[SOLVED] Radioactive decay


by timothy123
Tags: decay, radioactive, solved
timothy123
timothy123 is offline
#1
Jan4-08, 01:30 PM
P: 4
Hi guys, this is just a basic maths incompetency. I hope someone can guide me forwards! (Slack biology student, delving into physics)

Cosmic ray muons decay with a half-life of 1.53 x 10-6 s. During an experiment 563 +/- 10 muons are counted at the top of Mount Washington, an altitude of 6265 ft. When the same apparatus is moved to an altitude of 10 ft the number of muons detected falls to 408 +/- 9, over the same time period.

(a) Given that the speed of the muons is 0.992c determine the flight time, Tf, ( 1 metre = 3.28 ft )

(b) Using the appropriate decay law find the time of flight as measured from the muon ‘clock’, Tm, and determine the ratio Tm/Tf.

(c) Compare this with the value predicted by the appropriate Lorentz transformation and comment.


, where N(t) is the number left at time 't', No the original number and T the half life.

I can do (a) with no problems, and thus I can probably do the latter half of (b) and (c) without issue. My stumbling point is in actually rearranging the above equation to isolate the term I want. I have this:


- but I don't know how to get 't' where I want it! I've been googling how to manipulate powers properly, but no joy. What can I say... no maths since I was 16! :-)

Cheers all.

Tim
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Kurdt
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#2
Jan4-08, 01:41 PM
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You'll want to take the logarithm to base 2 of both sides.

More info: http://www.counton.org/alevel/pure/purtutloglaw.htm
timothy123
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#3
Jan5-08, 05:44 AM
P: 4
Quote Quote by Kurdt View Post
You'll want to take the logarithm to base 2 of both sides.

More info: http://www.counton.org/alevel/pure/purtutloglaw.htm
Thanks Kurdt. A much-appreciated pointer.

Does that mean that the following is true?:


It was this youtube video that took me there... http://youtube.com/watch?v=OS-EnAFKWpI
Seems like a great resource for basic maths tips.

Kurdt
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#4
Jan5-08, 07:06 AM
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PF Gold
P: 4,975

[SOLVED] Radioactive decay


Nice video. Yes you are correct.


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