Question

The half-life of a radioactive substance is 4 hours. If initially there are 256 grams, how much remains after 10 hours?

• A. 45.26 g
• B. 16 g
• C. 64 g
• D. 8 g

Hint:

Each half-life reduces the quantity to half. Use the formula: \( A = A_0 \left(\frac{1}{2}\right)^{t/T} \), where \( T \) is the half-life.

Answer 1

The Correct Option is A

Using the radioactive decay formula: \[ N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \] Given:

• Initial amount ($N_0$) = 256\,g
• Half-life ($t_{1/2}$) = 4\,h
• Elapsed time ($t$) = 10\,h

Calculation Steps
1. Compute the exponent: \[ \frac{t}{t_{1/2}} = \frac{10}{4} = 2.5 \] 2. Calculate the decay factor: \[ \left(\frac{1}{2}\right)^{2.5} = 2^{-2.5} \approx 0.1768 \] 3. Determine remaining quantity: \[ N(10) = 256 \times 0.1768 \approx 45.26\,g \] Verification
After each 4\,h half-life:

• At 4\,h: $256/2 = 128\,g$
• At 8\,h: $128/2 = 64\,g$
• At 12\,h: $64/2 = 32\,g$

Since 10\,h is 2.5 half-lives, the exact calculation shows: \[ 256 \times (0.5)^{2.5} \approx 45.26\,g \] Conclusion
The exact amount remaining after 10\,h is \(\boxed{45.26\,g}\).