Structure of Atom

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V
Vishal Baghel

Contributor-Level 10

nm = 2l + 1
As there are (2l + 1) number of permissible values of magnetic quantum number.
Hence, l = (nm - 1)/2

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Vishal Baghel

Contributor-Level 10

An Atom has three fundamental particles-electron, proton and neutron. The mass of the electron is 9.10939 * 10? ³¹ kg. Neutrons and protons, both are collectively known as nucleons. All the isotopes of a given element show same chemical properties. Dalton's atomic theory, regarded atom as an ultimate particle of matter.

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Vishal Baghel

Contributor-Level 10

For irreversible expansion of an ideal gas under isothermal condition
ΔU = 0, ΔS (Total) ≠ 0

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Vishal Baghel

Contributor-Level 10

The number of Body centred unit cells in all 14 types of Bravais lattice unit cells is 3.

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Vishal Baghel

Contributor-Level 10

105.8 * (2/4) = 52.9 pm ⇒ rLi+ + rX- = 52.9 * (3²/3) = 158.7 pm

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Vishal Baghel

Contributor-Level 10

Shapes of d? ²? ² and d? ² are not similar to each other.

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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Raj Pandey

Contributor-Level 9

Bohr's theory accounts for the line spectrum of single electron species but Li? has two electrons. Bohr's theory fails to explain splitting of spectral lines in presence of magnetic field i.e. Zeeman effect.

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Raj Pandey

Contributor-Level 9

Combustion of ethane: C? H? + 7/2 O? → 2CO? + 3H? O.

Moles of C? H? = 3g / 30 g/mol = 0.1 mol.

Moles of H? O produced = 0.3 mol.

Number of H? O molecules = 0.3 * 6.023 * 10²³ ≈ 18.06 * 10²².

So, x = 18.

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Raj Pandey

Contributor-Level 9

Quantum Numbers and Orbitals:

o   Total Node = n - 1 (l => Azimuthal Q.N)

o   Radial Node = n - l - 1

o   Angular Node: l

o   If Angular node = 0, then l = 0, i.e., S orbital.

o   If Radial nodes = 2, then n - l - 1 = 2.

o   Substituting l = 0 gives n - 0 - 1 = 2, so n = 3.

o   Therefore, the orbital is 3s.

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