The half-life for U 238 is 4.5x109 years. a) If five half-lives have gone by how many years have gone by?
b) If you start with 240 grams of U 238 and end up with 60 grams, how many years have gone by?
c) If you start with 240 grams of U 238 and 1.8 x 1010 years go by, how much U 238 is left?
d) If you start with 562 g and six half lives go by how many grams are left?

A. 2.25×10¹⁰ years

B. 9×10⁹ years

C. 15 g

D. 8.78 g

Explanation:

From the question given above, the following data were obtained:

Half-life (t₁/₂) = 4.5×10⁹ years

A. Determination of how many years has gone by.

Half-life (t₁/₂) = 4.5×10⁹ years

Number of half-lives (n) = 5

Time (t) =?

t = n × t₁/₂

t = 5 × 4.5×10⁹

t = 2.25×10¹⁰ years

Therefore, 2.25×10¹⁰ years has gone by.

B. Determination of how many years has gone by.

We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:

Original amount (N₀) = 240 g

Amount remaining (N) = 60 g

Number of half-lives (n) =?

N = 1/2ⁿ × N₀

60 = 1/2ⁿ × 240

Cross multiply

60 × 2ⁿ = 240

Divide both side 60

2ⁿ = 240/60

2ⁿ = 4

Express 4 in index form with 2 as the base

2ⁿ = 2²

n = 2

Finally, we shall determine how many years has gone by. This can be obtained as follow:

Half-life (t₁/₂) = 4.5×10⁹ years

Number of half-lives (n) =

Time (t) =?

t = n × t₁/₂

t = 2 × 4.5×10⁹

t = 9×10⁹years

Therefore, 9×10⁹ years has gone by.

C. Determination of the amount remaining.

We'll begin by calculating the number of half-lives that has elapse. This can be obtained as follow:

Half-life (t₁/₂) = 4.5×10⁹ years

Time (t) = 1.8×10¹⁰ years

Number of half-lives (n) = ?

t = n × t₁/₂

1.8×10¹⁰ = n × 4.5×10⁹

Divide both side by 4.5×10⁹

n = 1.8×10¹⁰ / 4.5×10⁹

n = 4

Finally, we shall determine the amount remaining. This can be obtained as follow:

Original amount (N₀) = 240 g

Number of half-lives (n) = 4

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2⁴ × 240

N = 1/16 × 240

N = 240 / 16

N = 15 g

Therefore, 15 g is remaining.

D. Determination of the amount remaining.

Original amount (N₀) = 562 g

Number of half-lives (n) = 6

Amount remaining (N) =?

N = 1/2ⁿ × N₀

N = 1/2⁶ × 562

N = 1/64 × 562

N = 562 / 64

N = 8.78 g

Therefore, 8.78 g is remaining.

found in the nucleus

has mass of one amu