JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Chemical Kinetics

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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

)

Answer: 75

Solution

Using the radioactive decay formula:

t = \frac{t _{1/2}}{0.3} lo g \frac{[ A ] _{0}}{[ A ] _{t}}

. Substituting

t = 83

days and

t_{1/2} = 200

days:

83 = \frac{200}{0.3} lo g \frac{[ A ] _{0}}{[ A ] _{t}}

Solving:

lo g \frac{[ A ] _{0}}{[ A ] _{t}} = 0.125

Taking antilog:

\frac{[ A ] _{0}}{[ A ] _{t}} = 1.333 = \frac{4}{3}

Therefore, percentage remaining =

\frac{[ A ] _{t}}{[ A ] _{0}} \times 100 = \frac{3}{4} \times 100 = 75%

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Integer answer

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