How many undecayed nuclei will remain?

[Physicsman788]

Homework Statement

A sample contains 500 undecayed nuclei initially. Its decay constant is 0.5 /s. After 5 second, how many undecayed nuclei will remain?

Relevant Equations

𝑁 = No𝑒−λ𝑡

Attempt at solution:

𝑁 = 500𝑒−5 = 3.3689735 undecayed nuclei remaining

Edit: Sorry guys, I found it out its 500e-0.5x5

 

[Richard R Richard]

Hello,
The formula you have tried to use only gives you the number of cores that have changed state or decayed.
To find those that have not yet decayed, you must subtract that value from the number of original nuclei.

Pd it is convenient that you express your formulas in LaTeX language, it is easy to learn, in half an hour you will achieve it and it will always be very useful to ask other topics in the future.

Edit thanks for correction @Steve4Physics

$N_{rest}=N_0 e^{-\lambda t}$

 

[Steve4Physics]

I think the OP was correct. Your formula gives the number that have decayed, not the numer that have 'not yet decayed'. E.g put t=0 and see what happens.

 

[Steve4Physics]

The number of undecayed nuclei is given by $N(t)=N_0(1-e^{-\lambda t})$. The answer "3.3689735 undecayed nuclei" can be safely rounded to "3 undecayed nuclei."

With t=0 this would give the number of undecayed nuclei as
$N_0(1-e^{-\lambda *0}) = N_0(1-e^0) = N_0(1-1) = 0$

But at t=0 the number of undecayed nuclei is $N_0$, not $0$.

The expression $N_0(1-e^{-\lambda t})$ is the number of nuclei that have decayed in time t.

 

[kuruman]

With t=0 this would give the number of undecayed nuclei as
$N_0(1-e^{-\lambda *0}) = N_0(1-e^0) = N_0(1-1) = 0$

But at t=0 the number of undecayed nuclei is $N_0$, not $0$.

The expression $N_0(1-e^{-\lambda t})$ is the number of nuclei that have decayed in time t.

You are absolutely correct. I will delete the post to avoid clutter.

 

[etotheipi]

Another way of looking at the same thing is to see that 2.5 lifetimes $(\tau = 2\text{s})$ have elapsed, so the fraction remaining is just $(e^{-1})^{2.5}$.

 

[Richard R Richard]

A proportional relationship between the rate of decay and the number of nuclei that can still decay has been experienced and seen.
This is
$$\dfrac {\partial N} {\partial t} = - K N$$
This is an easy differential equation to solve by applying as initial parameters that for t = 0 the number of available cores is $N_0$

$$\dfrac {dN} {N} = - K dt$$

$$\displaystyle {\int_ {N_0} ^ N \dfrac {dN} {N} = - K \int_0 ^ t dt}$$
Whose solution is
$$Ln (N) -Ln (N_0) = Ln \left (\dfrac {N} {N_0} \right)= - K (t-0)$$
From where operated mathematically
$\dfrac {N} {N_0} = e <- Kt}$

then
$$N = N_0 e^{ - Kt}$$
It is evident that $K$ is the decay constant is the quotient between the number of decays per second and the number of radioactive atoms ($λ = A / N$).

Staying
$$N = N_0 e^{- λt}$$
Either it is understood that
$$\dfrac {A} {\lambda} = \dfrac {A_0} {\lambda} e^{- \lambda t}$$
Because
$$N \lambda = N_0 \lambda e^{- \lambda t}$$
It is an exponential distribution
The time that elapses until the number of radioactive nuclei of a radioactive isotope is reduced to half of the initial amount is known as the half-life, period, half-period, half-life or half-life (not to be confused with the aforementioned life time ) ($t_{1/2} = ln2 / λ$). At the end of each period, the radioactivity is reduced to half of the initial radioactivity. Each radioisotope has a characteristic half-period, generally different from other isotopes. And it is that way easy to measure experimentally $\lambda$.