Question

Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours. Determine the half-life of a nuclide that loses 38.0% of its mass in 407 hours. 204 hours 568 hour 590 hours 291 hours 281 hours

Answer 1

Answer:

Use 62% - the equation is for the amount present at a given time. 0.62 = (1) e-kt -> ln(0.62)=-kt -> k = -ln(0.62)/t. I get k = .00117 hr-1 t(half) = 0.693/k = 590 hr.

HOPE THIS HELPS AND PLSSS MARK AS BRAINLIEST AND THNXX :)

Answer 2

The half-life is the time at which the substance's concentration is reduced by half of its initial amount. The half-life of a nuclide that lost its 38.0% mass is 590 hr. Thus, option C is correct.

What is half-life?

Half-life is the time required by a substance to get reduced to half of its initial concentration. The half-life of the substance can be determined by the rate constant.

Given,

The initial quantity of substance (A₀) = 100

Remaining quantity (At) = 10 - 38 = 62

Time elapse (t) = 407 hours

The rate constant (k) is calculated as:

ln (At ÷ A₀) = - kt

ln (62 ÷ 100) ÷ 407 hour = - k

-0.47803580094 ÷ 407 = - k

k = 0.00117453513

Now, half-life from rate constant (k) is calculated as:

$\rm t ^{\frac{1}{2}}$ = 0.693 ÷ k

$\rm t ^{\frac{1}{2}}$ = 0.693 ÷ 0.00117453513

$\rm t ^{\frac{1}{2}}$ = 590 hours

Therefore, option C. 590 hours is the half-life of the substance.

Learn more about half-life here:

https://brainly.com/question/16387602

#SPJ2