The activity of a radioactive sample is measured as 9750 counts per minute at t= 0 and as 975 counts per minute at t = 5 minutes. The decay constant is approximately

Question

The activity of a radioactive sample is measured as 9750 counts per minute at t= 0 and as 975 counts per minute at t = 5 minutes. The decay constant is approximately (A) 0.230 per minute (B) 0.461 per minute (C) 0.691 per minute (D) 0.922 per minute.

Tardigrade Admin · Accepted Answer

A = A0 e-λ t or 975 = 9750 e-λ × 5 Hence e5 λ = 10 . That is 5 λ = loge 10 = 2.3026 log10 10 = 2.3026 or λ = 0.461

Q. The activity of a radioactive sample is measured as 9750 counts per minute at t= 0 and as 975 counts per minute at t = 5 minutes. The decay constant is approximately

0.230 per minute

0.461 per minute

0.691 per minute

0.922 per minute.

$A = A_{0} e^{- λ t}$ or 975 = 9750 $e^{- λ \times 5}$ Hence $e^{5 λ}$ = 10 . That is $5 λ = l o g_{e}$ 10
= 2.3026 $l o g_{10}$ 10 = 2.3026 or $λ$ = 0.461

Q. The activity of a radioactive sample is measured as 9750 counts per minute at t= 0 and as 975 counts per minute at t = 5 minutes. The decay constant is approximately

0.230 per minute

0.461 per minute

0.691 per minute

0.922 per minute.

A=A0​e−λt or 975 = 9750 e−λ×5 Hence e5λ = 10 . That is 5λ=loge​ 10 = 2.3026 log10​ 10 = 2.3026 or λ = 0.461

$A = A_{0} e^{- λ t}$ or 975 = 9750 $e^{- λ \times 5}$ Hence $e^{5 λ}$ = 10 . That is $5 λ = l o g_{e}$ 10
= 2.3026 $l o g_{10}$ 10 = 2.3026 or $λ$ = 0.461