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.?

Relevance

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

• Log in to reply to the answers
• 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

• Log in to reply to the answers