Find the probability of the link in polymers where average values of links are (1) 10, (2) 50 and (3) 10
(a)(1) 0.99, (2) 0.98, (3) 0.90
(b)(1) 0.98, (2) 0.90, (3) 0.99
(c)(1) 0.90, (2) 0.98, (3) 0.99
(d)(1) 0.90, (2) 0.99, (3) 0.98
Answer
Answer: C ✓ checked by 4AB · confidence medium
Explanation
For linear step polymerisation ⟨N⟩ = 1/(1−p): 10 gives 0.90, 50 gives 0.98, and 100 gives 0.99. The third value is printed as '10' but must be 100 (c).