Answer:
Option B
Explanation:
It is known
$\frac{(t_{1/2})_{1}}{(t_{1/2})_{2}}=\left[\frac{a_{2}}{a_{1}}\right]^{(n-1)}$
Here, n= order of reaction
Given, $ (t_{1/2})_{1}=0.1 s, a_{1}=400$
$(t_{1/2})_{2}=0.8 s, a_{2}=50$
On putting the values
$\frac{0.1}{0.8}=\left[\frac{50}{400}\right]^{(n-1)}$
Taking log on both sides,
$\log \frac{0.1}{0.8}=(n-1) \log \frac{50}{400}$
$\log \frac{1}{8}=(n-1)\log \frac{1}{8}$
$n-1=1 \Rightarrow n=2$