Class 12 Physics

CBSE and UP Board

Atoms

Chapter-12 Exercise 12.13

NCERT Solutions for Class 12 Physics Chapter 12 Question-13

Additional Exercise

# 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.

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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:

E₁ = hν₁ = hme⁴ /[(4π )³ ε0²(h/2π )³] x (1/n²)—————- Eq-1

Where,

ν₁= Frequency of radiation at level n

h= Planck’s constant

m = Mass of hydrogen atom

e = Charge on an electron

ε0= Permittivity of free space

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

E2 = hν2 = hme⁴ /[(4π )³ ε0²(h/2π )³] x 1/[(n-1)²]————Eq-2

Where.

ν2 = Frequency of radiation at level (n-1)

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

E – E2-E1

hν = E2-E1 —————–Eq-3

Where,

ν = Frequency of radiation emitted

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

ν =me⁴ /[(4π )³ ε0²(h/2π )³] x{1/(n-1)² – 1/n² }

= me⁴ (2n-1)/[(4π )³ ε0²(h/2π )³] n² (n-1)²

For large n, we can write (2n-1) ≈ 2n and (n-1) ≈n.

Therefore, ν = me⁴/32 π³ ε0² (h/2π )³ n³—————-Eq-4

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

νc = ν/2πr,—————Eq-5

Where,

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

v = e²/[(4π ε0)(h/2π )ⁿ]—————-Eq-6

And radius of nth orbit is given as :

r= [4π ε0 (h/2π )²] n²/me² —————-Eq-7

Putting the values of Eq-6 and Eq-7 in Eq-5 we get

vc = me⁴/32 π³ ε0² (h/2π )³ n³—————-Eq-8

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