Question

An unknown radioactive substance has a half-life of 3.20 hours . If 23.9 g of the substance is currently present, what mass Ao was present 8.00 hours ago? Express your answer with the appropriate units. ► View Available Hint(s) ? Å Value o O Units A0 =

Answer 1

PART 1

We have, decay constant ( ) = 0.693 / t 1/2

Where, t _1/2 is a half-life radioactive element.

Decay constant of unknown substance =  = 0.693 / t _1/2 = 0.693 / 3.20 hours = 0.02166 hour ^-1

We have equation: t =2.303 /    log (N ₀ / N _t)

Where, is a decay constant, N ₀ is the initial amount of radioactive element, N _t is the amount of radioactive element after time t.

In this example, N ₀ = [A ] ₀ = ?

N _t = [A ] _t =23.9 g

t = 8 hours

Hence, 8 days = 2.303 / 0.02166 hour ^-1 log [A] ₀ / 23.9 g

log [A ] ₀ / 23.9 g = 8 days x 0.02166 days^-1 / 2.303 = 0.07524

A] ₀ / 23.9 g = 10^0.07524 =1.189

[A] ₀ =1.189 x 23.9 g = 28.4 g

ANSWER: Amount of A before 8 hours ago was 28.4 g

PART 2

Consider given reaction , CO (g) + Cl ₂ (g) $\rightleftharpoons$ COCl ₂ (g)

For above reaction, K p = P _{COCl 2} / (P _CO )( P _{Cl 2})

Where, P _{COCl 2} is pressure of COCl ₂ at equilibrium , P _CO is pressure of CO at equilibrium and P _{Cl 2}is pressure of Cl ₂ at equilibrium .

Therefore, K p = ( 0.250 ) / ( 0.840)(1.23) = 0.242

ANSWER : K p = 0.242

PART 3

We have, decay constant = 0.693 / t _1/2

Where, t _1/2 is a half life radioactive element.

Decay constant of ¹⁴ C =  = 0.693 / t _1/2

= 0.693 / 5715 y

= 1.21 x 10 ^-04 y ^-1

We have equation: t = - 1/ ln (N _t / N ₀)

Where N is a amount of radioactive element.

We can replace N by rate of disintegration.

Therefore, t = - 1/ ln (Rate _t / Rate ₀)

Where, Rate _t is a rate of disintegration of ¹⁴ C from wooden artifact and

  Rate ₀ is a rate of disintegration of ¹⁴ C from live wood.

Therefore, t = - 1/ 1.21 x 10 ^-04 y ^-1  ln ( 39.3 cpm / 58.2 cpm)   

t = - 1/ 1.21 x 10 ^-04 y ^-1 x ( - 0.3927 )

t = 3245.4 y

ANSWER : Age of wooden artifact = 3.2 $\times$ 10 ³ yr

PART 4 and PART 1 are same, hence part 4 not solved. Refer answer of PART 1.