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JEE Main levelSingle correct optionDifficulty 2 of 3as172

The half-life of a radon isotope is $3.8$ days. Using $ln 2 = 0.693$ and $1 day = 86400 s$ , its decay constant in $s^{- 1}$ is closest to

A $2.1 \times 1 0^{- 6}$
B $3.1 \times 1 0^{- 6}$
C $5.1 \times 1 0^{- 5}$
D $2.3 \times 1 0^{5}$

Answer: A

λ = 0.693/ t_{1/2}

with

t_{1/2} = 3.8 \times 86400 = 3.28 \times 1 0^{5} s

, so

λ = 2.11 \times 1 0^{- 6} s^{- 1}

. Omitting 0.693 gives

3.05 \times 1 0^{- 6}

, and using hours instead of seconds gives

5.07 \times 1 0^{- 5}

.

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