Question 7

(a) Using the Bohr’s model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels.

(b) Calculate the orbital period in each of these levels.

Answer

(a) Let ν_{1} be the orbital speed of the electron in a hydrogen atom in the ground state level, n_{1} = 1. For charge (e) of an electron, ν_{1} is given by the relation,

ν_{ 1} = e^{2}/n_{1}4πϵ_{0}(h/2π) = e^{2}/2ϵ_{0}h

Where, e = 1.6 × 10^{−19} C

ϵ_{0} = Permittivity of free space = 8.85 × 10^{-12} N^{−1} C^{2} m^{−2}

h = Planck’s constant = 6.62 × 10^{−34} Js

∴ ν_{1} = (1.6x10^{-19})2/2x8.85x10^{-12}x6.62x10^{-34} = 0.0218 x 10^{8} = 2.18 x 10^{6} *m/s*

For level n_{2} = 2, we can write the relation for the corresponding orbital speed as:

ν_{2} = e^{2}/n_{2}2ϵ_{0}h = (1.6x10^{-19})2/2x2x8.85x10^{-12}x6.62x10^{-34} = 1.09 x 10^{6} m/s

And, for n_{3} = 3, we can write the relation for the corresponding orbital speed as:

ν_{3} = e^{2}/n_{3}2ϵ_{0}h = (1.6x10^{-19})2/3x2x8.85x10^{-12}x6.62x10^{-34} = 7.27 x 10^{5} m/s

Hence, the speed of the electron in a hydrogen atom in n = 1, n=2, and n=3 is 2.18 × 10 ^{6} m/s, 1.09 × 10 ^{6} m/s, 7.27 × 10 ^{5} m/s respectively.

(b) Let T _{1} be the orbital period of the electron when it is in level n_{1} = 1.

Orbital period is related to orbital speed as:

T_{1} = 2πr_{1}/ν_{ 1}

Where, r_{1} = Radius of the orbit

= n_{1}^{2}h^{2}ϵ_{0}/πme^{2}

h = Planck’s constant = 6.62 × 10^{−34} Js

e = Charge on an electron = 1.6 × 10^{−19} C

ϵ_{0} = Permittivity of free space = 8.85 × 10^{−12} N^{−1} C^{2} m^{−2}

m = Mass of an electron = 9.1 × 10^{−31} kg

∴ T_{1} = 2πr_{1}/ν _{1}

= (2πx(1)^{2}x(6.62x10^{-34})2x8.85x10^{-12})/2.18x10^{6}xπx9.1x10^{-31}x(1.6x10^{-19})^{2}

= 15.27x10^{-17} = 1.527x10^{-16} s

For level n _{2} = 2, we can write the period as:

T_{2} = 2πr_{2}/ν _{2}

Where, r_{2} = Radius of the electron in n_{2} = 2

= (n_{2})^{2}h^{2}ϵ_{0}/πme^{2}

∴ T_{2} = 2πr_{2}/ν_{2}

= (2πx(2)^{2}x(6.62x10^{-34})^{2}x8.85x10^{-12})/1.09 x 10^{6} x π x 9.1 x 10^{-31} x (1.6 x 10^{-19})^{2}

= 1.22 x 10^{-15} s

And, for level n_{ 3} = 3, we can write the period as:

T_{3} = 2πr_{3}/ν _{3 }

Where, r _{3} = Radius of the electron in n_{ 3} = 3

= (n_{3})^{2}h^{2}ϵ_{0}/πme^{2}

∴ T_{3} = 2πr_{3}/ν _{3}

= (2πx(3)^{2}x(6.62x10^{-34})2x8.85x10^{-12})/7.27 x 10^{5} x π x 9.1 x 10^{-31} x (1.6 x 10^{-19})^{2}

= 4.12 x 10^{-15} s

Hence, the orbital period in each of these levels is 1.52 × 10 ^{−16} s, 1.22 × 10 ^{−15} s, and 4.12 × 10 ^{−15} s respectively.

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