Saved Bookmarks
| 1. |
A radio active substance has a half life period of 30 days. Calculate (i) time taken for `3//4` of original number of atoms to disintegrate, (ii) time taken for `1//8` of the original number of atoms to remain unchanged. |
|
Answer» Here, T=30days,t=? No. of atoms disitingrated =`(3//4)N_0` No. of atoms left after time t, `N=N_0-3/4N_0=1/4N_0` No. half lives in time t days, `n=t/T=t/30` No. of nuclei left after n half lives is given by `N=N_0(1/2)^n` `:. (N_0)/4=N_0(1/2)^(t//30) or (2)^(t//30)=4=2^2` or`t/30=2 or t=60 days` (ii) `t=? ,N=N_0//8` As `N=N_0(1/2)^n :. (N_0)/8=N_0(1/2)^(t//30)` or `(2)^(t//30)=8=(2)^3 or t/30 =3` or `t=90 days ` |
|