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Structure of Atom - Bohr's model-advanced

Grade 9CBSE

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

🔑Concepts

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Neils Bohr proposed that electrons revolve around the nucleus in fixed circular paths called discrete orbits or stationary shells. In these orbits, electrons do not radiate energy.

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These orbits or energy levels are designated by the letters K,L,M,NK, L, M, N or the integers n=1,2,3,4n = 1, 2, 3, 4.

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Energy is emitted or absorbed only when an electron jumps from one orbit to another. The energy difference is given by ΔE=E2−E1\Delta E = E_2 - E_1.

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Bohr-Bury Scheme: The maximum number of electrons that can be accommodated in a shell is calculated using the formula 2n22n^2.

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Octet Rule: The maximum number of electrons that can be accommodated in the outermost shell is 88, except for the KK shell, which is stable with 22 electrons (Duplet rule).

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Advanced Postulate: The angular momentum of an electron in a shell is quantized and is an integral multiple of h2π\frac{h}{2\pi}.

📐Formulae

Maximum electrons in a shell=2n2\text{Maximum electrons in a shell} = 2n^2

mvr=nh2πmvr = \frac{nh}{2\pi}

ΔE=hν=Efinal−Einitial\Delta E = h\nu = E_{final} - E_{initial}

💡Examples

Problem 1:

Calculate the maximum number of electrons that can be accommodated in the NN shell.

Solution:

For the NN shell, the orbit number nn is 44. Using the Bohr-Bury formula: Max electrons=2n2\text{Max electrons} = 2n^2 Max electrons=2×(4)2\text{Max electrons} = 2 \times (4)^2 Max electrons=2×16=32\text{Max electrons} = 2 \times 16 = 32

Explanation:

Based on the 2n22n^2 rule, where nn represents the shell number, the fourth shell can hold a maximum of 3232 electrons.

Problem 2:

An atom has 1313 electrons. Write its electronic configuration and identify the number of electrons in its MM shell.

Solution:

The atomic number Z=13Z = 13 (Aluminum). Distribution of electrons:

  1. KK shell (n=1n=1): 2(1)2=22(1)^2 = 2 electrons.
  2. LL shell (n=2n=2): 2(2)2=82(2)^2 = 8 electrons.
  3. MM shell (n=3n=3): Remaining electrons =13−(2+8)=3= 13 - (2 + 8) = 3. Electronic configuration: (2,8,3)(2, 8, 3). Number of electrons in MM shell is 33.

Explanation:

Electrons fill lower energy shells (KK then LL) before moving to higher energy shells (MM). The MM shell here contains the valence electrons.

Problem 3:

If the angular momentum of an electron in a certain Bohr orbit is hπ\frac{h}{\pi}, find the value of nn (shell number).

Solution:

According to Bohr's advanced postulate: mvr=nh2πmvr = \frac{nh}{2\pi} Given that the angular momentum mvr=hπmvr = \frac{h}{\pi}: hπ=nh2π\frac{h}{\pi} = \frac{nh}{2\pi} Cancelling hh and π\pi from both sides: 1=n21 = \frac{n}{2} n=2n = 2

Explanation:

The angular momentum is quantized as nh2π\frac{nh}{2\pi}. By comparing the given value to the formula, we determine that the electron is in the LL shell (n=2n=2).

Bohr's model-advanced Class 9 Notes & Examples