### Rutherford model to Bohr's model of the hydrogen atom

According to Rutherford's model, 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 model was proposed which is based on several novel postulates. To study questions and answer on Bohr's model following formula are widely used.

Bohr's model for the hydrogen atom |

The energy of an electron in the first energy level of the hydrogen atom = -13.6 eV. Calculate the energy of an electron in the excited level of helium ion.

Answer

The energy of an electron

En = (E₁/n²) × Z²

where E₁ = energy of an electron of the hydrogen atom.

En = (E₁/n²) × Z²

where E₁ = energy of an electron of the hydrogen atom.

Lithium-ion has atomic number = 3 and the excited energy level of lithium-ion means second energy level where n = 2.

∴ The energy of an electron in the excited energy level of lithium-ion

= (-13.6/4) × 3²

= - 30.6 eV.

Question= - 30.6 eV.

How to calculate the radius of the second orbit hydrogen atom if the radius of the first orbit of hydrogen atom 0.529 Ã…?

Answer

The radius of an electron of hydrogen

rn = r₁ × n²

where r₁ = radius of the first orbit.

∴ The radius of the second orbit of the hydrogen atom

r₂ = 0.529 × 2²

= 2.12 Ã…

rn = r₁ × n²

where r₁ = radius of the first orbit.

∴ The radius of the second orbit of the hydrogen atom

r₂ = 0.529 × 2²

= 2.12 Ã…

### Quantum numbers and orbitals of an atom

Study the size shape and orientation of an orbital, we need to know about the four quantum numbers.- Principal quantum number(n)
- Azimuthal quantum number(l)
- Magnetic quantum number(m)
- Spin quantum number(s).

The principal quantum number coming from Bohr's model of an atom and it has integral values from 1 to ∞

The azimuthal quantum number denoted by l.

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

The magnetic quantum number denoted by ml.

ml have values from +l to -l.

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

The magnetic quantum number denoted by ml.

ml have values from +l to -l.

What are the four quantum numbers for the valence shell electron of rubidium atom?

Answer

Rubidium atom has atomic number 37 and electronic configuration

1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3d¹⁰4P⁶5S¹

or, [Kr]³⁶ 5S¹.

- The principal quantum number for 5S¹ = 5.
- The azimuthal quantum number for 5S¹= 0.
- The magnetic quantum number for 5S¹ = 0.
- The spin quantum number for 5S¹ = +½.

Question

What are the four quantum numbers of the 19th electron of chromium?

Answer

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

Chemistry articles for school-college students

- What is Chemical equilibrium?
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- Properties of gases
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- Crystalline solid

How many electrons in an atom can have principal quantum number four and azimuthal quantum number one?

Answer

Principal quantum number = 4 and the azimuthal quantum number = 1. This designated the 4P subshells of an atom.

P subshell magnetic quantum number = -1, 0, +1 and each magnetic quantum number has two spin quantum numbers.

∴ The total number of electrons in 4P subshell = (2 × 3) = 6.

Question

How many possible orbitals are there when principal quantum number = 4?

Answer

When principal quantum number = 4, azimuthal quantum number = 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.

∴ Total number of possible orbitals = (1 + 3 + 5 + 7)

= 16

Question

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

_{l}= 0?

Answer

Principal quantum number = 3, azimuthal quantum number = 1, and magnetic quantum number = 0.

It designated a 3S subshell of an atom.

∴ The number of possible orbital = 1.

Question

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

Answer

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

Principal quantum number of this orbital = 3+1 = 4.

### The emission spectrum of the hydrogen atom

QuestionWhat is the wavelength of the HÉ‘ line of the Balmer series of the hydrogen atom?

Answer

The wavenumber of the Balmer lines of the spectrum

⊽ = 1/Î» = 109677[(1/2²) - (1/n

_{₂}²)].HÉ‘ line of Balmer series n₂ = 3.

∴ 1/Î» = 109677 [(1/2²) - (1/3²)]

= 109677 × (5/36)

∴ Î» = 36/(5 × 109677)

= 6.564 × 10⁻⁵ cm

= 656.5 nm.

∴ 1/Î» = 109677 [(1/2²) - (1/3²)]

= 109677 × (5/36)

∴ Î» = 36/(5 × 109677)

= 6.564 × 10⁻⁵ cm

= 656.5 nm.

Question

The lowest wavelength of the Lyman series of the hydrogen atom is x. What is the highest wavelength of the Paschen series of the helium ion?

Answer

The highest wave number 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 wave number of the Paschen series of helium ion

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

or, Î»max = (9 × 16)/(4 × 7R)

∴ Î»max = 36x/7.

or, Î»max = (9 × 16)/(4 × 7R)

∴ Î»max = 36x/7.

Question

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

Answer

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?

Answer

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.

Electronic configuration of M is

[Ar]¹⁸ 3d³

∴ x = 4

[Ar]¹⁸ 3d³

∴ x = 4