Question

The amount of polonium-210 remaining, p(t), after t days in a sample can be modeled by the exponential function p(t) = 100e−0.005t, where 100 represents the initial number of grams in the sample. what is an equivalent expression, written as a percentage rate of polonium-210 lost, and how much polonium-210 remains (rounded to the nearest whole number) after 63 days?

Answer 1

% Po lost = 100[1 - e^(-0.005t)] %; 73.0 g

Explanation:

p(t) = 100e^(-0.005t)

Initial amount: p(0) = 100

Amount remaining: p(t) = 100e^(-0.005t)

Amount lost: p(0) – p(t) = 100 - 100e^(-0.005t) = 100[1 - e^(-0.005t)]

% of Po lost = amount lost/initial amount × 100 %

= [1 - e^(-0.005t)] × 100 % = 100[1 - e^(-0.005t)] %

p(63) = 100e^(-0.005 × 63) = 100e^(-0.315) = 100 × 0.730 = 73 g

The mass of polonium remaining after 63 days is 73 g.