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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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firefighter118 firefighter118
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10 years ago
The question states that a source of Beta radiation falls to 85% of its initial activity in 52 seconds.
Thus far I applied the equation A=A0e^(-lambda x t)
I applied that to get as far as ln85=ln100 x lambda x 52
But when I rearrange that I get the wrong answer. Firstly are these numbers correct and if so, how do I rearrange them to get the right answer.
Thanks for any help. Slight Smile
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smith123smith123
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10 years ago
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#2
10 years ago
You've put the values into the equation as:

85 = 100 x e^(-lambda x 52)

But then you've taken the natural log of both sides and made a mistake in taking the log of the right hand side. The ln of 100 x e^(-lambda x 52) is not ln(100) x ln(e^(-lambda x 52).

The log of (a times b) is log (a) + log (b), not log (a) times log(b).
wrote...
#3
10 years ago
A=A?e^(-?t)

When you take logarithms you should do this:
ln(A)=ln[A?e^(-?t)]

On the right hand side you are taking the logarithm of A? multiplied by e^(-?t). This gives:
ln(A)=ln(A?) + ln(e^(-?t))

Which in turn gives:
ln(A)=ln(A?) - ?t
_____________________________ __________
The neatest method is this:

85 = 100e^(-52?)

e^(-52?) = 0.85

ln(e^(-52?)) = ln(0.85)

-52? = -0.1625

? = 0.0031s?¹
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