### Bohr's atomic theory

- According to Rutherford's atomic theory, the entire mass of an atom was concentrated in a tiny positively charged nucleus around which the extranuclear electrons are moving in constant motion in the orbit.

- But a moving charged particle emits radiation, will then loss kinetic energy and eventually hit the nucleus.

- To resolve the anomalous position Bohr's atomic theory was proposed which is based on several novel postulates.

Atomic theory |

- The following formulas will be widely used in solving the questions of Bohr's atomic theory.

- The energy of an electron in the first Bohr orbit of the hydrogen atom is -13.6 eV. What is the energy value of the electron in the excited state of Li⁺²?

The energy of an electron in nth orbit,

En = (E₁/n²) × Z²

For Li⁺², Z = 3 and in the excited state of Li⁺², n = 2

Putting the values on the above equation we have,

Energy of Li⁺² = (-13.6/4) × 3²

= - 30.6 eV

QuestionEn = (E₁/n²) × Z²

For Li⁺², Z = 3 and in the excited state of Li⁺², n = 2

Putting the values on the above equation we have,

Energy of Li⁺² = (-13.6/4) × 3²

= - 30.6 eV

- What is the radius of the second orbit of a hydrogen atom? Give radius of first orbit = 0.529 Ã….

The radius of an electron in nth orbit,

rn = r₁ × n²

For second orbit n = 2

Thus the radius of the second orbit of a hydrogen atom,

r₂ = 0.529 × 2²

= 2.12 Ã…

rn = r₁ × n²

For second orbit n = 2

Thus the radius of the second orbit of a hydrogen atom,

r₂ = 0.529 × 2²

= 2.12 Ã…

### Quantum number orbitals

- For the size shape and orientation of an orbital, we need to know about the four quantum numbers and these quantum numbers are designed as, principal quantum number(n), azimuthal quantum number(l), magnetic quantum number(m) and spin quantum number(s).

- The following rules will be widely used in solving the questions of quantum number orbitals.

Principal quantum number is denoted by n

It can have integral values from 1 to ∞

The azimuthal quantum number denoted by l

For a given value of n

l = 0, 1, 2, 3, ........ (n -1)

Total number of different values of l equal to n

The magnetic quantum number is denoted by ml

For a given value of l

ml = +l to -l

Total number of different values of ml equal to (2l + 1)

QuestionIt can have integral values from 1 to ∞

The azimuthal quantum number denoted by l

For a given value of n

l = 0, 1, 2, 3, ........ (n -1)

Total number of different values of l equal to n

The magnetic quantum number is denoted by ml

For a given value of l

ml = +l to -l

Total number of different values of ml equal to (2l + 1)

- Write the correct set of four quantum numbers for the valence electron of Rubidium.

- The atomic number of Rubidium atom is 37. Thus the electronic configuration of rubidium is,

1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3d¹⁰4P⁶5S¹

or, [Kr]³⁶ 5S¹

or, [Kr]³⁶ 5S¹

- So the valence electron of rubidium means 5S¹ electron of a rubidium atom. Thus for 5S¹ electron, principal quantum number = 5, azimuthal quantum number = 0, magnetic quantum number = 0, and spin quantum number = +½.

- So the correct set of four quantum numbers for the valence electron of Rubidium atom are,

5, 0, 0, +½

Question- What are the four quantum numbers of the 19th electron of Cr atom?

- The four quantum numbers of the 19th electron of Cr atom are 4, 0, 0, +½

- How many electrons in an atom can have the following quantum numbers n=4 and l=1?

- n = 4 means principal quantum number 4 and l = 1 means the azimuthal quantum number is 1, that is P orbital.

- Thus when n =1 and l = 0 we can describe 4P orbital. For P orbital magnetic quantum numbers = -1, 0, +1 and each magnetic quantum number has two spin quantum numbers.

- So (2 × 3) = 6 electrons in an atom can have the following quantum numbers n =4 and l =1.

- How many possible orbitals are there for n = 4?

- n = 4 means principal quantum number 4.

