The fraction of a radioactive sample that will decay during half of its half life period is

Question

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Answer 1

Let the half life be $t_{\frac{1}{2}}$ .

Then, $λ = \frac{0 . 6 9 3}{t _{\frac{1}{2}}} = \frac{l n 2}{t _{\frac{1}{2}}}$

$N = N_{o} e^{- ⎝ ⎛ \frac{l n 2}{t _{\frac{1}{2}}} ⎠ ⎞ \times \frac{t _{\frac{1}{2}}}{2}}$
Ratio of decay $= \frac{N _{o} ⎝ ⎜ ⎜ ⎛ 1 - e ^{- \frac{l n 2}{2}} ⎠ ⎟ ⎟ ⎞}{N _{o}} = 1 - e^{l n \frac{1}{2}} = 1 - \frac{1}{2} = \frac{2 - 1}{2}$

Answer 2

Let the half life be

t_{\frac{1}{2}}

.

Then,

λ = \frac{0 . 6 9 3}{t _{\frac{1}{2}}} = \frac{l n 2}{t _{\frac{1}{2}}}

N = N_{o} e^{- ⎝ ⎛ \frac{l n 2}{t _{\frac{1}{2}}} ⎠ ⎞ \times \frac{t _{\frac{1}{2}}}{2}}

Ratio of decay

= \frac{N _{o} ⎝ ⎜ ⎜ ⎛ 1 - e ^{- \frac{l n 2}{2}} ⎠ ⎟ ⎟ ⎞}{N _{o}} = 1 - e^{l n \frac{1}{2}} = 1 - \frac{1}{2} = \frac{2 - 1}{2}