Radium (Ra 226 ,half life is 1622 years) decay into radon (Em 222, half life is 3.82 days), which is a monoatomic gas .A sample of 100 mC initially pure radium is allowed to come to equilibrium with its decay products for several months. Compute the number of radon atoms present at that time.What volume would this gas occupy at NTP?

Question

irishboy123 · Answer

Can you have a go at this first?!

Happy to provide a second opinion.  :-))

muayad99 · Answer

can you help me to solve this problem please?

osprey · Answer

radio active EXPONENTIAL DECAY LAW is maybe going to be PART of the solution here ? http://perendis.webs.com

osprey · Answer

"several months" in 1622 YEARS ????? Je ne comprends pas.

irishboy123 · Answer

I am assuming you are familiar with the usual population growth/decay model used in radioactive decay.....

$\dot N = - \lambda N  \implies N = N_o e^{- \lambda t}$

Where N is the unit you choose, say number of atoms, or total mass of element

Here, you have a sequence of decay ....  ie substance 1 decaying into substance 2 decaying into substance 3, so we need to modify that model


we can say that the amount  of  substance 1 (Radium) at any time is as above:

$N_1 = N_{o1} e^{- \lambda_1 t}$, and thus $  \dot N_1 = -\lambda_1 N_{o1} e^{- \lambda_1 t} \quad \star$


But, for substnce  2 (Radon), it is both being created and decaying itself, we we say that in terms of number of atoms at any point in time:

$\dot N_2 = -\dot N_1 - \lambda_2 N_2  $ ...  adding the decayed Radium ($\dot N_1$ is negative so negative sign means we're adding it here), and substracting the decayed Radon.

Once here, it's process until we get a useful model.  From $\star$

$\dot N_2 =  \lambda_1 N_{o1} e^{- \lambda_1 t}  - \lambda_2 N_2  $ 

Then write it like this....

$\dot N_2  + \lambda_2 N_2  =  \lambda_1 N_{o1} e^{- \lambda_1 t}    $ 

.....from which you have integrating factor $\exp (\int ~ \lambda_2 dt ) = e^{\lambda_2 t}  $

.....and the useful IV that $N_{o2} = 0$!!

And 4 or 5 more lines you end up here:

$N_2(t) = \dfrac{\lambda_1}{\lambda_2 - \lambda_1} N_{o1} (e^{- \lambda_1 t}- e^{- \lambda_2 t} )$

Finishing it off requires a litle bit of chemistry but be very careful with the units.  You can convery Curie into grammes and number of atoms, if you like.  They want the latter so let thet be the units of N.  

In terms of time, the parameter $\lambda$ connects with half life $\tau$ as follows

$\dfrac{1}{2} = e ^ {-\lambda \tau} \implies \lambda = \dfrac{ln(2)}{\tau}$

You could use days or minutes, depends how granular you want to be, but be sure to use the same units for time and half lives throughout.

muayad99 · Answer

thank you very much

osprey · Answer

Bravo encore @IrishBoy123, mais je ne comprends pas encore le question. (Rubbish french, but you may get the gist of it ... 1622 years and 3.82 days  ???)

irishboy123 · Answer

@osprey 

it means that you will get small amounts of Radon as the Radium half-life is so long, and that the Radon then decays relatively v quickly so you end up with v little Radon too.

The problem then lies with the horrible by-products of Radon decay.  

 Loads of alpha's killed Mme Curie, i think  :-((

osprey · Answer

@IrishBoy123 
The q says pure radium. So that's an awful long half life to wait a few months to get some radon. 
Yes, M Professor Curie, Nobel, and maybe Professor Lisa Meitner, unknowingly were killed by "their best friends" so to speak ?.

irishboy123 · Answer

@osprey

you got me wondering about the point of all this so i got on my horse :-)

We agreed that:

$N_2(t) = \dfrac{\lambda_1}{\lambda_2 - \lambda_1} N_{o1} (e^{- \lambda_1 t}- e^{- \lambda_2 t} ) $

Now, if we look at what happens over time,

$\lim\limits_{t \to \infty} N_2(t) \ = \lim\limits_{t \to \infty}  \dfrac{\lambda_1}{\lambda_2 - \lambda_1} N_{o1} (e^{- \lambda_1 t}- e^{- \lambda_2 t} ) \ =   \dfrac{\lambda_1}{\lambda_2 - \lambda_1} N_{o1} $

So there's a steady-state here.

Plotting $N_2(t) $ shows that it happens very quickly !!!

|dw:1479682300952:dw|

Look at the (very poorly formatted) x -axis, which is in days.  It is reached after only 30-40 days, so the 2 month experiment is more than adequate to relate a barely-decaying source amount to a daily background level.

the y axis is normalised Radium, ie we set y=1 $ \implies $ 100mC, so now we know that you will have a **constant** background level of 6.45184E-06 mC.    

PS: Baby Boycott got a ball worth of an Irish track today :-))  i reckon they switched balls for that one !!

osprey · Answer

@IrishBoy123 Today is the last day of "cheese grater voiced" Baby B and 2nd test. He wears me out, just trying to listen through the rasping implied arrogance of his "I'm right" voice. Problem is, though, his wealth of technical expertise for me as a total cricket novice. "Line, length and corridor of uncertainty" are all v interesting, as is his refs to Sir D Bradman, Brian Lara, and Harold Larwood's speed and accuracy. So, what AM I to do ?
On the current problem, I'll have a noss maybe after India have blown it with a draw/loss or actually got 8 wickets. That needs wading through another cricket comm. As entertaining as they are, they're also pretty taxing.
Baby B also declined, flatly, the invite and listener encouragement to got to "Happy Street" and embrace the local culture, as JA puts it. Miserable ...

irishboy123 · Answer

in another exchange, you mentioned Meitner. and this sprung to mind. https://www.youtube.com/watch?v=mvW93cxd2o8 scroll to 1:24:20 you may have seen it before. i find it totally fascinating: just pen and paper and genius. re- the corky: if Trump gets to be President of the free world, then England win in **Visakhapatnam**. wondering why "big games at new venues"? i hope it's Enlightenment and not Religion.

osprey · Answer

@IrishBoy123
Hahn and Meitner (M after a discussion with Bohr, Nobel) realise that they have fissioned the nucleus. Hahn gets Nobel. Meitner (female and Jewish) isn't cited. And the citation isn't in the physics list of Nobel winners. It's in the chemistry section. Why (I asked myself). Answer = Radiochemistry ! Assaying the lethal stuff to work out what was in it and to her surprise there two heavyish but lighter than before nuclides present. 
And, in so doing (and being able to sneak off to see Pr Bohr in Denmark in wartime ish) they had unlocked the "gates of hell" which is probably how it felt in Hiroshima and Nagasaki after Truman's decision. 
Professor Meitner suffered a similar "reputational fate" to that of Mileva Maric - Einstein who seems to have done quite a lot of hubby's "mathematical legwork".
Wow, we ARE in August company here !