Afreshly prepared sample of curium-243 undergoes 3312 disintegrations per second. after 6.00 yr, the activity of the sample declines to 2755 disintegrations per second. the of is

I took the liberty of finding for the complete question. And here I believe that the problem asks for the half life of Curium. Assuming that the radioactive decay of Curium is of 1st order, therefore the rate equation is in the form of:

A = Ao e^(-kt)

where,

A = amount after t years = 2755

Ao = initial amount = 3312

k = rate constant

t = number of years passed = 6

Therefore the rate constant is:

2755/3312 = e^(-6k)

-6k = ln (2755/3312)

k = 0.0307/yr

The half life, t’, can be calculated using the formula:

t’ = ln 2 / k

Substituting the value of k:

t’ = ln 2 / 0.0307

t’ = 22.586 years

or

t’ = 22.6 years

just take the area of the square and then the area of the circle then subtract the circle area from the squares area

step-by-step explanation: