Q. 80 kg of a radioactive material reduces to 10 kg in 1 h. The decay constant of the material is
A) $5.80\ \times \ 10^{- 4}\ s^{- 1}$
B) $1.16\ \times \ 10^{- 3}\ s^{- 1}$
C) $2.32\ \times 10^{- 3}\ s^{- 1}$
D) $4.64 \times 10^{- 3}\ s^{- 1}$
Explanation
(a) Fraction of material that remains undecayed
$\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{10}{80} = \left( \frac{1}{2} \right)^{\frac{1h}{T_{1/2}}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3 = \frac{1h}{T_{1/2}}$
or $T_{1/2}\ \ \ = \ \frac{60}{3}\ min\ = \ 20\ min\ = 1200s$
Now, $\ \ \ \ \ \lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1200} = \ 5.8\ \times \ 10^{- 4}\ s^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.