The $t_{\frac{1}{2}}$ of a radioactive element is 20 min. The time interval between the stages of it's $34 %$ and $67 %$ decay is

40 min

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20 min

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

30 min

No worries! We‘ve got your back. Try BYJU‘S free classes today!

25 min

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 20 min
Radioactive decay follows first order kinetics.
So,
$k = \frac{l n 2}{20} m i n^{- 1}$
$∴ \frac{l n 2}{20} Δ t = l n \frac{66}{33}$
$\Rightarrow Δ t = 20$ min

Suggest Corrections

Similar questions

t_{\frac{1}{2}}

34 %

67 %

The half - life of a radioactive element is 30 min. The time interval between the stages of 33.3% and 67.7% decay will be ?

7.2

33.33

66.66

(t_{2} - t_{1})

t_{2}

\frac{2}{3}

t_{1}

\frac{1}{3}

Join BYJU'S Learning Program