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Structure of Atom - Achievement of Bohr Model-advanced

Grade 9CBSE

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

🔑Concepts

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Neils Bohr's model of the atom was a major achievement as it successfully explained the stability of an atom, which Rutherford's model could not.

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Electrons revolve around the nucleus in certain discrete orbits called 'Stationary States' or 'Energy Shells'. These are designated as K,L,M,N,…K, L, M, N, \dots or n=1,2,3,4,…n = 1, 2, 3, 4, \dots

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As long as an electron revolves in a particular discrete orbit, it does not radiate energy. This explains why electrons do not spiral into the nucleus.

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Bohr's Postulate of Quantization: The angular momentum of an electron is quantized and can only be an integral multiple of h2π\frac{h}{2\pi}.

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Energy is emitted or absorbed only when an electron jumps from one energy level to another. If an electron jumps from a higher level (E2E_2) to a lower level (E1E_1), energy is released as radiation of frequency ν\nu.

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The energy difference is given by ΔE=E2−E1=hν\Delta E = E_2 - E_1 = h\nu, where hh is Planck's constant.

📐Formulae

mvr=nh2πmvr = n\frac{h}{2\pi}

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

ΔE=Ehigher−Elower\Delta E = E_{higher} - E_{lower}

2n2 (Maximum number of electrons in a shell)2n^2 \text{ (Maximum number of electrons in a shell)}

💡Examples

Problem 1:

Calculate the angular momentum of an electron moving in the second orbit (LL shell) of a hydrogen atom according to Bohr's model.

Solution:

Given, orbit number n=2n = 2. According to Bohr's quantization rule: mvr=nh2πmvr = \frac{nh}{2\pi} Substituting n=2n = 2: mvr=2h2π=hπmvr = \frac{2h}{2\pi} = \frac{h}{\pi} If we take h≈6.626×10−34 J sh \approx 6.626 \times 10^{-34} \text{ J s}: mvr=6.626×10−343.141≈2.1×10−34 kg m2s−1mvr = \frac{6.626 \times 10^{-34}}{3.141} \approx 2.1 \times 10^{-34} \text{ kg m}^2 \text{s}^{-1}

Explanation:

Bohr's second postulate states that angular momentum is an integral multiple of h2π\frac{h}{2\pi}. For the second shell, n=2n=2.

Problem 2:

Determine the energy of an electron in the third energy level (n=3n=3) of a hydrogen atom.

Solution:

The energy of an electron in the nthn^{th} orbit is given by: En=−13.6n2 eVE_n = -\frac{13.6}{n^2} \text{ eV} For the third energy level, n=3n = 3: E3=−13.632E_3 = -\frac{13.6}{3^2} E3=−13.69E_3 = -\frac{13.6}{9} E3≈−1.51 eVE_3 \approx -1.51 \text{ eV}

Explanation:

The energy of orbits in a hydrogen atom is quantized and decreases (becomes less negative) as the distance from the nucleus increases.