In the visible region of the spectrum,the rot | vedclass

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Question

(B) Given the equation for optical rotation: $	heta = a + \frac{b}{\lambda^2}$.At $\lambda_1 = 5000 \ \mathring{A}$,$	heta_1 = 30^\circ$ per mm: $30

Vedclass · Accepted Answer

$- \frac{50^\circ}{9}$ per mm

Vedclass · Answer

(B) Given the equation for optical rotation: $	heta = a + \frac{b}{\lambda^2}$.At $\lambda_1 = 5000 \ \mathring{A}$,$	heta_1 = 30^\circ$ per mm: $30 = a + \frac{b}{(5000)^2} \quad (1)$.At $\lambda_2 = 4000 \ \mathring{A}$,$	heta_2 = 50^\circ$ per mm: $50 = a + \frac{b}{(4000)^2} \quad (2)$.Subtracting equation $(1)$ from equation $(2)$:$50 - 30 = b \left( \frac{1}{4000^2} - \frac{1}{5000^2} ight) = b \left( \frac{5000^2 - 4000^2}{4000^2 	imes 5000^2} ight)$.$20 = b \left( \frac{(5000-4000)(5000+4000)}{16 	imes 10^6 	imes 25 	imes 10^6} ight) = b \left( \frac{1000 	imes 9000}{400 	imes 10^{12}} ight) = b \left( \frac{9 	imes 10^6}{400 	imes 10^{12}} ight) = b \left( \frac{9}{400 	imes 10^6} ight)$.$b = \frac{20 	imes 400 	imes 10^6}{9} = \frac{8000 	imes 10^6}{9}$.Substitute $b$ into equation $(1)$:$a = 30 - \frac{8000 	imes 10^6}{9 	imes 25 	imes 10^6} = 30 - \frac{8000}{225} = 30 - \frac{320}{9} = \frac{270 - 320}{9} = - \frac{50^\circ}{9} 	ext{ per mm}$.

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