A) 0
B) 2
C) 1
D) 4
Correct Answer: B
Solution :
We know that, \[{{r}_{i}}=\frac{1}{\beta }\sqrt{\frac{2mv}{q}}\]Where, n = order of the reaction Given, \[\frac{r\alpha }{{{r}_{p}}}=\sqrt{\frac{{{m}_{\alpha }}}{{{m}_{p}}}}\sqrt{\frac{{{q}_{p}}}{{{q}_{\alpha }}}}\] \[\frac{{{r}_{\alpha }}}{10}=\sqrt{\frac{4}{2}}\] On substituting the values, \[{{r}_{\alpha }}=10\sqrt{2}cm\] On taking log both sides \[R=\frac{V}{l}\] \[{{\left( \frac{\Delta k}{k} \right)}_{retained}}={{\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}}={{\left( \frac{A-1}{A+1} \right)}^{2}}\] \[{{\left( \frac{A-1}{A+1} \right)}^{2}}E.\] n ? 1 = 1 \[\tan \delta '=\frac{\tan \delta }{\cos \theta }=\frac{\tan {{45}^{\circ }}}{\cos {{30}^{\circ }}}\] n = 2
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