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Question: At the point of intersection of the two curves shown, the conc. of B is given by.......for, A \(\rig...

At the point of intersection of the two curves shown, the conc. of B is given by.......for, A →\rightarrownB :

A

nA02\frac{nA_{0}}{2}

B

A0n−1\frac{A_{0}}{n - 1}

C

nA0n+1\frac{nA_{0}}{n + 1}

D

(n−1n+1)A0\left( \frac{n - 1}{n + 1} \right)A_{0}

Answer

nA0n+1\frac{nA_{0}}{n + 1}

Explanation

Solution

A→nBA \rightarrow nB

t=0A0t = 0A_{0}

t = t A0-x nx\text{t = t } A_{0}\text{-x } \text{nx}

From graph

A0-x = nxA_{0}\text{-x = nx}

x=A0n+1x = \frac{A_{0}}{n + 1}

∣B∣=n∣A0∣n+1|B| = \frac{n|A_{0}|}{n + 1}