Question

A meteorite contains 80 grams of radioactive chemical 'Q' when it lands on the Graeters in
Worthington. By the time NASA gets to the site (18 hours later), the sample is only 10 grams. What is the half-life of Q

Answer 1

Answer:

$\huge\boxed{\sf 6 \ hours}$

Explanation:

Half life is when half of the total material gets decayed.

During first half life:

Radioactive chemical Q = 80 g / 2 = 40 g

During second half life:

Radioactive chemical Q = 40 g / 2 = 20 g

During third half life:

Radioactive chemical Q = 20 g / 2 = 10 g

$\rule[225]{225}{2}$

Hence, we get to know that three half lives had been passed when NASA gets to the site.

If:

3 half lives = 18 hours (as indicated)

Divide both sides by 3

1 half live = 18 / 3 hours

1 half life = 6 hours

$\rule[225]{225}{2}$

Hope this helped!

~AH1807

Answer 2

$\\ \tt\Rrightarrow A=A_o(0.5)^{t/t/2}$

• A is amount remains
• Ao is initial amount
• t is time taken
• t/2 is half life(Let it be h)

$\red{\rule{50pt}{2pt}}\blue{\rule{50pt}{2pt}}\green{\rule{50pt}{2pt}}\pink{\rule{50pt}{2pt}}$

$\\ \tt\Rrightarrow 10=80(0.5)^{18/h}$

$\\ \tt\Rrightarrow 0.125=(0.5)^{18/h}$

$\\ \tt\Rrightarrow log0.125=log_{10}(0.5)^{18/h}$

$\\ \tt\Rrightarrow -0.9=18/h(log5-1)$

$\\ \tt\Rrightarrow -0.9=18/h(-0.3)$

$\\ \tt\Rrightarrow 18/h=3$

$\\ \tt\Rrightarrow h=18/3=6hours$