whats the exponential decay equation for 100 atoms that decay 1/20 per day
1/20 = .05
dn/dt = -.05 n
dn/n = -.05 dt
ln n = -.05 t + c
n = C e^-.05 t
n = 100 e^-.05 t
Hmm. Since the amount present on day n is 0.95 times the amount on day (n-1) I think the equation should be
n = 100*0.95t
= 100 * eln.95 t
ln.95 = -.05129
I keep thinking this is absorbed into the c of the integral, but that must certainly be 100.
How to reconcile?
To find the exponential decay equation for a given situation, we need to use the formula:
N(t) = N₀ * e^(-kt)
Where:
- N(t) represents the number of atoms remaining at time t
- N₀ represents the initial number of atoms
- e is the base of natural logarithm (approximately equal to 2.71828)
- k is the decay constant
In this case, you have 100 atoms that decay at a rate of 1/20 per day. To find the decay constant (k), we need to determine the value where N(t) decreases by 1/20 (0.05) per day.
So, we can set up the equation:
N(t) = N₀ * e^(-kt)
N(t + 1 day) = N₀ * e^(-k(t + 1))
From the given information, we know that N(t + 1 day) = N₀ - 1/20 * N₀ = (20 - 1)/20 * N₀ = 19/20 * N₀
Now, let's substitute these values into the equation:
N₀ * e^(-k(t + 1)) = 19/20 * N₀
Canceling out N₀ on both sides:
e^(-k(t + 1)) = 19/20
Taking the natural logarithm (ln) of both sides:
ln(e^(-k(t + 1))) = ln(19/20)
Simplifying:
-k(t + 1) = ln(19/20)
Solving for k:
k = -ln(19/20)/(t + 1)
Given that t = 0 (initial time), we can further simplify:
k = -ln(19/20)/1
k = ln(20/19)
Therefore, the exponential decay equation for this scenario is:
N(t) = 100 * e^(-ln(20/19) * t)