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Graph shows the variation of the number of radioactive atoms left undecayed with time. Find the time corresponding to N=N_0//3? |
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Answer» SOLUTION :We KNOW that `N=N_0e^(-lambdat)` `(2N_0)/3 = N_0 e^(-lambdat_0) rArr e^(lambdat_0) =3/2 rArr lambdat_0=ln(3//2)`…(1) ALSO, `N_0/3 =N_0 e^(-lambdat_1) rArr lambdat_1`=ln 3 `rArr t_1=1/lambda ln(3) = (t_0 ln(3))/(ln (3//2))` [From (i)] |
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