Cobalt-57 is radioactive, emitting `beta-`particles.The half-life for this is 270 days. If 100 mg of this is kept in an open container, then what mass (in mg) of Cobalt-57 will remain after 540 days ?

Answer

To solve the problem of how much Cobalt-57 will remain after 540 days, we can follow these steps:

### Step 1: Understand the half-life concept
The half-life of a radioactive substance is the time required for half of the substance to decay. In this case, the half-life of Cobalt-57 is given as 270 days.

### Step 2: Determine the number of half-lives that have passed
To find out how many half-lives have passed in 540 days, we can use the formula:
\[ n = \frac{t}{t_{1/2}} \]
where:
- \( t \) is the total time elapsed (540 days)
- \( t_{1/2} \) is the half-life (270 days)

Calculating \( n \):
\[ n = \frac{540 \text{ days}}{270 \text{ days}} = 2 \]

### Step 3: Calculate the remaining mass
The remaining mass of a radioactive substance after \( n \) half-lives can be calculated using the formula:
\[ R = R_0 \times \left(\frac{1}{2}\right)^n \]
where:
- \( R_0 \) is the initial mass (100 mg)
- \( n \) is the number of half-lives (2)

Substituting the values:
\[ R = 100 \text{ mg} \times \left(\frac{1}{2}\right)^2 \]
\[ R = 100 \text{ mg} \times \frac{1}{4} \]
\[ R = 25 \text{ mg} \]

### Final Answer
After 540 days, the mass of Cobalt-57 remaining is **25 mg**.

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