A sample of francium- 212 will decay to one sixteenth its original amount after \(80 \mathrm{~min}\). What is the half-life of francium-212?

Short Answer

Expert verified
The half-life of francium-212 is 4 minutes.

Step by step solution

01

Understand the problem

Given that a sample of francium-212 will decay to one-sixteenth its original amount after 80 minutes. Our aim is to find the half-life of this radioisotope. Francium-212 follows exponential decay, so we can use the formula \[ N(t) = N(0) \times (1/2)^{t/T} \] where \( N(t) \) is the remaining amount after time \( t \), \( N(0) \) is the original amount, \( T \) is the half-life we're looking for, and \( t \) is the lapse time, which is 80 minutes.
02

Set up equation

Since our remaining amount is one-sixteenth of the original after 80 minutes, we can plug in the values into our equation. So, our equation becomes \[ 1/16 = (1/2)^{80/T} \].
03

Simplify the equation

Using properties of exponents, we rewrite the left-hand side of the equation, \( 1/16 = 1/2^4 = (1/2)^{4 \cdot 1} = (1/2)^{80/(T/4)} \]. So now we have our equation as \[ (1/2)^{80/(T/4)} = (1/2)^{80/T} \]. When the bases are equal, the exponents must also be equal, thus we get \( 80/(T/4) = 80/T \).
04

Solve for T

Cross multiply to solve for \( T \), we get \( 80T = 4 \cdot 80 \) or \( T = 4 \cdot 80 / 80 = 4 \) minutes. This implies that the half-life of francium-212 is 4 minutes.

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