The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is: (Take log 10 1.88=0.274 )

Question

The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is: (Take log 10 1.88=0.274 ) (A) 0.02 min -1 (B) 2.7 min -1 (C) 0.063 min -1 (D) 6.3 min -1

Tardigrade Admin · Accepted Answer

At t =0 disintegration rate =4250 dpm At t =10 disintegration rate =2250 dpm A=Ao e-λ t 2250=4250 e -λ(10) ⇒ λ(10)= ln ((4250/2250)) ⇒ λ=0.063 min -1

Q. The disintegration rate of a certain radioactive sample at any instant is $4250$ disintegrations per minute. $10$ minutes later, the rate becomes $2250$ disintegrations per minute. The approximate decay constant is : (Take $lo g_{10} 1.88 = 0.274$ )

$0.02 min^{- 1}$

$2.7 min^{- 1}$

$0.063 min^{- 1}$

$6.3 min^{- 1}$

At $t = 0$ disintegration rate $= 4250 d p m$
At $t = 10$ disintegration rate $= 2250 d p m$
$A = A_{o} e^{- λ t}$
$2250 = 4250 e^{- λ (10)}$
$\Rightarrow λ (10) = ln (\frac{4250}{2250})$
$\Rightarrow λ = 0.063 min^{- 1}$

Q. The disintegration rate of a certain radioactive sample at any instant is 4250 disintegrations per minute. 10 minutes later, the rate becomes 2250 disintegrations per minute. The approximate decay constant is : (Take log10​1.88=0.274 )

0.02min−1

2.7min−1

0.063min−1

6.3min−1