Question

In Bohr 's atomic model,an electron is revolving in n=3 level. Calculate the speed and time period of revolution of electron?

Answer 1

Answer:

0.5 mv Square

because electron carry a net charge the value which is 1.6*10

Answer 2

Concept:

The orbital speed can be defined as the v = e²/ (n₁4π∈₀ (h/2π))

where n₁ is the value of energy level, e is the charge of electron and h is Planck's constant.

And, the time period of the revolution can be defined as, T = 2πr/v

where v is the velocity.

Given:

The electron is revolving in n = 3 energy level.

Find:

The speed and time period of the revolution of the electron.

Solution:

Calculating orbital speed, for n = 3,

v = e²/ (n₁4π∈₀ (h/2π))

where e is 1.6× 10⁻¹⁹ coulomb, ∈₀ is 8.85×10⁻¹² F⋅m⁻¹ and h is 6.6 × 10⁻³⁴ m² kg / s.

v = e²/ (3×4π∈₀ (h/2π))

Substituting the values,

v = 7.3 ×10⁵ m/s

As the inner radius of the atom is 0.53 ×10⁻¹⁰ m.

Radius, rₙ = n²r

r₃= 3²× r = 9 × 0.53 ×10⁻¹⁰ m

Substituting the calculated value of v = 7.3 ×10⁵ m/s in T = 2πr/v,

T = 2πr/v = (2π × 9 × 0.53 ×10⁻¹⁰ m  )/ 7.3 ×10⁵ m/s

T = 4.12 × 10⁻¹⁵ s

Hence, the speed of the electron is 7.3 ×10⁵ m/s and the time period of the electron is 4.12 × 10⁻¹⁵ s.

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