Question

A radioactive sample has a half-life of 1770 years
At some instant 20% of the sample has decayed.
Taking this instant as t = ty, it is found that at
t = tz, 20% of the sample is left. Then, tz - t, is
(1) 1770 years
(2) 885 years
(3) 2655 years
(4) 3540 years​

Answer 1

your correct answer is (4) 3540

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Answer 2

Answer:

The answer will be 3540 years i.e. option (4).

Explanation:

Here in this question, we will use the law of radioactive decay which states that "the number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample". If N is the number of nuclei in the sample and dN undergo decay in time dT then:

dN/dT∝Nd

N/dT=-λN

dN/N=-λdt

λ=disintegration constant

let's integrate the equation ,

∫dN/N=-λ∫dT

the final equation will be,

N=N0$e^{-λt}$(1)

λ=ln2/T1/2 (2)

T1/2=half life

case I:

if 20% of the sample is decayed then we are left with only 80%of the sample

so the equation will be

 0.8N0=N0$e^{-λty}$

0.8=  $e^{-λty}$           (3)

case II:

only 20% of the sample is left

0.2N0=N0$e^{-λtz}$

0.2=$e^{-λtz}$             (4)

by taking the ratio of equations 3 and 4 we get:

4=$e^{λ(tz-ty)}$

ln4=λ(tz-ty)

now let's use equation 2

 2ln2=ln2/1770 x (tz-ty)

(tz-ty)=2 x 1770=3540 years