Anonymous
Anonymous asked in Science & MathematicsChemistry · 4 weeks ago

Calculate the half-life (in minutes) for a radioactive substance which loses 69.3% of its original intensity in 2.00 hours.?

2 Answers

Relevance
  • 4 weeks ago
    Favourite answer

    2 hours = 120 minutes 

    Let N(t) denote the amount of material present at time t. 

    N= Ce^(kt)

    N= Ni at t=0

    Ni= Ce^(k0)=C

    The solution becomes  N= Nie^(kt)

    at t=120 min 69.3 % of the original material has decayed, so 30.7 % remains.

    Hence N=0.307Ni at t=120 min. So, 0.307Ni= Nie^k(120)

    0.307= e^k(120)

    ln(0.307)=k*120

    k=−0.0098408960949591

    Ni/2= Nie^(−0.0098408960949591t)

    −0.0098408960949591t=ln(1/2)

    t=70.43 min

    That is your answer. 70.43 min at half life

    • Commenter avatarLog in to reply to the answers
  • 4 weeks ago

    100% total - 69.3% lost = 30.7% remains

    Let z be the half-life (in minutes) to be found:

    0.307 = (1/2)^((2.00 hours x 60 min/hour) / z)

    Solve for z algebraically:

    log 0.307 = ((2.00 hours x 60 min/hour) / z) x log (1/2)

    log 0.307 /  log (1/2) = (120 min) / z

    z = (120 min) / (log 0.307 /  log (1/2)) = 70.4 minutes

    • Commenter avatarLog in to reply to the answers
Still have questions? Get answers by asking now.