Question #ec74b

Question

1607 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-24T16:51:20

1/512

Let #m_0# be the mass at t = 0 day and m the mass, at time t days.

Then the form of the decay equation is #m =m_0 e^(-kt)#.

When #t = 1/3# day, #m =m_0/2#,

So, #m_0/2==m_0 e^(-k/3#. Solve for k.

#1/2=e^(-k/3)#. So, #e^(k/3)=2# and, therefore, #k=3 ln 2#.

Now the decay equation becomes

#m=m_9e^(-3t ln 2)#

#=m_0e^ln(2^(-3t))#

#=m_0 2^((-3t)#.

Now, when t = 3 days, #m=m_0 2^-9=m_0/512#

= 1 in 512 of initial value..

,