1.

You have 0.1gram-atom of a radioactive isotope ""_(Z)^(A)X (half-life =5 days). How many number of atoms will decay during the eleventh day?

Answer»

Solution :Let the amount of X after 10 days be x moles
We have, `lamda= (2.303)/(t) "log" (N^(0))/(N)= (0.6932)/(t_((1)/(2)))`
or `(2.303)/(10)"log" (0.1)/(x)= (0.6932)/(5)`
or x= 0.025 moles
`therefore` number of atoms `=0.025 xx 6.022 xx 10^(23)`
`=1.505 xx 10^(22)`
THUS the number of atoms on the ELEVENTH day is `1.505 xx 10^(22)`
Now, no. of atoms DECAYING on the eleventh day = no. of DISINTEGRATION per day
`= -(d(N))/(dt)= lamda (N)= (0.6932)/(t_((1)/(2))) xx N`
or `-(d(N))/(dt)= (0.6932)/(5) xx 1.505 xx 10^(22)= 2.086 xx 10^(21)`


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