Question

Open in App

Updated on : 2022-09-05

Solution

Verified by Toppr

Correct option is A)

$N_{0}$ = $λ−dN/dt $ = $2.95×10_{−8}7.4×10_{4} $ = $2.51×10_{12}$ nuclei

$N_{t}$ $=N_{0}e_{−λt}$ $=2.51×10_{12}e_{−2.95×10_{−8}×3.156×10_{7}}$

$=0.394(2.52×10_{12})$

Activity $=λN(t)$ = $.394(2.5×10_{12})×2.95×10_{−8}$

$=0.788$ $μ$ $Ci$

$=0.788$ $μ$ $Ci$

Alternative method:

$dtdN(t) $ = $dtdN(0) e_{λt}$$=(2μCi)$ $(e_{−2.95×10−8×3.156×10_{−7}})$$=2(.394)$ = $0.788$ $μ$ $Ci$

$dtdN(t) $ = $dtdN(0) e_{λt}$$=(2μCi)$ $(e_{−2.95×10−8×3.156×10_{−7}})$$=2(.394)$ = $0.788$ $μ$ $Ci$

Was this answer helpful?

0

0