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**(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.

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#### Solution

**(a)** Let v_{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, v_{1 }is given by the relation,

`"v"_1 = "e"^2/("n"_1 4pi in_0("h"/(2pi))) = "e"^2/(2in_0"h")`

Where,

e = 1.6 × 10^{−19} C

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

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

∴ `"v"_1 = (1.6 xx 10^(-19))^2/(2 xx 8.85 xx 10^(-12) xx 6.62 xx 10^34)`

= 0.0218 × 10^{8}

= 2.18 × 10^{6} m/s

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

`"v"_2 = "e"^2/("n"_2 2 in_0 "h")`

= `(1.6 xx 10^(-19))^2/(2 xx 2 xx 8.85 xx 10^(-12) xx 6.62 xx 10^(-34))`

= 1.09 × 10^{6} m/s

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

`"v"_3 = "e"^2/("n"_3 2 in_0 "h")`

= `(1.6 xx 10^(-19))^2/(3 xx 2 xx 8.85 xx 10^(-12) xx 6.62 xx 10^(-34))`

= 7.27 × 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 = (2pi"r"_1)/"v"_1`

Where

r_{1} = Radius of the orbit

`("n"_1^2 "h"^2 in_0)/(pi"me"^2)`

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

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

∈_{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 = (2pi"r"_1)/"v"_1`

= `(2pi xx (1)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(2.18 xx 10^6 xx pi xx 9.1 xx 10^(-31) xx (1.6 xx 10^(-19))^2`

= 15.27 × 10^{−17}

= 1.527 × 10^{−16} s

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

`"T"_2 = (2pi"r"_2)/"v"_2`

Where,

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

`= (("n"_2)^2 "h"^2 in_0)/(pi"me"^2)`

∴ `"T"_2 = (2pi"r"_2)/"v"_2`

= `(2pi xx (2)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(1.09 xx 10^6 xx pi xx 9.1 xx 10^(-31) xx (1.6 xx 10^(-19))^2)`

= 1.22 × 10^{−15} s

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

`"T"_3 = (2pi"r"_3)/"v"_3`

Where,

r_{3} = Radius of the electron in n_{3} = 3

= `(("n"_3)^2"h"^2 in_0)/(pi"me"^2)`

∴ `"T"_3 = (2pi"r"_3)/"v"_3`

= `(2pi xx (3)^2 xx (6.62 xx 10^(-34))^2 xx 8.85 xx 10^(-12))/(7.27 xx 10^5 xx pi xx 9.1 xx 10^(-31) xx (1.6 xx 10^(-19)))`

= 4.12 × 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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