Suppose an unknown radioactive substance produces 8000 counts per minute on a Geiger counter at a certain time, and only 500 counts per minute 13 days later. Assuming that the amount of radioactive substance is proportional to the number of counts per minute, determine the half-life of the radioactive substance.

The half-life of the radioactive substance is of 3.25 days.

Step-by-step explanation:

The amount of radioactive substance is proportional to the number of counts per minute:

This means that the amount is given by the following differential equation:

In which k is the decay rate.

The solution is:

In which Q(0) is the initial amount:

8000 counts per minute on a Geiger counter at a certain time

This means that

500 counts per minute 13 days later.

This means that . We use this to find k.

So

Determine the half-life of the radioactive substance.

This is t for which Q(t) = 0.5Q(0). So

The half-life of the radioactive substance is of 3.25 days.

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

yes

step-by-step explanation: