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Atoms - Hydrogen Line Spectra

Grade 12CBSEPhysics

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

πŸ”‘Concepts

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Spectral series are formed when an electron in a hydrogen atom makes a transition from a higher energy state nin_i to a lower energy state nfn_f.

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Lyman Series: Transitions to nf=1n_f = 1 from ni=2,3,4,…n_i = 2, 3, 4, \dots. These lines lie in the Ultraviolet (UV) region.

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Balmer Series: Transitions to nf=2n_f = 2 from ni=3,4,5,…n_i = 3, 4, 5, \dots. These lines lie in the Visible region.

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Paschen Series: Transitions to nf=3n_f = 3 from ni=4,5,6,…n_i = 4, 5, 6, \dots. These lines lie in the Infrared (IR) region.

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Brackett Series: Transitions to nf=4n_f = 4 from ni=5,6,7,…n_i = 5, 6, 7, \dots. These lines lie in the Far-Infrared region.

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Pfund Series: Transitions to nf=5n_f = 5 from ni=6,7,8,…n_i = 6, 7, 8, \dots. These lines lie in the Far-Infrared region.

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The energy of an electron in the nthn^{th} orbit is given by En=βˆ’13.6n2Β eVE_n = -\frac{13.6}{n^2} \text{ eV}. The negative sign indicates the electron is bound to the nucleus.

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The shortest wavelength in a series (series limit) occurs when ni=∞n_i = \infty.

πŸ“Formulae

1Ξ»=R(1nf2βˆ’1ni2)\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)

En=βˆ’me48Ο΅02h2n2β‰ˆβˆ’13.6n2Β eVE_n = -\frac{me^4}{8\epsilon_0^2 h^2 n^2} \approx -\frac{13.6}{n^2} \text{ eV}

hΞ½=Eniβˆ’Enfh\nu = E_{n_i} - E_{n_f}

R=me48Ο΅02ch3β‰ˆ1.097Γ—107Β mβˆ’1R = \frac{me^4}{8\epsilon_0^2 ch^3} \approx 1.097 \times 10^7 \text{ m}^{-1}

L=mvr=nh2Ο€L = mvr = \frac{nh}{2\pi}

πŸ’‘Examples

Problem 1:

Calculate the wavelength of the first line (HΞ±H_\alpha line) of the Balmer series of the hydrogen spectrum. (Given R=1.097Γ—107Β mβˆ’1R = 1.097 \times 10^7 \text{ m}^{-1})

Solution:

For the first line of the Balmer series, nf=2n_f = 2 and ni=3n_i = 3. Using the Rydberg formula: 1Ξ»=R(122βˆ’132)\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) 1Ξ»=R(14βˆ’19)=R(536)\frac{1}{\lambda} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) Ξ»=365R=365Γ—1.097Γ—107β‰ˆ6.563Γ—10βˆ’7Β m=656.3Β nm\lambda = \frac{36}{5R} = \frac{36}{5 \times 1.097 \times 10^7} \approx 6.563 \times 10^{-7} \text{ m} = 656.3 \text{ nm}

Explanation:

The first line of any series corresponds to the transition from the immediately higher orbit. For Balmer, it starts at n=2n=2, so the first line is from n=3n=3 to n=2n=2.

Problem 2:

Determine the ionization energy of a hydrogen atom in its ground state.

Solution:

Ionization energy is the energy required to remove the electron from n=1n = 1 to n=∞n = \infty. E1=βˆ’13.612=βˆ’13.6Β eVE_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV} E∞=0Β eVE_{\infty} = 0 \text{ eV} Ξ”E=Eβˆžβˆ’E1=0βˆ’(βˆ’13.6)=13.6Β eV\Delta E = E_{\infty} - E_1 = 0 - (-13.6) = 13.6 \text{ eV}

Explanation:

Ionization refers to the transition from the ground state (n=1n=1) to the free state (n=∞n=\infty). The energy difference is the ionization potential.