- Thus when n =4, l = 0, 1, 2, 3. That is 4S, 4P, 4d, and 4f orbitals. S orbital has one orientation of space, p orbitals have three orientation in space, d orbitals have five orientation in space, and f orbitals have seven orientation in space.

- Thus the number of possible orbitals when n = 4 is (1 + 3 + 5 + 7) = 16

- How many possible orbitals are there for n = 3, l =1, and m

_{l}=0 ?

- n = 3 means principal quantum number 3, l = 1 means azimuthal quantum number = 0, and ml = 0 means magnetic quantum number zero that is one orientation in space.

- Thus the number of orbitals is 1, 3S orbital.

- An electron with the highest value of the magnetic quantum number as 3, what is its principal quantum number?

- If the highest value of the magnetic quantum number is 3, then the highest value of the azimuthal quantum number is 3.

- Thus the principal quantum number of this orbital is (l+1) = 3+1 = 4.

- What is the wavelength of the HÉ‘ line of the Balmer series of hydrogen?

- The wavenumber of the Balmer lines, ⊽ = 1/Î» = 109677[(1/2²) - (1/n

_{₂}²)]

Again for HÉ‘ line n2 = 3

Thus the wavelength for HÉ‘ - line, 1/Î» = 109677 [(1/2²) - (1/3²)]

= 109677 × (5/36)

∴ Î» = 36/(5 × 109677)

= 6.564 × 10⁻⁵ cm

= 656.5 nm

QuestionThus the wavelength for HÉ‘ - line, 1/Î» = 109677 [(1/2²) - (1/3²)]

= 109677 × (5/36)

∴ Î» = 36/(5 × 109677)

= 6.564 × 10⁻⁵ cm

= 656.5 nm

- The lowest wavelength of the Lyman series of the hydrogen atom is x. What is the highest wavelength of the Paschen series of the He⁺² ion?

- The highest wavenumber of the Lyman lines, ⊽max = 1/Î»min = R[(1/1²) - (1/∞²)]

or, 1/x = R

or, R = 1/x

or, R = 1/x

- The lowest wavenumber of the Paschen series of He⁺² ion is,

⊽min = 1/Î»max = R Z² [(1/3²) - (1/4²)]

or, Î»max = (9 × 16)/(4 × 7R)

∴ Î»max = 36x/7

Questionor, Î»max = (9 × 16)/(4 × 7R)

∴ Î»max = 36x/7

- How many photons of light having the wavenumber a is necessary to provide 3 J of energy?

- From the plank theory,

E = nhÎ½

Where n is the number of photons and E is the energy of the photon source.

3 = n h c ⊽ where ⊽ = wave number

∴ n = 3/hca

QuestionWhere n is the number of photons and E is the energy of the photon source.

3 = n h c ⊽ where ⊽ = wave number

∴ n = 3/hca

- What is the De-Broglie wavelength of an electron orbiting in the n = 6 state of a hydrogen atom? (r₀ = Bohr's radius)

We know that 2Ï€r = nÎ»

or, Î» = 2Ï€r/n

where r = 6² × r₀ =36 r₀

where r = 6² × r₀ =36 r₀

∴ Î» = (2Ï€ ×36 r₀)/6 = 12Ï€r₀

Question- Find out the number of unpaired electrons of an ion M

^{+x}(Z of M = 25) with the magnetic moment √15 B. M. and also find out the value of x?

We know that magnetic moment = √n(n+2)

Thus √15 = √n(n+2)

or, √3(3+2) = √n(n+2)

∴ n = 3

Thus √15 = √n(n+2)

or, √3(3+2) = √n(n+2)

∴ n = 3

Again the electronic configuration of M is [Ar]¹⁸ 4S² 3d⁵

- We find out that the number of unpaired electrons of M

^{+x}ion is 3. For this electronic configuration, we need to remove four electrons from its valence shell.

Thus the electronic configuration of M is

[Ar]¹⁸ 3d³

Thus x = 4

[Ar]¹⁸ 3d³

Thus x = 4