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

the answer is b, have negligible mass

which three statements are true about all alloys?