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

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

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)