In an experiment on the transport of nutrients in a plant's root structure, two radioactive nuclides X and Y are used. Initially, 2.70 times more nuclei of type X are present than of type Y. At a time 2.50 d later, there are 4.60 times more nuclei of type X than of type Y. Isotope Y has a half-life of 1.50 d. What is the half-life of isotope X?

Explanation:

From radioactive equation, it is know that,

N = No•exp(-λt)

Where λ is the decay constant

N is is the material remaining after time t

No is the original material at the beginning I.e. at t = 0

Half-life is given as

T½ = In(2) / λ

Now

Given that,

The initial number of nuclides of type X to that of type Y is

No(X) / No(Y) = 2.7

The final number of nuclides of type X to that of type Y is

N(X) / N(Y) = 4.7

The time elapsed between this period is

t = 2.5d

the half life of isotope Y is

T½(Y) = 1.5d

What is the half life of isotope X

From first equation at t = 2.5d

For type X

N = No•exp(-λ•t)

N(X) = No(X) • exp(-2.5•λx)

For Type Y

N = No•exp(-λ•t)

N(Y) = No(Y) • exp(-2.5•λy)

Divide N(X) by N(Y)

Then, we have

N(X) / N(Y) = [No(X) / No(Y)]•exp(-2.5λx) / exp(-2.5λy)

Let substitute the given ratio

4.7 = 2.7 exp(-2.5λx+2.5λy)

4.7 / 2.7 = exp(-2.5λx+2.5λy)

1.741 = exp(-2.5λx+2.5λy)

Take In of both sides

In(1.741) = -2.5λx+2.5λy

0.5543 = -2.5(λx-λy)

(λx-λy) = 0.5543/-2.5

(λx-λy) = -0.2217 equation 1

From the second equation

T½ = In2 / λ

For Type Y

T½(Y) = In2 / λy

λy = In2 / T½(Y)

λy = In2 / 1.5d

λy = 0.4621 /d

From equation 1

λx-λy = -0.2217

λx - 0.4621 = -0.2217

λx = -0.2217 + 0.4621

λx = 02204 /d

Finally,

T½(X) = In2 / λx

T½(X) = In2 / 0.2204 /d

T½(X) = 2.8833 d

So the half life of Element X is 2.8833d

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explanation:

i had this question and the answer was 3.15 miles.

martina travels from school to her mom's office. from there, she walks to school. after school, she walks to the library and then to a friends house before finally ending back at home. find martina's total distance traveled.

a) 0 miles  

b) 0.4 miles  

c) 1 mile  

d) 3.15 miles