In an experiment on the transport of nutrients in the root...

Name: In an experiment on the transport of nutrients in the root structure of a plant, two radioactive nuclides X and Y are used. Initially 2.40 times more nuclei of type X are present than of type Y. Just three days later there are 4.20 times more nuclei of type X than of type Y. Isotope Y has a half-life of 1.60 d. What is the half-life of isotope X?
Uploaded: 2022-03-18T03:34:00-08:00
Duration: 8 min 37 s
Channel: Eduard Sanchez
Description: In an experiment on the transport of nutrients in the root structure of a plant, two radioactive nuclides X and Y are used. Initially 2.40 times more nuclei of type X are present than of type Y. Just three days later there are 4.20 times more nuclei of type X than of type Y. Isotope Y has a half-life of 1.60 d. What is the half-life of isotope X?

In an experiment on the transport of nutrients in the root structure of a plant, two radioactive nuclides X and Y are used. Initially 2.40 times more nuclei of type X are present than of type Y. Just three days later there are 4.20 times more nuclei of type X than of type Y. Isotope Y has a half-life of 1.60 d. What is the half-life of isotope X?

Transcript

00:01 Hi there, so for this problem we are told that in an experiment on the transport of nutrients in the root structure of a plan, to do radioactive nucleides x and y are being used.

00:16 Now, initially there are 2 .40 times more nucleate of type x and than of types y.

00:25 So we can write that initially the number of x is equal to 2 .4 times initial value of the element y.

00:42 Now remember that this is when the time is zero days.

00:47 Now, just three days later, there are 4 .20 times more nuclei of type x.

00:55 So we can write that this is the following equals to 4 .2 times the nuclei of y, but this corresponds when the time is equal to three days after.

01:11 Now, we are also given the information about the isotope white of its half -life.

01:19 So the half -life for the y element it is given, and then, that is 1 .6 days.

01:30 And what we need to determine is the half -life of the isotope x.

01:39 Oh, sorry, the half -life of this.

01:43 Now, to calculate this, we need first to write the equations for the decay for both components.

01:56 So we know that the general form for the x component is that the number or the amount of the x component is equal to the initial amount times the exponential of minus where what we are going to call lambda x to identify this with the element x times the time.

02:18 Now we also going to have something similar for the element y.

02:27 And now what we can do is to take the difference between these two, sorry, to take the quotient between these two expressions.

02:43 That is the first one divided by the second one and then use the condition for the time equals to three days.

02:56 Because if we do that, we will have nx divided by n y.

03:02 And this after three days, we know that that is going to be.

03:07 You can solve that from this, so that will give us 4 .2.

03:13 And then this is equal to n sub zero, divided by n y sub zero.

03:23 And this times the s penitial.

03:24 This is the difference between minus the times.

03:28 Lambda x minus lambda y.

03:32 Now remember that in this case we are going to use the condition for the time equals to after three days.

03:41 So in here we can also subsolve from this.