the activity A0 is stated with 2% error. we hav to get the time for 10% ka error. means original se 8% error. if original is A0, then after 8% error, it will be 0.92A0. we will consider the expression: A = A0* e^(-lambda*time)
lambda will be 0.693/half life
lambda = 0.693/(5.27*365*24*3600)
A = 0.92 A0
so the equation will be:
0.92A0 = A0 * e^(-lambda*time)
A0 will be cancelled
0.92 = e^ (-lambda*time)
ln (0.92) = -lambda*time