Question

A certain radioactive material is known to decay at a rate proportional to the amount present. If after one hour, it is observed that 10% of the material has decayed, find the half-life of the material.

Answer 1

Can someone help me solve this?

Answer 2

do you have answer choices??

Answer 3

Let initial amount of radioactive material is No then N(0)=No , N(1)=(1−0.1)No=0.9No .

Answer 4

i have answer.
It is 6.6hrs

Answer 5

We need to find such t that N(t)=No/2 .

Answer 6

you are correct ^-^ @BizPro

Answer 7

hi again @BizPro

Answer 8

no sarah, that's the answer given, i'm just not sure how to do it.

Answer 9

ok Im tellin you step by step .
got first two step ???

Answer 10

why No?

Answer 11

or is that just short for number?

Answer 12

we are supposing

Let initial amount of radioactive material is No then N(0)=No , N(1)=(1−0.1)No=0.9No .
We need to find such t that N(t)=No/2 .
So, as was shown above, if N(0)=No then N(t)=Noe^kt .
Since N(1)=0.9No then 0.9No=Noe^k⋅1 or k=ln(0.9) .
Now equation has form N(t)=Noe^ln(0.9)t .
Now, find t, such that N(t)=0.5No : 0.5No=Noe^ln(0.9)t or t=ln(0.5)/ln(0.9)≈6.58 hours.

Answer 13

im sorry just a bit confused. Need read through it again.

Answer 14

you know na that half-life is period of time it takes for the amount of material to decrease by half . So 10% or .1 is halfed then 5% or .5

Answer 15

Oh Noe^kt is like Ae^kt

Answer 16

No is initial amount that we suppose .

Answer 17

These questions always confuse me :(

Answer 18

initial amount is 1 , how much decayed as mentioned in question 10 % means 10/100 = .1 So subtract .1 from 1(initial amount) we get .9 left

Answer 19

yeah i understand up till Noe^kt

Answer 20

then u need to know this how to find growth and decay formula .
Let N(t) denote amount of substance (or population) that is either growing or decaying. If we assume that dN/dt , the time rate of change of this amount of substance, is proportional to the amount of substance present, then dN/dt=kN where k is the constant of proportionality.
When k is positive, population grows, when k is negative - population decays.
We are assuming that N(t) is a differentiable, hence continuous, function of time. For population problems, where N(t) is actually discrete and integer-valued, this assumption is incorrect. Nonetheless, above model still provides a good approximation to she physical laws governing such a system.
This differential equation is separable. Rewriting it we obtain dN/N=kdt .
Integration of both sides gives: ln(N)=kt+c1 or N=Ce^kt where C=e^c1 .
If we are given that N(0)=No thenNo=Ce^k⋅0 or C=no .
So, N=Noe^kt .

Answer 21

If still you dun get it then .-.

Answer 22

Oh ok thnx :)

Answer 23

yw ^-^