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  • If is the wave vector of incident light <img alt="\mathrm{|\vec{k}|=\frac{2 \pi}{\l

If \mathrm{\vec{k}} is the wave vector of incident light \mathrm{|\vec{k}|=\frac{2 \pi}{\lambda}, \lambda} is the wave length of light and \mathrm{\vec{G}} is reciprocal lattice vector, then the bragg's law can be written as:-

Option: 1

\mathrm{\vec{k}+\vec{G}=0}


Option: 2

\mathrm{2 \vec{k} \cdot \vec{G}+G^2=0}


Option: 3

\mathrm{2 \vec{k} \cdot \vec{G}+k^2=0}


Option: 4

\mathrm{\vec{K} \cdot \vec{G}=0}


Answers (1)

best_answer

Bragg's law in terms of reciprocal lattice vector \mathrm{(\vec{G})}
\mathrm{\therefore \quad \vec{k}^{\prime}=\vec{k}+\vec{G}}
\mathrm{\begin{aligned} & \Rightarrow \quad \vec{k}^{\prime} \cdot \overrightarrow{k^{\prime}}=(\vec{k}+\vec{G}) \cdot(\vec{k}+\vec{G}) \\ & \Rightarrow\left|\overrightarrow{k^{\prime}}\right|^2=|\vec{k}|^2+2 \vec{k} \cdot \vec{G}+|\vec{G}|^2 \end{aligned}}
\mathrm{\begin{aligned} & \because|\vec{k}|^2=|\vec{k}|^2 \\ & \therefore \quad 2 \vec{k} \cdot \vec{G}+G^2=0 \end{aligned}}.

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Rishabh

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