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What is the half-life period of a radioactive material if its activity drops to 1/16^{th} of its initial value of 30 years?

#### Options

9.5 years

8.5 years

7.5 years

10.5 years

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#### Solution

**7.5 years**

**Explanation:**

The activity is given as,

A = A_{o} e^{-λt}

⇒ -λt = ln `("A"/"A"_"o")`

⇒ `ln 2/"t"_(1//2)xx30=ln(1/16)`

⇒ `ln 2/"t"_(1//2)xx30=(ln2)xx-4`

⇒ t_{1/2} = `30/4`

= 7.5 years

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