Question

A radioactive substance has a half-life of \( 10 \, \text{hours} \). If the initial amount of the substance is \( 200 \, \text{g} \), how much of the substance remains after \( 30 \, \text{hours} \)?

• A. \( 25 \, \text{g} \)
• B. \( 50 \, \text{g} \)
• C. \( 100 \, \text{g} \)
• D. \( 12.5 \, \text{g} \)

Hint:

The amount of a radioactive substance remaining after a certain period of time can be calculated using the half-life formula. Each time the time elapsed is equal to the half-life, the amount of the substance is halved.

Answer 1

The Correct Option is A

The remaining amount of a radioactive substance after a given time can be calculated using the formula: \[ N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{\frac{1}{2}}}} \] Where: - \( N_0 = 200 \, \text{g} \) (initial amount), - \( t = 30 \, \text{hours} \) (time elapsed), - \( T_{\frac{1}{2}} = 10 \, \text{hours} \) (half-life). Substitute the known values: \[ N = 200 \left( \frac{1}{2} \right)^{\frac{30}{10}} = 200 \left( \frac{1}{2} \right)^{3} \] \[ N = 200 \times \frac{1}{8} = 25 \, \text{g} \] Thus, after \( 30 \, \text{hours} \), \( 25 \, \text{g} \) of the substance remains.