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Structure of the Atom - Bohr’s Model of Atom

Grade 9CBSE

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

🔑Concepts

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Neils Bohr proposed the model of the atom in 19121912 to overcome the objections raised against Rutherford's model regarding the stability of an atom.

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Electrons revolve around the nucleus in certain discrete orbits known as 'Discrete Orbits of Electrons' or 'Energy Levels'.

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While revolving in these discrete orbits, the electrons do not radiate energy, which explains why the atom remains stable.

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These orbits or shells are represented by the letters K,L,M,N,…K, L, M, N, \dots or the numbers n=1,2,3,4,…n = 1, 2, 3, 4, \dots starting from the nucleus.

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The maximum number of electrons that can be accommodated in a shell is given by the formula 2n22n^2, where nn is the orbit number or energy level index.

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The maximum number of electrons that can be accommodated in the outermost shell is 88, except for the KK shell which can hold a maximum of 22 electrons.

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Electrons are not accommodated in a given shell unless the inner shells are filled; that is, the shells are filled in a stepwise manner.

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The atomic number (ZZ) of an element is equal to the number of protons in the nucleus of its atom.

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The mass number (AA) is the sum of the total number of protons and neutrons present in the nucleus of an atom: A=Z+nA = Z + n.

📐Formulae

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

Atomic Number (Z)=Number of protons (p)\text{Atomic Number } (Z) = \text{Number of protons } (p)

Mass Number (A)=Number of protons (p)+Number of neutrons (n)\text{Mass Number } (A) = \text{Number of protons } (p) + \text{Number of neutrons } (n)

Number of neutrons (n)=A−Z\text{Number of neutrons } (n) = A - Z

💡Examples

Problem 1:

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

Solution:

For the MM shell, the orbit number is n=3n = 3. Using the formula 2n22n^2, we get 2×(3)2=2×9=182 \times (3)^2 = 2 \times 9 = 18.

Explanation:

According to Bohr-Bury scheme, the third energy level (MM shell) can hold up to 1818 electrons.

Problem 2:

An atom of an element has 1111 protons and 1212 neutrons. Find its atomic number, mass number, and electronic configuration.

Solution:

Atomic number Z=p=11Z = p = 11. Mass number A=Z+n=11+12=23A = Z + n = 11 + 12 = 23. Electronic configuration: K=2,L=8,M=1K=2, L=8, M=1.

Explanation:

The atomic number is defined by the number of protons (1111). The mass number is the sum of protons and neutrons (2323). The electrons are distributed as 22 in the 1st1^{st} shell (KK), 88 in the 2nd2^{nd} shell (LL), and the remaining 11 in the 3rd3^{rd} shell (MM).

Problem 3:

If KK and LL shells of an atom are full, then what would be the total number of electrons in the atom?

Solution:

Max electrons in KK shell (n=1n=1) =2(1)2=2= 2(1)^2 = 2. Max electrons in LL shell (n=2n=2) =2(2)2=8= 2(2)^2 = 8. Total electrons =2+8=10= 2 + 8 = 10.

Explanation:

When both KK and LL shells are completely filled, the atom reaches a stable configuration with a total of 1010 electrons, which corresponds to the element Neon (NeNe).