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Two elements $A$ and $B$ with atomic numbers $Z_{A}$ and $Z_{B}$ are used to produce characteristic X-rays with frequencies $v_{A}$ and $v_{B}$ respectively. If $Z_{A}: Z_{B}=1: 2,$ then $v_{A}: v_{B}$ will be
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The correct answer is:
$1: 4$
Given $Z_{A}: Z_{B}=1: 2$ We know
$v \propto Z^{2}$
Hence,
$\begin{array}{l}
\frac{v_{A}}{v_{B}}=\frac{(1)^{2}}{(2)^{2}} \\
v_{A}: v_{B}=1: 4
\end{array}$
$v \propto Z^{2}$
Hence,
$\begin{array}{l}
\frac{v_{A}}{v_{B}}=\frac{(1)^{2}}{(2)^{2}} \\
v_{A}: v_{B}=1: 4
\end{array}$
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