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    Initial sample activity= 3450Bq
    Half life= 3.89x10^8s
    decay constant= 1.78x10^-9

    Show that the time taken for the activity of the sample to fall to 10% of its initial value is about 40 years.

    don't understand this question even with the markscheme
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    (Original post by AndyOC)
    Initial sample activity= 3450Bq
    Half life= 3.89x10^8s
    decay constant= 1.78x10^-9

    Show that the time taken for the activity of the sample to fall to 10% of its initial value is about 40 years.

    don't understand this question even with the markscheme
    Turn the seconds (in half life etc) into more useful numbers, then use the numbers. I don't know the question, but if there's space to draw a graph that'd be a good way to prove it.
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    (Original post by Vicky628)
    Turn the seconds (in half life etc) into more useful numbers, then use the numbers. I don't know the question, but if there's space to draw a graph that'd be a good way to prove it.
    Answer

    My phone isn't working bear with me.
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    (Original post by AndyOC)
    Initial sample activity= 3450Bq
    Half life= 3.89x10^8s
    decay constant= 1.78x10^-9

    Show that the time taken for the activity of the sample to fall to 10% of its initial value is about 40 years.

    don't understand this question even with the markscheme
    A = A0 * e^-λt

    A = 100
    A0 = 10
    λ = 1.78x10^-9

    Rearrange for t. It will be in seconds. Divide by the number of seconds in a year. It should give ~40.

    (A could = 10, and A0 = 1. As long as the ratio is the same – 10:1. I used 100 and 10 so that you can see them as percentages.)
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    (Original post by Thediplomat56)
    Answer
    Since you asked so nicely...

    Don't you have the mark scheme??
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    This is the answer sorry didn't show u on the last post
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    (Original post by ombtom)
    A = A0 * e^-λt

    A = 100
    A0 = 10
    λ = 1.78x10^-9

    Rearrange for t. It will be in seconds. Divide by the number of seconds in a year. It should give ~40.

    (A could = 10, and A0 = 1. As long as the ratio is the same – 10:1. I used 100 and 10 so that you can see them as percentages.)
    Don't i have to log both sides to get t the formula?
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    (Original post by Thediplomat56)
    Answer

    My phone isn't working bear with me.
    1.29x10^9s
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    (Original post by Thediplomat56)
    This is the answer sorry didn't show u on the last post
    If you use the formula you can see that the activity will be one tenth, because activity is proportional to the amount of substance left.

    After you know that the rest is algebra and converting from seconds to years

    The conversion is 60 seconds in a minute, 60 minutes in an hour

    24 hours in a day and 365 days in a year

    60 x 60 x 24 x 365

    I omitted numbers but feel free to check
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    (Original post by Thediplomat56)
    This is the answer sorry didn't show u on the last post
    ah brilliant, understand it now thanks for the help!
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    (Original post by AndyOC)
    Don't i have to log both sides to get t the formula?
    You will have to use the natural log, yes. If you are not doing A-level maths then it might take a while to get used to it.

    I assume that you've got this far:

    A/A0 = e^-λt

    From here you can take the natural log to both sides. ln(e^x) = x. (Not sure how much you know about logs; remember that ln is log in base e.) So ln(e^-λt) = -λt.

    So ln(A/A0) = -λt.

    Then carry on as I said before.

    Edit: someone else has explained it, but I'll leave this incase the stuff about logs helps you.
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    you're doing unit 5 already =O, we aren't even half way through unit 4 yet.
 
 
 
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