# Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n − 1). For large n, show that this frequency equals the classical - Physics

Numerical

Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n − 1). For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

#### Solution

It is given that a hydrogen atom de-excites from an upper level (n) to a lower level (n − 1).

We have the relation for energy (E1) of radiation at level n as:

"E"_1 = "hv"_1 = ("hme"^4)/((4pi)^3 in_0^2 ("h"/(2pi))^3) xx (1/"n"^2)  ...........(i)

Where

v1 =  Frequency of radiation at level n

h = Planck's constant

m = Mass of hydrogen atom

e = Charge on an electron

in_0 = Permittivityof free space

Now, the relation for energy (E2) of radiation at level (n − 1) is given as:

"E"_2 = "hv"_2 = ("hme"^4)/((4pi)^3 in_0^2 ("h"/(2pi))^3) xx 1/("n" - 1)^2 ...........(ii)

Where

v2 =  Frequency of radiation at level (n − 1)

Energy (E) released as a result of de-excitation:

E = E2 − E1

hv = E2 − E1 …............(iii)

Where,

v = Frequency of radiation emitted

Putting values from equations (i) and (ii) in equation (iii), we get:

"v" = ("me"^4)/((4pi)^3 in_0^2("h"/(2pi))^3) [ 1/("n" - 1)^2 - 1/"n"^2]

= ("me"^4 (2"n" - 1))/((4pi)^3 in_0^2 ("h"/(2pi))^3 "n"^2("n" - 1)^2)

For large n, we can write (2n − 1) ≃ 2n and (n − 1) ≃ n.

∴ v = ("me"^4)/(32pi^3 in_0^2 ("h"/(2pi))^3 "n"^3 ...............(iv)

Classical relation of frequency of revolution of an electron is given as:

"v"_"c" = "v"/(2pi"r") ...........(V)

Where,

Velocity of the electron in the nth orbit is given as:

"v" = "e"^2/(4pi in_0 ("h"/(2pi)) "n") .........(vi)

And, radius of the nth orbit is given as:

"r" = (4pi in_0 ("h"/(2pi))^2)/("me"^2) "n"^2 ............(vii)

Putting the values of equations (vi) and (vii) in equation (v), we get:

"v"_"c" = ("me"^4)/(32pi^3 in_0^2 ("h"/(2pi))^3 "n"^3) ................(Viii)

Hence, the frequency of radiation emitted by the hydrogen atom is equal to its classical orbital frequency.

Concept: De Broglie’s Explanation of Bohr’s Second Postulate of Quantisation
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 12 Atoms
Exercise | Q 12.13 | Page 436
NCERT Class 12 Physics Textbook
Chapter 12 Atoms
Exercise | Q 13 | Page 436

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