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Question: Given that the amplitude A of scattered light is : (A<sub>0</sub>) of incident light. (ii) Directly...

Given that the amplitude A of scattered light is : (A0) of incident light.

(ii) Directly proportional to the volume (V) of the scattering particle

(iii) Inversely proportional to the distance (r) from the scattered particle

(iv) Depend upon the wavelength (λ\lambda) of the scattered light. then:

A

A1λA \propto \frac{1}{\lambda}

B

A1λ2A \propto \frac{1}{\lambda^{2}}

C

A1λ3A \propto \frac{1}{\lambda^{3}}

D

A1λ4A \propto \frac{1}{\lambda^{4}}

Answer

A1λ2A \propto \frac{1}{\lambda^{2}}

Explanation

Solution

Let A=KA0VλxrA = \frac{KA_{0}V\lambda^{x}}{r}

By substituting the dimension of each quantity in both sides

[L]=[L].[L3][Lx][L]\Rightarrow \lbrack L\rbrack = \frac{\lbrack L\rbrack.\lbrack L^{3}\rbrack\lbrack L^{x}\rbrack}{\lbrack L\rbrack}

[L]=[L3+x]\therefore\lbrack L\rbrack = \lbrack L^{3 + x}\rbrack; 3+x=1\Rightarrow 3 + x = 1 or x=2x = - 2

Aλ2\therefore A \propto \lambda^{- 2}