A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is___________. (Nearest integer)

(Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)

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Alok Kumar Singh | Contributor-Level 10

6 months ago

$t_{1 / 2} = 200 d a y s = \frac{0.693}{k}$

$t = \frac{2.303}{k} l o g_{10} \frac{N_{0}}{N}$

83 = $\frac{2.303}{0.693} * 200 l o g \frac{N_{0}}{N}$

$83 = \frac{200}{0.3010} l o g \frac{N_{0}}{N}$

0.125 = $l o g \frac{N_{0}}{N}$

$\frac{N_{0}}{N} = a n t i l o g (0.125)$

= 1.333

Activating remaining = $\frac{N}{N_{0}} * 100$

$= \frac{\frac{N_{0}}{1.333}}{N_{0}} * 100$

= 75%

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Kindly consider the following Image

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A radioactive element has a half life of 200 days. The percentage of original activity remaining after 83 days is___________. (Nearest integer) (Given : antilog 0.125 = 1.333, antilog 0.693 = 4.93)