Calculate the half-life period of a radioactive element which remains only $1 / 16$ of it's original amount in $4740$ years:

1185

years

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2370

years

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52.5

years

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none of these

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Solution

The correct option is A $1185$ years
In four half times it remains $1 / 16$ of it's original amount so $4 t_{1 / 2} = 4740; t_{1 / 2} = 1185$ years

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