Question

Half life of a radioactive substance is 20 minutes the time between 20% and 80% decay will be

Answer 1

Answer :- 40 Min

Half Life = 20 mins

So,  t 1/2 = 20

Now,

80 N0 / 100 = N0 $e^{-lt(20) }$ -(1) -- 20% decay where 80% remain

20 N0 / 100 = N0 e$e^{-lt(80) }$ -(2)-- 80% decay where 20% remain

here l is lemda (λ)

Dividing (1) and (2) we get,

4 = $e^{l ( t 80 - t 20)}$

4=l(t80−t20)  (taking log on both sides)

2 ln 2 = 0.693 / t 1/2 x (t 80 - t 20)

(t 80 - t 20) / 2 = 20

So, t 80 - t 20 = 40 Min

Answer 2

Given that the half-life of a radioactive substance is 20 min. So, t

2

1

=20 min.

For 20% decay, we have 80% of the substance left, hence

100

80N

0

=N

0

e

−λl

20

…(i)

Where N

0

=initial undecayed substance andt

20

is the time taken for 20% decay.

For 80% decay, we have 20% of the substance left, hence

100

20N

0

=N

0

e

−λt

80

…(ii)

Dividing eq. (i) and eq. (ii), we get

4=e

λ(t

80

−t

20

)

⇒ In 4=λ(t

80

−t

20

)

Taking log on both sides

⇒2ln2=

t

2

1

0.693

(t

80

−t

20

)

⇒t

80

−t

20

=2×t

2

1

=40 min.