Q132 / 160 · WBJEE 2012 · Physics
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
- A$1: \sqrt{2}$
- B$1: 8$
- C$4: 1$
- Correct answerD$1: 4$
Solution
Given $Z_{A}: Z_{B}=1: 2$ We know
$v \propto Z^{2}$
Hence,
$\begin{array}{l}<br />\frac{v_{A}}{v_{B}}=\frac{(1)^{2}}{(2)^{2}} \\<br />v_{A}: v_{B}=1: 4<br />\end{array}$
$v \propto Z^{2}$
Hence,
$\begin{array}{l}<br />\frac{v_{A}}{v_{B}}=\frac{(1)^{2}}{(2)^{2}} \\<br />v_{A}: v_{B}=1: 4<br />\end{array}$
Hence, the correct answer is option: 'D'.