How long will it take for 10% of the u−238 atoms in a sample of u−238 to decay?

The half-life the radioactive decay of U−238 is 4.5 billion
many years and it is separate of preliminary focus. A) the length of time does it simply take for 10per cent of this U−238 atoms in an example
of U−238 to decay? Express your solution making use of two considerable numbers. B) If an example of U−238 at first included 1.4×1018
atoms once the world had been created 13.8 billion years back, exactly how many
U−238 atoms does it include these days? 18 Express your solution making use of two considerable numbers.

(A)
For first-order decay procedure:
ln(Ct/Co) = -kt
……………….(1)
in which
Co= preliminary focus
Ct = focus of U-238 staying at time t
k = decay constant
we all know for first-order effect
k = 0.693 / t
k = 0.6931 / 4.5 x 109 = 1.5403 x 10-10
year-1
10per cent decayed implies 90per cent keeps
and Ct/Co x 100per cent = 90per cent
therefore
Ct/Co = 0.9
ln(0.9) = -1.5403 x 10-10 x t
– 0.10536 = -1.5403 x 10-10 x t
t = – 0.10536 / -1.5403 x 10-10
t = 0.68 x 10-9 many years
t = 0.68 billion many years
(B) offered Co = 1.4 × 1018
atoms
t = 13.8 billion many years = 13.8 x 109 many years
Level of U-238 staying Ct = Co
e(-kt)
Ct = 1.4 × 1018 x e(-1.5403 x
10-10 x 13.8 x 10 9)
Ct = 1.4 × 1018 x
e-2.125614
Ct = 1.4 × 1018 x 0.11935965893247
Ct = 1.67 x 1017 ≈ 1.7 x 1017
atoms
therefore
Ct = 1.7 x 1017 atoms

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