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

Respuesta :

Answer:

option a is the right answer

The decay factor is the decay constant subtracted from 1.

The decay factor is (c) [tex]0.87[/tex]

We have:

[tex]T_\frac{1}{2} = 5.26[/tex]

First, we calculate the decay constant [tex](\lambda)[/tex]

This is calculated using

[tex]\lambda = \frac{0.693}{T_\frac{1}{2}}[/tex]

So, we have:

[tex]\lambda = \frac{0.693}{5.26}[/tex]

[tex]\lambda = 0.13175[/tex]

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

[tex]b =1 - \lambda[/tex]

So, we have:

[tex]b =1 - 0.13175[/tex]

[tex]b =0.86825[/tex]

Approximate to 2 decimal places

[tex]b =0.87[/tex]

Hence, the decay factor is (c) [tex]0.87[/tex]

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