The half-life of $^{65}$Zn is 245 days. After $x$ days, 75% of the original activity remained. The value of $x$ in days is __________ (Nearest integer).}
(Given: $\log 3 = 0.4771$ and $\log 2 = 0.3010$)
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If less than one half-life has passed, the remaining activity will be greater than 50%.
The half-life of a radioactive isotope is the time required for half of the original sample to decay. Here, for $^{65}$Zn, the half-life is 245 days. The decay process can be mathematically described by the formula: \( N = N_0 \times (1/2)^{t/T_{1/2}} \) where \( N \) is the remaining quantity, \( N_0 \) is the initial quantity, \( t \) is the time elapsed, and \( T_{1/2} \) is the half-life.
Given that after \( x \) days, 75% of the original activity remains, we have \( N = 0.75N_0 \). Substituting these into the equation gives: \( 0.75N_0 = N_0 \times (1/2)^{x/245} \) Simplifying this, we have: \( 0.75 = (1/2)^{x/245} \) Taking the logarithm of both sides, we obtain: \( \log 0.75 = \left(\frac{x}{245}\right) \log(1/2) \) Using the given values \( \log 3 = 0.4771 \) and \( \log 2 = 0.3010 \), we find: \( \log 0.75 = \log\left(\frac{3}{4}\right) = \log 3 - \log 4 = 0.4771 - 2(0.3010) = 0.4771 - 0.6020 = -0.1249 \) Also, \( \log(1/2) = -0.3010 \). Substituting back, we have: \( -0.1249 = \left(\frac{x}{245}\right)(-0.3010) \) Solving for \( x \), we get: \( x = \frac{-0.1249 \times 245}{-0.3010} \approx 101.65 \) Rounding to the nearest integer, \( x = 102 \). This value fits within the given range 102,102, thus ensuring accuracy. Therefore, the value of \( x \) is 102 days.
