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Structure of Atom - Line Spectrum of Hydrogen-advanced

Grade 9CBSE

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

🔑Concepts

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According to Bohr's model, electrons in a Hydrogen atom revolve in fixed circular orbits called energy levels or shells, represented by n=1,2,3,…n = 1, 2, 3, \dots.

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Energy is quantized, meaning an electron can only exist in these specific energy levels and not in between them.

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When an electron absorbs energy, it jumps from a lower energy level (ground state) to a higher energy level (excited state). This is called absorption.

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When an excited electron falls back to a lower energy level, it emits energy in the form of electromagnetic radiation (photons). The energy of the photon matches the difference between the two levels: ΔE=Ehigher−Elower\Delta E = E_{higher} - E_{lower}.

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Because the energy levels are fixed, the emitted photons have specific wavelengths, creating a 'line spectrum' rather than a continuous rainbow.

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The Balmer Series consists of the spectral lines emitted when electrons transition from higher levels (n>2n > 2) down to the second energy level (n=2n = 2). These lines fall in the visible light spectrum.

📐Formulae

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

ΔE=hν=hcλ\Delta E = h\nu = \frac{hc}{\lambda}

1λ=RH(1n12−1n22)\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

RH≈1.09677×107 m−1R_H \approx 1.09677 \times 10^7 \text{ m}^{-1}

Angular Momentum (L)=mvr=nh2π\text{Angular Momentum } (L) = mvr = \frac{nh}{2\pi}

💡Examples

Problem 1:

Calculate the energy required to excite an electron in a Hydrogen atom from the ground state (n=1n=1) to the first excited state (n=2n=2).

Solution:

The energy of the nthn^{th} shell is given by: En=−13.6n2 eVE_n = -\frac{13.6}{n^2} \text{ eV} For the ground state (n=1n=1): E1=−13.612=−13.6 eVE_1 = -\frac{13.6}{1^2} = -13.6 \text{ eV} For the first excited state (n=2n=2): E2=−13.622=−13.64=−3.4 eVE_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \text{ eV} The energy required (ΔE\Delta E) is the difference: ΔE=E2−E1\Delta E = E_2 - E_1 ΔE=−3.4−(−13.6)\Delta E = -3.4 - (-13.6) ΔE=10.2 eV\Delta E = 10.2 \text{ eV}

Explanation:

To move an electron to a higher orbit, energy must be absorbed. The calculated value of 10.2 eV10.2 \text{ eV} represents the specific 'quantum' of energy needed for this transition.

Problem 2:

Determine the wavelength (in nm) of the photon emitted when an electron in a Hydrogen atom transitions from n=3n=3 to n=2n=2. (Take RH≈1.1×107 m−1R_H \approx 1.1 \times 10^7 \text{ m}^{-1})

Solution:

Using the Rydberg formula: 1λ=RH(1n12−1n22)\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) Given n1=2n_1 = 2 and n2=3n_2 = 3: 1λ=1.1×107(122−132)\frac{1}{\lambda} = 1.1 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) 1λ=1.1×107(14−19)\frac{1}{\lambda} = 1.1 \times 10^7 \left( \frac{1}{4} - \frac{1}{9} \right) 1λ=1.1×107(9−436)\frac{1}{\lambda} = 1.1 \times 10^7 \left( \frac{9 - 4}{36} \right) 1λ=1.1×107×536\frac{1}{\lambda} = 1.1 \times 10^7 \times \frac{5}{36} λ=365.5×107 m\lambda = \frac{36}{5.5 \times 10^7} \text{ m} λ≈6.545×10−7 m=654.5 nm\lambda \approx 6.545 \times 10^{-7} \text{ m} = 654.5 \text{ nm}

Explanation:

This transition corresponds to the first line of the Balmer series, which appears as red light in the visible spectrum.