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If 10% of a radioactive material decay in 5 days, then the amount of original material left after 20 days is approximately :

#### Options

50

60

65

70

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#### Solution

**65**

**Explanation - **

N = N00e^{-λt}

When 10% decay N= 90%

So, 9 N_{0} = N_{0} `e^(-λ^5)`

⇒ log . 9 = -5λ ...(i)

Let assume after 20 days fraction in decayed be x.

xN_{0} = N_{0} e - 20λ

log x = -20λ ... (ii)

from eqn (i) & (ii)

`1/4 = ("log".9)/("log" x)`

log x = 4 log . 9

x=(9)^{4} = 65

% of N = 65%

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