Question

The half-life period of a radioactive element is 140 days. After 650 days, one gram of the element will reduce to

• A. $\frac{1}{2}g$
• B. $\frac{1}{4}g$
• C. $\frac{1}{8}g$
• D. $\frac{1}{16}g$

Answer 1

The Correct Option is D

560 daya $=\frac{560}{140}$ =4 half-lives
Amount of reactant remaining after /z-half-lives
$=\bigg(\frac{1}{2}\bigg)^n\times $ initial amount $=\bigg(\frac{1}{2}\bigg)^4\times1.0g=\frac{1}{16}g$

Answer 2

Let's correct and restate the calculation using the correct elapsed time of 560 days:

Given:
- Half-life period (\( t_{1/2} \)) = 140 days
- Elapsed time (\( t \)) = 560 days
- Initial quantity (\( N_0 \)) = 1 gram

First, determine the number of half-lives that have passed:

\[ \text{Number of half-lives} = \frac{560}{140} = 4 \]

Next, use the formula for the remaining amount after \( n \) half-lives:

\[ N(t) = N_0 \left( \frac{1}{2} \right)^n \]

Where \( n \) is the number of half-lives. Substituting the values:

\[ N(560) = 1 \left( \frac{1}{2} \right)^4 \]

Calculate \( \left( \frac{1}{2} \right)^4 \):

\[ \left( \frac{1}{2} \right)^4 = \frac{1}{16} \]

Thus, the remaining quantity of the element after 560 days is:

\[ N(560) = \frac{1}{16} \, \text{grams} \]

Therefore, the amount of the element remaining after 560 days is:

\[ \boxed{\frac{1}{16} \, \text{grams}} \]
So Correct Answer is Option D :\(\frac{1}{16}gm\)