Question

$G{d}^{153}$, which is used to detect osteoporosis (porous bones), has a half life of 242 days. Which value is closest to the percentage of $G{d}^{153}$ left in a patient's system after 2 years

• A. 33%
• B. 26%
• C. 12.5%
• D. 6.25%

Answer 1

The correct option is C 12.5%

$t_{\frac{1}{2}}$ or $t_{50\%}$

for $1st$ order

$t_{75\%}=2\times t_{\frac{1}{2}}$ or $\frac{t_{75\%}}{t_{\frac{1}{2}}}=2$

$t_{87.5\%}=3\times t_{\frac{1}{2}}$

$t_{93.75\%}=4\times t_{\frac{1}{2}}$

$t_{\frac{1}{2}}=242days$

for $2$ years

$\frac{t_{x\%}}{t_{\frac{1}{2}}}=\frac{730days}{242days}=\mathrm{3.01......}$

$=3approx$

$\therefore t_{x\%}=3\times t_{\frac{1}{2}}$

$\therefore x\%=87.5\%$

therefore percentage left in patient's system is

$=100-87.5\%$

$=12.5\%$