Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of each element's decay is described by its half-life, which is the amount of time it takes for the number of radioactive atoms of this element to be reduced by half. The half-life of the isotope dubnium- 263 is 29 seconds. A sample of dubnium- 263 was first measured to have 1024 atoms. After t seconds, there were only 32 atoms of this isotope remaining. Write an equation in terms of t that models the situation.

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Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of each element's decay is described by its half-life, which is the amount of time it takes for the number of radioactive atoms of this element to be reduced by half. The half-life of the isotope dubnium- 263 is 29 seconds. A sample of dubnium- 263 was first measured to have 1024 atoms. After  t seconds, there were only 32 atoms of this isotope remaining. Write an equation in terms of  t that models the situation.

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Question Prompt: Question prompt: Write an equation in terms of t that models the decay of dubnium-263 atoms over time.Identify Initial Quantity and Half-life: Identify the initial quantity (a) and the half-life (b). The initial quantity of dubnium-263 atoms is given as 1024. The half-life of dubnium-263 is 29 seconds. a = 1024 b = 29 secondsDetermine Decay Factor: Determine the decay factor per half-life. Since the quantity of atoms is halved every half-life, the decay factor is 1/2.Write Exponential Decay Formula: Write the exponential decay formula. The general formula for exponential decay is y = a(1/2)^(t/b), where y is the remaining quantity after time t, a is the initial quantity, and b is the half-life.Substitute Known Values: Substitute the known values into the formula. We know that a = 1024 and b = 29. We want to find the equation that models the situation after t seconds when the remaining quantity is y. y = 1024(1/2)^(t/29)Verify Equation: Verify the equation with the given condition. We are given that after t seconds, there are only 32 atoms remaining. Let's check if the equation works for this condition. 32 = 1024(1/2)^(t/29) (1/2)^(t/29) = 32/1024 (1/2)^(t/29) = 1/32 Since 32 is 2^5 and 1024 is 2^10, we can rewrite the equation as: (1/2)^(t/29) = (1/2)^5 t/29 = 5 t = 5 * 29 t = 145 The equation holds true for the given condition, so there is no math error.

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