JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Chemical Kinetics

Assuming

1 μ g

of trace radioactive element X with a half life of 30 years is absorbed by a growing tree. The amount of X remaining in the tree after 100 years is ____

\times 1 0^{- 1} μ g

.
[Given :

ln 10 = 2.303; lo g 2 = 0.30

]

Answer: 1

Solution

For radioactive decay, we use the first-order kinetics equation:

t = \frac{t _{1/2}}{ln 2} ln (\frac{a}{a - x})

where

a

is the initial amount and

a - x

is the remaining amount.

Given:

t_{1/2} = 30

years,

t = 100

years,

a = 1 μ g

100 = \frac{30}{ln 2} ln (\frac{1}{w})

ln (\frac{1}{w}) = \frac{100 ln 2}{30} = \frac{10 ln 2}{3}

Using

ln 2 = 0.693

ln (\frac{1}{w}) = \frac{10 \times 0.693}{3} \approx 2.31 \approx ln 10

Therefore,

\frac{1}{w} = 10

, which gives

w = 0.1 μ g = 1 \times 1 0^{- 1} μ g

.

The answer is 1.

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Integer answer

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