Question

A radio active material has half life of 600 years and 300 years for alpha and beta emission resp. The material decays when simultaneously alpha and beta emission. what is the time in which 3/4th of material will decay??

Answer 1

If the decays are alternate modes for decay of the same material they you also need what is know as the branching ratio. the relative amount of alpha decay vs beta decay. If alpha decay predominates then the decay will take longer than if the beta decay predominates.

Answer 2

@gleem
The question does mention simultaneous emission.. so I guess the ration is 1:1.

Let t be the amount of time taken for the 3/4 of material to decay. That means 1/4th of the material would remain.
If I only consider alpha decay..
Then we know the decay constant is \[\lambda_{\alpha} = \frac{0.693}{600} (year^{-1})\]

then the time for decay would be
\[N = N_o e^{-\lambda_{\alpha}t}\]
where \[N = \frac{N_o}{4}\]
Then you can do the same if it was beta emission alone..

Then take a weighted average.. giving you the equation
\[\frac{N_o}{4} = \frac{N_oe^{-\lambda_{\alpha}t}}{2} + \frac{N_oe^{-\lambda{\beta}t}}{2}\]

and solve for t .. I think that should work... xD
But I want someone to confirm that

Answer 3

Outside of a numerical solution how would you solve for t?

Answer 4

I think it is more like solving the question

if a tap A fills bucket in 600 years, and tap B fills the same bucket in 300 years..
how long before the bucket is filled, when both taps are switched on simultaneously?

Answer 5

except the flows decrease at a different rate as time goes on.

Answer 6

ermm yea..
And that is why I think we need the exponential equation to solve.. right?

Answer 7

I think so .

Answer 8

So the equation I put up.. is that right one?

Answer 9

I keep getting similar equations with different assumptions/ But I am not finished

Answer 10

@aryandecoolest @Mashy

This is my final proposal assuming equal probability of decaying by Alpha and Beta emission \[dN _{\alpha} =\frac{ ln2 }{ \tau _{\alpha} } Ndt\] where I have taken the liberty of using the half-llife tau from the start
\[dN _{_{\beta}}=\frac{ ln 2 }{ \tau _{\beta} } Ndt\]
But \[dN _{\alpha}+dN _{b}= dN\]
Solving I get \[N = N _{0}e ^{-t*ln2(1/\tau _{\beta}+1/\tau _{\alpha})}\] subing for the taus we get\[N = N _{0}e ^{-t*ln2/200}\] so if N/No =.25 then

t=400 days

Are any of you having troubles writing and editing your responses I lose my ability to cut and past or copy and there is a large lag between keystrokes and seeing the results.

Answer 11

ok thanks a ton @gleem @Mashy I figured it out. Thanks again