IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry - chemistry problem -urgent

sirisha

10% of a radioactive element disintegrates in 1 hr.the % of the element remaining at the end of 2 hrs.

rishabh

ans=81%
d(A)/dt=-k(A)
ln(A/Ai)=-kt
ln(0.9/1)= - k*1
ln(x/1)= -k *2
divide & solve

sirisha

sorry i didnt get u
can u repeat it

smriti.mathur

We know that,

=2.303/t log a ₀/a

where lembda= disintegration constant

t = time

a ₀= amount of radioactive substance at 0 time(in our case 100%)

a = substance after time t

2.303/1 log 100/90 = 2.303/2 log 100/a

cancelling 2.303 on both sides ,we get

2 log 100 - 2 log 90 = log 100 -log a

4 - 2 log 90 = 2 - log a

log a= 2 - 4 + 2 log 90

= -2 +2 (1.9542)

= -2 +3.9084

= 1.9084

a = antilog of 1.9084

= 80.98

hence, the pecentage of radioactive substance remaining at the end of two hours is 80.98%.