Question

The half-life of an isotope known as colbalt-60 is 5.26 years. What is the
decay factor of this substance? (Hint: use the equation y= ca” ) Round your
answer to 2 decimal places.

A) 1.14
B) 0.026
C) 0.877
D) 2

Answer 1

Answer:

option a is the right answer

Answer 2

The decay factor is the decay constant subtracted from 1.

The decay factor is (c) $0.87$

We have:

$T_\frac{1}{2} = 5.26$

First, we calculate the decay constant $(\lambda)$

This is calculated using

$\lambda = \frac{0.693}{T_\frac{1}{2}}$

So, we have:

$\lambda = \frac{0.693}{5.26}$

$\lambda = 0.13175$

Next, we calculate the decay factor (b) using:

$b =1 - \lambda$

So, we have:

$b =1 - 0.13175$

$b =0.86825$

Approximate to 2 decimal places

$b =0.87$

Hence, the decay factor is (c) $0.87$

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