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Atoms - Atomic Spectra

Grade 12CBSEPhysics

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

🔑Concepts

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Atomic spectra are of two types: emission spectra and absorption spectra. An emission spectrum is produced when an atom undergoes a transition from a higher energy state n2n_2 to a lower energy state n1n_1, emitting a photon of energy hν=E2−E1h\nu = E_2 - E_1.

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An absorption spectrum is produced when an atom in a lower energy state absorbs a photon and moves to a higher energy state. Only certain frequencies are absorbed, appearing as dark lines in a continuous spectrum.

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The spectral lines of the Hydrogen atom are grouped into several series named after their discoverers: Lyman (n1=1n_1=1), Balmer (n1=2n_1=2), Paschen (n1=3n_1=3), Brackett (n1=4n_1=4), and Pfund (n1=5n_1=5).

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The Balmer series is the only series that lies in the visible region of the electromagnetic spectrum. The Lyman series lies in the ultraviolet region, while Paschen, Brackett, and Pfund lie in the infrared region.

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Bohr's third postulate states that an atom radiates energy only when an electron jumps from one outer stationary orbit of higher energy EiE_i to an inner stationary orbit of lower energy EfE_f. The frequency of the emitted radiation is given by ν=Ei−Efh\nu = \frac{E_i - E_f}{h}.

📐Formulae

1λ=R(1n12−1n22)\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) where RR is the Rydberg constant ≈1.097×107 m−1\approx 1.097 \times 10^7 \text{ m}^{-1}

En=−13.6n2 eVE_n = -\frac{13.6}{n^2} \text{ eV}

hν=hcλ=En2−En1h\nu = \frac{hc}{\lambda} = E_{n_2} - E_{n_1}

L=mvr=nh2πL = mvr = \frac{nh}{2\pi}

R=me48ϵ02h3cR = \frac{me^4}{8\epsilon_0^2 h^3 c}

💡Examples

Problem 1:

Calculate the wavelength of the HαH_{\alpha} line in the Balmer series of hydrogen spectrum. (R=1.097×107 m−1R = 1.097 \times 10^7 \text{ m}^{-1})

Solution:

For the HαH_{\alpha} line in the Balmer series, n1=2n_1 = 2 and n2=3n_2 = 3. Using the Rydberg formula: 1λ=R(122−132)\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) 1λ=R(14−19)\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) 1λ=R(9−436)=5R36\frac{1}{\lambda} = R \left( \frac{9 - 4}{36} \right) = \frac{5R}{36} λ=365R=365×1.097×107\lambda = \frac{36}{5R} = \frac{36}{5 \times 1.097 \times 10^7} λ≈6.563×10−7 m=656.3 nm\lambda \approx 6.563 \times 10^{-7} \text{ m} = 656.3 \text{ nm}

Explanation:

The HαH_{\alpha} line corresponds to the first member of the Balmer series, which occurs when an electron transitions from the n=3n=3 to the n=2n=2 energy level.

Problem 2:

Calculate the energy difference between n=2n=2 and n=1n=1 levels in a Hydrogen atom and the corresponding frequency of emitted radiation.

Solution:

Energy at n=1n=1: E1=−13.612=−13.6 eVE_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV} Energy at n=2n=2: E2=−13.622=−3.4 eVE_2 = -\frac{13.6}{2^2} = -3.4 \text{ eV} Energy difference ΔE\Delta E: −3.4−(−13.6)10.2\begin{array}{r} -3.4 \\ -(-13.6) \\ \hline 10.2 \end{array} ΔE=10.2 eV=10.2×1.6×10−19 J=1.632×10−18 J\Delta E = 10.2 \text{ eV} = 10.2 \times 1.6 \times 10^{-19} \text{ J} = 1.632 \times 10^{-18} \text{ J} Frequency ν=ΔEh\nu = \frac{\Delta E}{h}: ν=1.632×10−186.63×10−34≈2.46×1015 Hz\nu = \frac{1.632 \times 10^{-18}}{6.63 \times 10^{-34}} \approx 2.46 \times 10^{15} \text{ Hz}

Explanation:

The energy of the emitted photon is exactly equal to the difference in energy levels of the electron transition.