A substance decays according to the formula \( Q=36(0.5)^{t / 36.7} \), where \( t \) is in seconds. (a) What is its half-life? (Do not round off.) The half-life is Number seconds (b) By what percent does it decay each second? (Report to 2 decimal places.) It decays by Number \% per second

1. **Half-life** The decay formula is given by \[ Q = 36\left(0.5\right)^{t/36.7}. \] The half-life is the time \( t \) when \( Q \) becomes half of its initial value. Since the initial value is \( 36 \), we have \[ \frac{36}{2} = 18. \] Substitute \( t = T_{1/2} \) in the formula: \[ 18 = 36\left(0.5\right)^{T_{1/2}/36.7}. \] Dividing both sides by \( 36 \) gives: \[ 0.5 = \left(0.5\right)^{T_{1/2}/36.7}. \] Since the bases are the same, the exponents must be equal: \[ \frac{T_{1/2}}{36.7} = 1 \quad \Longrightarrow \quad T_{1/2} = 36.7. \] Therefore, the half-life is \( 36.7 \) seconds. 2. **Percent Decay per Second** The decay per second is found by considering the decay factor when \( t = 1 \). For \( t = 1 \), the formula becomes: \[ Q(1) = 36\left(0.5\right)^{1/36.7}. \] The factor of decay per second is: \[ r = \left(0.5\right)^{1/36.7}. \] Expressing the exponentiation in terms of the natural exponential function: \[ r = \exp\left(\frac{1}{36.7}\ln(0.5)\right). \] Since \(\ln(0.5) \approx -0.693147\), we have: \[ r \approx \exp\left(\frac{-0.693147}{36.7}\right) = \exp(-0.0188824) \approx 0.98129. \] The fraction decayed in one second is: \[ 1 - r \approx 1 - 0.98129 = 0.01871. \] Therefore, the percent decay per second is: \[ 0.01871 \times 100\% \approx 1.87\%. \] - (a) The half-life is \( 36.7 \) seconds. - (b) It decays by approximately \( 1.87\% \) per second.