Question

$80 \, kg$ 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. $4.64\times 10^{- 3} \, s^{- 1}$

Answer 1

Answer: (A)

Fraction of material that remains undecayed
$\frac{10}{80}=\left(\frac{1}{2}\right)^{\frac{1 h}{T_{1 / 2}}}$
$3=\frac{1 h}{T_{1 / 2}}$
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}$