Radioactivity of a radioactive element remains` 1//10` of the original radioactivity after 2.303 seconds. The half life period of the element is:

Options

A. `2.303`
B. `0.2303`
C. `693`
D. `0.693`

Answer

D

Explanation

`K=(2.303)/(t)log``(a)/(a-x)` `[{:(t=2.303(given)),((a-x)=(1)/(10)a_(0)):}]`
`=(2.303)/(t)log``(1)/(a//10)=1`
`:. t_(1//2)=(0.693)/(K)=(0.693)/(1)=0.693`