Question

The activity of a radioactive material is 2.56 x 10^-3 Ci. If the half life of the material is 5 days, after how many days the activity will become 2 x 10^-5 Ci ?

पर्याय

• 30 days

• 35 days

• 40 days

• 25 days

Answer 1

35 days

Explanation:

Given, at t = 0 ⇒ Activity of Radioactive Sample = 2.56 × 10^–3 Ci

Thus, R_o = 2.56 × 10^–3 Ci

Half life of sample, T_½ = 5 days

As, Radioactive decay is of first order reaction. Thus, Rate of decay or activity decreases exponentially.

By Radioactive Decay law,

R = R_oe ^–λt

⇒ 2 × 10^–5 = 2.56 × 10^–3 e ^–λt

Where, R ⇒ Activity at time t

λ ⇒ Activity constant of Radioactive sample

λ  ⇒ `(ln2)/"T"_(1/2)`

Taking logarithm on both sides

ln (2 × 10^–5) = ln (2.56 × 10^–3) + ln (e^–λt)

ln (2 × 10^–5) – ln (2.56 × 10^–3) = –λt

ln`((2xx10^-5)/(2.56xx10^-3))` = –λt

ln `(1/128)` =  –λt

– ln 128 = –λt

⇒ ln 2⁷ = = `(In2)/(T_(1/2))t`

⇒ 7ln 2 =   `(In2)/(T_(1/2))t`

⇒ t = 7 `"T"_(1/2)`

⇒ t = 7 × 5 = 35days