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If the background countrate is 1 per second and the countrate with the beta source present is only 3 per second, how long would you have to measure the background counts and how long would you have to measure the beta source counts in order to have 1% precision in the difference between the two countrates? x=3...P(m)=((x^m)/m!)(e^-x)

2007-01-27 18:09:03 · 1 answers · asked by chijliak 1 in Science & Mathematics Physics

1 answers

for a random variable the error is the sqrt(n)/n and to have 1% difference each must have an error of .5% so 1/sqrt(n)=.005 therefore n=1/.005^2=40000
to get n counts you must measure for n sec back ground and n/3 sec for beta

2007-01-27 19:42:38 · answer #1 · answered by meg 7 · 1 0

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