Question

In four years, 40% of a radioactive element decays. Find its half-life. Round to one decimal place.

Answer 1

Answer:

T(Half Life) = 5.42 s

Step-by-step explanation:

Given :

t = 4years

[R]₀(Initial Concentration) = 100(say)

[R](Concentration Left after time t=4yrs) = 100 - 40 = 60

We know that,

k = 2.303 / t (log{[R]₀/[R]})

where k = rate constant

k = 2.303 / t (log[100/60])

k = 2.303 / 4 (log[10] - log[6])

k = 2.303 / 4 (1 - 0.7781)

k = 2.303 / 4 (0.2219)

k = 0.5110 / 4

k = 0.1277          --------(i)

Now, to find its half life,

We know that,

T(Half Life) = 0.693 / k

T = 0.693 / 0.1277

T = 5.42 years

Answer 2

To find the answer, use $k = \frac{2.303 }{ 4} (log\frac {100 }{ 40})$ , and then use the equation $t=\frac{.693}{k}$ To find it's half life.