For a certain radioactive process the graph between ln(R) and t(sec) is obtained as shown in the figure. Then the value of half life for the unknown radioactive material is approximately :
(A)2.62 sec
(B)9.15 sec
(C)4.62 sec✓ Correct
(D)6.93 sec
Answer: (C)
Solution
The decay rate falls as R=R0e−λt, so lnR=lnR0−λt and the plot is a straight line of slope −λ. The line drops from lnR=6 at t=0 to lnR=0 at t=40 s, so
λ=406−0=0.15s−1
t1/2=λln2=0.150.693
The half life of the material is about 4.62 sec.
JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option