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  • Suppose in imaginary values of Bragg's equation which one is the correct output of intern of <img alt="\mathrm{\lambda}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clamb

Suppose in imaginary values of Bragg's equation which one is the correct output of intern of \mathrm{\lambda} will be - 

Option: 1

\mathrm{e^{i \theta}=2 i \lambda-e^{-i \theta}}


Option: 2

\mathrm{e^{i \theta}=2 i \lambda+e^{-i \theta}}


Option: 3

\mathrm{e^{2 i \theta}=2 i \lambda+e^{-i \theta}}


Option: 4

\mathrm{d\left(e^{i \Theta }-e^{-i \Theta }\right)= \lambda\ 2i\ }


Answers (1)

best_answer

According to Bragg's law - 
\mathrm{2d\ sin \Theta =n\lambda ,}
                        where \mathrm{n=1,2,3....}
\mathrm{\begin{aligned} & 2 d \sin \theta=2 \lambda \\ & d \sin \theta=\lambda\ \ \ .......\left ( 1 \right ) \end{aligned}}
but, \mathrm{e^{i \theta}=\cos \theta+i \sin \theta \text\ { then }\ \sin \theta=\frac{e^{i\Theta }-e^{-i\Theta}}{2 i}}
where \mathrm{i } is imaginory number. 
from \mathrm{-\left ( 1 \right ) }
             \mathrm{ \begin{aligned} d\left(\frac{e^{i \theta}-e^{-i \theta}}{2 i}\right) & =\lambda \\ d\left(e^{i \theta}-e^{-i \theta}\right) & =\lambda 2 i \end{aligned} } 

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Riya

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