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