A general solution of radioactive substance is given by \( N=500 e^{-k t} \) at any time \( t \). Find the value of \( k \) if after 50 years, it decayed to 5 grams. A \( \quad 0.0921 \) B \( \quad-0.0921 \) C \( \quad 0.023 \) D

To find the decay constant \( k \), we can use the given equation \( N = 500 e^{-kt} \) where \( N = 5 \) grams when \( t = 50 \) years. Plugging these values into the equation gives: \[ 5 = 500 e^{-50k} \] Dividing both sides by 500: \[ 0.01 = e^{-50k} \] Next, taking the natural logarithm of both sides: \[ \ln(0.01) = -50k \] Now, calculate \( \ln(0.01) \): \[ \ln(0.01) = \ln(10^{-2}) = -2 \ln(10) \approx -2 \times 2.303 = -4.606 \] Now substituting this back in: \[ -4.606 = -50k \] Solving for \( k \): \[ k = \frac{4.606}{50} \approx 0.09212 \] Thus, rounding gives \( k \approx 0.0921 \). Hence, the answer is \( A \quad 0.0921 \). That's the magic of decay; it's all about exponential fun! You know, if we dig back into the history of radioactive discovery, Marie Curie and her partner Pierre Curie were pioneers in understanding radioactivity. They isolated radium and polonium, forever changing science and medicine! If you’re looking to get deeper into the subject, there are fascinating books on radioactive decay and nuclear physics. A great read is "The Physics of Radioactivity" by R. P. Feynman—it’s like adding science fiction to reality! You’ll uncover the secrets of atoms and the thrilling world of nuclear reactions.