1=1n_1=1), Balmer (n1=2n_1=2), Paschen (n1=3n_1=3), etc.

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Absorption Spectra: Produced when white light passes through a sample. It consists of dark lines in a continuous spectrum, corresponding to the wavelengths absorbed by the substance.

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Rydberg Formula: Used to calculate the wavenumber (νˉ\bar{\nu}) of lines in the hydrogen spectrum: νˉ=RH(1n12−1n22)\bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right).

📐Formulae

c=νλc = \nu \lambda

νˉ=1λ\bar{\nu} = \frac{1}{\lambda}

E=hν=hcλE = h\nu = \frac{hc}{\lambda}

hν=W0+12mev2 (where W0=hν0 is the work function)h\nu = W_0 + \frac{1}{2}m_e v^2 \text{ (where } W_0 = h\nu_0 \text{ is the work function)}

νˉ=109,677(1n12−1n22) cm−1\bar{\nu} = 109,677 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ cm}^{-1}

c=3.0×108 m s−1,h=6.626×10−34 J sc = 3.0 \times 10^8 \text{ m s}^{-1}, h = 6.626 \times 10^{-34} \text{ J s}

💡Examples

Problem 1:

Calculate the energy of one mole of photons of radiation whose frequency is 5×1014 Hz5 \times 10^{14} \text{ Hz}.

Solution:

Energy of one photon: E=hνE = h\nu. For one mole of photons: E=NAhνE = N_A h \nu. Given ν=5×1014 s−1\nu = 5 \times 10^{14} \text{ s}^{-1}, h=6.626×10−34 J sh = 6.626 \times 10^{-34} \text{ J s}, and NA=6.022×1023 mol−1N_A = 6.022 \times 10^{23} \text{ mol}^{-1}. E=(6.022×1023)×(6.626×10−34)×(5×1014)E = (6.022 \times 10^{23}) \times (6.626 \times 10^{-34}) \times (5 \times 10^{14}) E=199.51×103 J mol−1=199.51 kJ mol−1E = 199.51 \times 10^3 \text{ J mol}^{-1} = 199.51 \text{ kJ mol}^{-1}

Explanation:

We use Planck's equation and multiply by Avogadro's number to find the energy per mole.

Problem 2:

When electromagnetic radiation of wavelength 300 nm300 \text{ nm} falls on the surface of sodium, electrons are emitted with a kinetic energy of 1.68×105 J mol−11.68 \times 10^5 \text{ J mol}^{-1}. What is the minimum energy needed to remove an electron from sodium?

Solution:

Energy of incident photon (EE): E=hcλ=6.626×10−34×3×108300×10−9=6.626×10−19 JE = \frac{hc}{\lambda} = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} = 6.626 \times 10^{-19} \text{ J} Energy per mole of photons: Emole=6.626×10−19×6.022×1023=3.99×105 J mol−1E_{mole} = 6.626 \times 10^{-19} \times 6.022 \times 10^{23} = 3.99 \times 10^5 \text{ J mol}^{-1} Minimum energy (W0W_0) per mole: W0=Emole−K.E.W_0 = E_{mole} - K.E. 399000−168000231000\begin{array}{r} 399000 \\ - 168000 \\ \hline 231000 \end{array} W0=2.31×105 J mol−1W_0 = 2.31 \times 10^5 \text{ J mol}^{-1} Minimum energy per atom: 2.31×1056.022×1023=3.84×10−19 J\frac{2.31 \times 10^5}{6.022 \times 10^{23}} = 3.84 \times 10^{-19} \text{ J}

Explanation:

Calculated the total energy of incoming light and subtracted the kinetic energy of emitted electrons to find the threshold energy (work function).

Problem 3:

What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n=4n=4 to an energy level with n=2n=2?

Solution:

Using Rydberg formula: νˉ=109,677(122−142) cm−1\bar{\nu} = 109,677 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \text{ cm}^{-1} νˉ=109,677(14−116)=109,677(316)=20564.44 cm−1\bar{\nu} = 109,677 \left( \frac{1}{4} - \frac{1}{16} \right) = 109,677 \left( \frac{3}{16} \right) = 20564.44 \text{ cm}^{-1} λ=1νˉ=120564.44 cm≈4.86×10−5 cm=486 nm\lambda = \frac{1}{\bar{\nu}} = \frac{1}{20564.44} \text{ cm} \approx 4.86 \times 10^{-5} \text{ cm} = 486 \text{ nm}

Explanation:

The transition belongs to the Balmer series since n1=2n_1=2. The resulting wavelength is in the visible region (blue-green light).