Solve the problem.
-The half-life of Cesium $134 \mathrm {~m}$ is $3.0$ hours. If the formula $\mathrm { P } ( \mathrm { t } ) = \left( \frac { 1 } { 2 } \right) ^ { \mathrm { t } / 3.0 }$ gives the percent (as a decimal) remaining after time $\mathrm { t }$ (in hours), sketch P versus $\mathrm { t }$.

Question

Quizplus · Accepted Answer

We know that half-life of Cesium ($134 \mathrm {~m}$) is $3.0$ hours. This means that after every $3$ hours, the amount of Cesium ($134 \mathrm {~m}$) reduces to half of its original amount.

The given formula $\mathrm { P } ( \mathrm { t } ) = \left( \frac { 1 } { 2 } \right) ^ { \mathrm { t } / 3.0 }$ represents the percent (as a decimal) remaining after time $\mathrm { t }$ (in hours).

For example, after $3$ hours, we plug in $\mathrm { t } = 3$ in the formula to get:

$\mathrm { P } ( 3 ) = \left( \frac { 1 } { 2 } \right) ^ { 3 / 3.0 } = \left( \frac { 1 } { 2 } \right) ^ { 1 } = \frac { 1 } { 2 } = 0.5$

This means that after $3$ hours, only $50\%$ of the original Cesium ($134 \mathrm {~m}$) remains.

Similarly, after $6$ hours, we plug in $\mathrm { t } = 6$ in the formula to get:

$\mathrm { P } ( 6 ) = \left( \frac { 1 } { 2 } \right) ^ { 6 / 3.0 } = \left( \frac { 1 } { 2 } \right) ^ { 2 } = \frac { 1 } { 4 } = 0.25$

This means that after $6$ hours, only $25\%$ of the original Cesium ($134 \mathrm {~m}$) remains.

Thus, the graph of P versus $t$ should start at $1$ and decrease exponentially to $0$ as $t$ increases in multiples of $3$ (since that is the half-life).

The only graph that matches this description is option B.

Solve the Problem 134 m134 \mathrm {~m}134 m Is 3.03.03.0 Hours If the Formula P(t)=(12)t/3.0\mathrm { P } ( \mathrm { t } ) = \left( \frac { 1 } { 2 } \right) ^ { \mathrm { t } / 3.0 }P(t)=(21​)t/3.0

Solve the Problem $134 \mathrm {~m}$ Is $3.0$ Hours If the Formula $\mathrm { P } ( \mathrm { t } ) = \left( \frac { 1 } { 2 } \right) ^ { \mathrm { t } / 3.0 }$