Nuclide a Has a Half-Life of 4 Days, and Nuclide$2 ^ { 18 }$

Question

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The Answer is The radioactive decay law gives

N _ { A } ( t ) = N _ { A } ( 0 ) \exp \left( - \lambda _ { A } t ight)

where

\lambda _ { A } = \frac { \ln ( 2 ) } { t _ { A 1 / 2 } }

and

t _ { A 1 / 2 } = 4 \mathrm {~d} , N _ { A } ( 0 ) = 2 ^ { 18 }

. For the other nuclear species

N _ { B } ( t ) = N _ { B } ( 0 ) \exp \left( - \lambda _ { B } t ight)

and,

t _ { B 1 / 2 } = 16 \mathrm {~d} , N _ { B } ( 0 ) = 2 ^ { 12 }

. We want

N _ { A } ( t ) = N _ { B } ( t )

so that

N _ { A } ( 0 ) \exp \left( - \lambda _ { A } t ight) = N _ { B } ( 0 ) \exp \left( - \lambda _ { B } t ight)

56 and we want to solve for

t

. Then

\frac { N _ { A } ( 0 ) } { N _ { B } ( 0 ) } = \exp \left( \left( \lambda _ { A } - \lambda _ { B } ight) t ight)

or

\ln \left( \frac { N _ { A } ( 0 ) } { N _ { B } ( 0 ) } ight) = \left( \lambda _ { A } - \lambda _ { B } ight) t ightarrow t = \left[ \frac { 1 } { \lambda _ { A } - \lambda _ { B } } ight] \ln \left( \frac { N _ { A } ( 0 ) } { N _ { B } ( 0 ) } ight)

Substituting in values from above gives

t = \left[ \frac { 1 } { \frac { \ln 2 } { 4 } - \frac { \ln 2 } { 16 } } ight] \ln \left( \frac { 2 ^ { 18 } } { 2 ^ { 12 } } ight) = \left[ \frac { 1 } { 0.1733 \mathrm {~d} ^ { - 1 } - 0.0433 \mathrm {~d} ^ { - 1 } } ight] imes 4.159 = 32 \mathrm {~d}

Nuclide a Has a Half-Life of 4 Days, and Nuclide $2 ^ { 18 }$