Two elements A and B with atomic numbers Z_{A | vedclass

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(D) According to Moseley's law for characteristic $X$-rays, the frequency $v$ of the emitted $X$-rays is related to the atomic number $Z$ as $v \propt

Vedclass · Accepted Answer

$1: 4$

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(D) According to Moseley's law for characteristic $X$-rays, the frequency $v$ of the emitted $X$-rays is related to the atomic number $Z$ as $v \propto (Z - b)^2$, where $b$ is a screening constant. For the $K_{\alpha}$ line, $b = 1$. Assuming the same series for both elements, we have $v \propto Z^2$. Given the ratio of atomic numbers $Z_{A} : Z_{B} = 1 : 2$. Therefore, the ratio of frequencies is: $\frac{v_{A}}{v_{B}} = \left( \frac{Z_{A}}{Z_{B}} \right)^2$ $\frac{v_{A}}{v_{B}} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$ Thus, $v_{A} : v_{B} = 1 : 4$.

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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