lynton2312
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If the activity of a source is 70 Bq after 2 years, and 35 Bq after 15 years. How long will be until the activity is 8.75 Bq?​

Respuesta :

Answer:

N = N0 e^-kt          calling k the decay constant

70 = N0 e^-2 kt

35 = N0 e^-15 k

2 = e^(-2 + 15) k

.693 = 13 k

k = .0533

N0 = N e^kt

N0 = 70 e^(.0533 * 2) = 77.87

8.75 = 77.87 e^-(.0533 t)

ln .1124 = -.0533 t

t = -ln .1124 / .0533 = 41 yr

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