(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. - Physics

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Numerical

(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 v1 be the orbital speed of the electron in a hydrogen atom in the ground state level, n1 = 1.

For charge (e) of an `electron, vis 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 C2 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 × 108

= 2.18 × 106 m/s

For level n2 = 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 × 106 m/s

And, for n3 = 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 × 105 m/s

Hence, the speed of the electron in a hydrogen atom in n = 1, n = 2, and n = 3 is 2.18 × 106 m/s, 1.09 × 106 m/s, 7.27 × 105 m/s respectively.

(b) Let T1 be the orbital period of the electron when it is in level n1 = 1.

Orbital period is related to orbital speed as:

`"T"_1 = (2pi"r"_1)/"v"_1`

Where

r1 = 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

= Permittivity of free space = 8.85 × 10−12 N−1 C2 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 n2 = 2, we can write the period as:

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

Where,

r2 = Radius of the electron in n2 = 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 n3 = 3, we can write the period as:

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

Where,

r3 = Radius of the electron in n3 = 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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Chapter 12: Atoms - Exercise [Page 436]

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Chapter 12 Atoms
Exercise | Q 12.7 | Page 436
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Chapter 12 Atoms
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