Americium -235 has a half-life of 10.3 minutes. The mass of americium- 235 in a sample decays from 1.6 mg to 0.1 mg. How long will this take (in minutes)?

We start with the exponential decay formula \[ N(t)=N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}, \] where: - \( N(t) \) is the mass at time \( t \), - \( N_0 \) is the initial mass, - \( T_{1/2} \) is the half-life. Given: - \( N_0 = 1.6 \) mg, - \( N(t) = 0.1 \) mg, - \( T_{1/2} = 10.3 \) minutes. Substitute these values into the equation: \[ 0.1 = 1.6\left(\frac{1}{2}\right)^{\frac{t}{10.3}}. \] Divide both sides by 1.6: \[ \left(\frac{1}{2}\right)^{\frac{t}{10.3}} = \frac{0.1}{1.6} = 0.0625. \] Recognize that \[ 0.0625 = \frac{1}{16} = \left(\frac{1}{2}\right)^4. \] Thus, we have: \[ \left(\frac{1}{2}\right)^{\frac{t}{10.3}} = \left(\frac{1}{2}\right)^4. \] Since the bases are the same, the exponents must be equal: \[ \frac{t}{10.3} = 4. \] Now, solve for \( t \): \[ t = 4 \times 10.3 = 41.2 \text{ minutes}. \] Therefore, the mass will decay from 1.6 mg to 0.1 mg in \( 41.2 \) minutes.