Given :
\(\lambda=\frac{h}{mv}\)
Now, \(\frac{\lambda_2}{\lambda_1}=\frac{m_1v_1}{m_2v_2}=\frac{4}{1}\)
\(=\frac{9\times v_2}{1\times v_1}\)
\(\frac{v_2}{1}=\frac{9}{4}\)
K. E = 12 mv2
Now, Ratio of K.E is :
\(\frac{K.E_1}{K.E_2}=\frac{16}{9}\)
So, the correct option is (D) : 16 : 9.
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: