William asked in Science & MathematicsChemistry · 1 month ago

# The starting mass of a radioactive isotope is 20.0g The half life period of this iso- is 2 days. The sample is observed for 14 days?

What percentage of the original amount remains after 14 days?

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• Dr W
Lv 7
1 month ago

memorize this equation

.. A(t) = A(o) * (1/2)^(t / half life)

where

.. A(t) = amount remaining after time = t has elapsed

.. A(o) = starting amount

.. .. t = elapsed time

.. half life =... you guessed it, half life

**********

.. (A(t) / A(o)) * 100%

so let's start by rearranging that equation

.. (A(t) / A(o)) = (1/2)^(t / half life)

.. (A(t) / A(o)) * 100% = (1/2)^(t / half life) * 100%

solving

.. % remaining = (1/2)^(14day / 2 day) = 0.781%

• Anonymous
1 month ago

A(t) = A(o) * (1/2)^(t / half life)

A(14) = 20.0g * 1/2^(7)

A(14) = 20.0 * 0.0078125 = 0.15625g

Round to 3 sigfigs: 0.156 g

• 1 month ago

At

2day = 10 g

4 day = 5g

6 day = 2.5 g

8day = 1.25 g

10 day = 0.625 g

12 day = 0.3125 g

14 day = 0.15625 g

percentage remaining

0.15625 x 100 / 10 = 1.5625 %

• Dr W
Lv 7
1 month agoReport

no.  at day 14 you have this % remaining (0.15625g / 20.0) * 100%

• Anonymous
1 month ago

Nt/No = exp(-kt)

k = ln2/t0.5 = 0.3466 per day

Nt/No = exp(-0.3466 x 14) = exp(-5.8524) = 0.0078

• Bo1 month agoReport

0.0078125 is the FRACTION remaining.  The PERCENT remaining is 0.781%