Question

A radioactive substance decays exponentially: The mass at time t is m(t) = m(0)ekt, where m(0) is the initial mass and k is a negative constant. The mean life M of an atom in the substance is M = −k [infinity] tekt dt. 0 For a radioactive isotope, the value of k is −0.000192. Find the mean life of the isotope's atom. (Round your answer to one decimal place.) yr

Answer 1

Answer:

M = 1/0.000121 = 8264.5 years

Step-by-step explanation:

M = − k ∫∞₀ teᵏᵗdt

To obtain this mean life, we'll use integration by parts to integrate the function ∫ teᵏᵗdt

∫udv = uv - ∫ vdu

u = t

du/dt = 1

du = dt

∫ dv = ∫ eᵏᵗdt

v = eᵏᵗ/k

∫udv = ∫ teᵏᵗdt

uv = teᵏᵗ/k

∫ vdu = eᵏᵗ/k

∫ teᵏᵗdt = (teᵏᵗ/k) - ∫eᵏᵗ/k

But, ∫eᵏᵗ/k = (1/k) ∫eᵏᵗ = (1/k²) eᵏᵗ = eᵏᵗ/k²

∫ teᵏᵗdt = (teᵏᵗ/k) - eᵏᵗ/k²

The rest of the calculation is done on paper in the image attached to this question

Answer 2

The required mean life is $[\dfrac{1}{0.000192} ]^{2}$.

Half-life and mean life

Mean life is the time taken for the radioactivity substance to fall to half its original value whereas the mean life is the average lifetime of all the nuclei of particular unstable atomic species.

Given

$\rm M =- k \int \infty _{o} te^{kt} dt$

How to calculate mean life?

To get the mean life, we will integrate $\rm \int te^{kt} dt$.

$\rm \int\limits^\infty_0 { te^{kt}} \, dt\\\\\rm t \int\limits^\infty_0 { e^{kt}} \, dt - \int\limits^\infty_0 { \dfrac{d}{dt} t(\int e^{kt dt})} \, dt\\\\\rm \dfrac{te^{kt} }{k} ]_{0}^{\infty } - \int\limits^\infty_0 {1 \dfrac{e^{kt} }{k} } \, dt\\\\\rm \dfrac{te^{kt} }{k} ]_{0}^{\infty } - \dfrac{e^{kt} }{k^{2} } ]_{0}^{\infty }$

Here the value of k is -0.000192.

$\rm \dfrac{1}{-0.000192} [({te^{-0.000192t} } )_{0}^{\infty } - ( \dfrac{e^{-0.000192t} }{-0.000192} } )_{0}^{\infty } ]\\\\\rm \dfrac{1}{-0.000192} [({ \infty e^{-0.000192* \infty} } - 0e^{-0.000192*0} )- ( \dfrac{e^{-0.000192* \infty} }{-0.000192} - \dfrac{e^{-0.000192* 0} }{-0.000192}} )]\\\\\rm \dfrac{1}{-0.000192} [ ( \infty *e^{- \infty} - 0*e^{0} ) + ( \dfrac{e^{-0.000192} }{0.000192} -\dfrac{e^{0} }{0.000192}) ]$

$\rm \dfrac{-1}{0.000192} [ (0 - 0) + (\dfrac{0}{0.000192} -\dfrac{1}{0.000192} ) ]$

$\dfrac{-1}{0.000192} *\dfrac{-1}{0.0192} \\\\[\dfrac{1}{0.000192} ]^{2}$

Thus, the required mean life is $[\dfrac{1}{0.000192} ]^{2}$.

More about the mean life link is given below.

https://brainly.com/question/7184917