Question

The half-life of radio active ${}_{}{}^{14}\mathrm{C}$is $5760$ years. In how many years $200\mathrm{mg}$ sample will reduce to $25\mathrm{mg}$?

• A. 23,040 years
• B. 17,280 years
• C. 11,520 years
• D. 5, 760 years

Answer 1

The correct option is C

11,520 years

The half life of radioactive element ${}_{}{}^{14}\mathrm{C}$ is $5760$ years.

Mathematically: Time taken (t) by the sample to reduce from $200\mathrm{mg}$ to $25\mathrm{mg}$ is calculated by-

Formula used: $\mathrm{N}\left(\mathrm{t}\right)={\mathrm{N}}_0{\left(\frac{1}{2}\right)}^{{}^{\frac{\mathrm{t}}{{\mathrm{t}}^{1/2}}}}$

Where,

$\mathrm{N}\left(\mathrm{t}\right)=$Remaining quantity of substance after time t.

${\mathrm{N}}_0$$=$Initial quantity of substance.

${\mathrm{t}}^{\frac{1}{2}}$$=$Half-life period

Substitute the given values in formula used-

$\begin{array}{rcl}\mathrm{N}\left(\mathrm{t}\right)& =& {\mathrm{N}}_0{\left(\frac{1}{2}\right)}^{\frac{t}{t^{1/2}}}\\ \frac{25}{200}& =& {\left(\frac{1}{2}\right)}^{\frac{\mathrm{t}}{5760}}\\ \log \frac{1}{4}& =& \frac{\mathrm{t}}{5760}\log \frac{1}{2}\\ \log 0.25& =& \frac{\mathrm{t}}{5760}\times \log 0.30\\ 0.30\times \mathrm{t}& =& 5760\times 0.60\\ \mathrm{t}& =& \frac{5760\times 0.60}{0.30}\\ \mathrm{t}& =& 11520\mathrm{years}\\ & & \end{array}$

Final answer:

Hence, the option (C) is correct.