The activity of a radioactive sample is measured as $9750$ counts ${(m i n u t e)}^{- 1}$ at $t = 0$ and $975$ counts ${(m i n u t e)}^{- 1}$ at $t = 5$ minute. The decay constant is nearly

Question

The activity of a radioactive sample is measured as 9750 counts -minute-1- t-0 and 975 counts -minute-1- t-5 minute- The decay constant is nearly0-922-min-1-0-691-min-1-0-461-min-1-0-230-min-1-

Toppr Admin · Accepted Answer

Activity after timet is-xA0- N-N0e-x2212-x3BB-twhere N0 is activity at t-0 putting the above values we get-xA0-decay constant -x3BB-0-461min-x2212-1-xA0-use -x3BB-1-t-logeN0N and loge10-2-303

The activity of a radioactive sample is measured as $9750$ counts ${(m i n u t e)}^{- 1}$ at $t = 0$ and $975$ counts ${(m i n u t e)}^{- 1}$ at $t = 5$ minute. The decay constant is nearly
$0.922 {m i n}^{- 1}$
$0.691 {m i n}^{- 1}$
$0.461 {m i n}^{- 1}$
$0.230 {m i n}^{- 1}$

activity after time $t$ is $N = N_{0} e^{- λ t}$
where $N_{0}$ is activity at $t = 0$ putting the above values we get
decay constant $λ = 0.461 m i n^{- 1}$
use $λ = (1 / t) l o g_{e} \frac{N_{0}}{N}$ and $l o g_{e} 10 = 2.303$

The activity of a radioactive sample is measured as 9750 counts (minute)−1 at t=0 and 975 counts (minute)−1 at t=5 minute. The decay constant is nearly0.922min−10.691min−10.461min−10.230min−1

activity after timet is N=N0e−λtwhere N0 is activity at t=0 putting the above values we get decay constant λ=0.461min−1 use λ=(1/t)logeN0N and loge10=2.303

The activity of a radioactive sample is measured as $9750$ counts ${(m i n u t e)}^{- 1}$ at $t = 0$ and $975$ counts ${(m i n u t e)}^{- 1}$ at $t = 5$ minute. The decay constant is nearly
$0.922 {m i n}^{- 1}$
$0.691 {m i n}^{- 1}$
$0.461 {m i n}^{- 1}$
$0.230 {m i n}^{- 1}$

activity after time $t$ is $N = N_{0} e^{- λ t}$
where $N_{0}$ is activity at $t = 0$ putting the above values we get
decay constant $λ = 0.461 m i n^{- 1}$
use $λ = (1 / t) l o g_{e} \frac{N_{0}}{N}$ and $l o g_{e} 10 = 2.303$