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The nuclide %Nb decays by first-order process with rate constant of 2.96 x [0-2 h-'. How long will it take for 86.0% of the initial amount of #Nb be consumed?...

Question

The nuclide %Nb decays by first-order process with rate constant of 2.96 x [0-2 h-'. How long will it take for 86.0% of the initial amount of #Nb be consumed?33.7 h30 h66.4 h0 5.09 h4.72 h

The nuclide %Nb decays by first-order process with rate constant of 2.96 x [0-2 h-'. How long will it take for 86.0% of the initial amount of #Nb be consumed? 33.7 h 30 h 66.4 h 0 5.09 h 4.72 h



Answers

A certain radioactive nuclide has a half-life of 3.00 hours.
a. Calculate the rate constant in $\mathrm{s}^{-1}$ for this nuclide.
b. Calculate the decay rate in decays/s for 1.000 mole of this nuclide.

Mhm. Okay. So with this problem Um were given some substance, we don't know what it is but we're told that it has a half life of 80.9 years And that there is 87.5% of the substance left. And we're trying to figure out how old this um this sample is. So we're going to use this half life equation. We've got um X. I. Or X. Initial is going to be the amount of the thing that you started with before any decay happened. And then X. For X final is how much those left right now and then t. Is going to be the number of half lives. So We're told that there is 87.5% left. So in other words are sorry I wrote this wrong, There is 87.5 that has decayed this. In other words that means that there is only 12.5% left. So we can write this equation. We don't know what the initial amount is but we know there's some initial amount and since 87.5% has decayed, there's 12.5% left and percent as we know, it can be written as decimals. So Since there's 12.5% left we can rewrite XF as .125 X. initial. Then from there with our algebra skills we can cancel the X. Is the X. I. On each side. And we're left with this one half to the power of t equals .125. So now we're gonna pull out our calculators and plug in log Base one half of 0.125 wow. And that's gonna give us tea. Think about it. Yes. When you plug that into your calculators you get that T equals three. Now T. Is the number of half life set have occurred. So we're going to use how long half life is that were given 80.9 years. And we're going to multiply that by three. So we're going to do 80.9 years Times 3/2 lives. And that gives us 2 42 0.7 years. So that right there is your answer. That's how old the sample is, 242.7.

Question 15 deals with the decay of gold. 1 98 were given its half life initial activity moving to find how many nuclei decay in this time frame from 10 hours to 12 hours. So between these two times, how many gold? One. Any Adams already. Essentially. Right. So? So they need to do as we want to determine moment and use relationship as is so that really activity is equivalent to the number of particles started to take on sometimes the number of atoms where our end goal is assault for the number of atoms at each time frame to be the activity divided by Medicaid constant. That means we need to find our at both of these times. So this activity at time one is given by the equation. That's a bill decay. Uh, equation t one, Go. Go ahead and find this initial. Are so the activity after 10 hours? I believe it In terms of Micro Curie. For now, that's fine. So you initially did you 40 my security I would replace my decay constant with my half life So natural love of two times t one richest 10 hours divided by the half life of 64.8 hours. We can find this activity at this time because it's not my final answer. I'm to leave a little more precision with 35.94 Micro Curie of the same calculation for the 12 hours. So I'm not gonna read Write it out. But the same equation is changing this value from 10 to 12. With that, we find the activity is 35 point 18 Micro Curie Great. So now we confined the number Adams present at each time using this these values. So the number Adams at this first time and T one is simply just the activity of t one, a TV active in zari over the Tiki Constant. So we're at the living of conversion here because notice, because they're the key constants given, obviously in some sort of time frame. No need to convert this into backgrounds. So if 35.94 times 10 to the sixth Curie and then converting from half life too, so decay constant 1/2 life, 64.8 hours. So if I just can't pay this, I would have units of Curie hours, which is not useful Of course, the first thing Jews convert my activity into back. Carell's says 3.7 times 10 to the 10 becquerel per curie. I'll take my units after, Remember becquerel that decay per second. So I need to get rid of my unit of ours. So the in one hour there are 3600 seconds, So just check my units here. My curious cats is out here. My hours cancel out that girl's K per second. So becquerel time seconds just indicates the number of the keys or number of atoms, so maintenance aren't good now. So captivating this out, that seasoning I would find. But the number of atoms present at this time would be four point for 75 I have sent to the 11 Adams. Great. I'll perform the same calculation, I would show it, but that same stuff again, this change in my value for my activity. And I find that the murder Adams president at this time it will be 4.381 10%. The love Adams, of course. And so that changed in end, or the number of particles that decayed in that time frame and t one minus end of t two. The tracking these values, you would find that the number of atoms that have decayed to three significant figures are nine 0.46 times 10 to the nine had of okay, Yeah, So recognizing we need to find activity at these two different time points and relating the activity to number of atoms leads us to our solution.

Uh, using the concept off half life period off the radioactive substances in terms of decay constant and knowing the value value off initial radio activity off 1 98 you isotope weaken didn't mind the number of nuclear that they decay and they're given developed time. So the number of nuclei and one and 22 and one that the case in the interval between the time, um, to a 92 off the radioactive sample 1 98. You is computed by using the formula, and to two. And one is a girl too. Are not behalf A bond 0.6 93 b to the power minus a 0.693 up on behalf they to minus a to the power minus 0.693 upon behalf Day one. Um, here have life period off 19. So 1 98 au is and the initial radio activity off when I did a u isotope is are not so ah, half life off the radioactive isotope, Um, 1 19 a U. S. 64.8 hours. Um, there is equal. That is 2.33280 in tow. Tend to the power five seconds and the initial radio activity off. 1 98. You is 40 Micros EI. That is equal to 1.48 in 10 to the power six. The case four second and the number of nuclei that became that I'm intervals between the times. Um, t one is 10 hour and two is 12 hours. So, um um but the formula for high fly spirit off the radioactive substances in terms off decay constant. Lambda is given by t half is 0.693 upon Lambda. So, lambda is there a 0.693 upon p half. So, um, we have, um 693 upon t half is 2.3 3 to 18 to 10 to the power. Five seconds. So lambda is there a point to 97 into 10 to deeper, minus five per second. And we know the form. Love for the number of nuclear that decay. In the interval between the times t one and T to work the radio active sample I won 98 au is and to Penis and one is are not behalf Upon 0.693 multiplied by eight to the power minus 0.693 or four. Behalf multiplied by 81 32 minus a to the power 0.693 upon t half multiplied by 81 on for N one minus and two is are not upon multiplied by a half upon 3.693 a to the power 0.693 upon t half Do you want minus e to the power minus 0.693 20 half ti toa eso From this we have, uh, by putting the values off are not anti half. We will get the value as, um nine 0.47 and to tend to the power ninth nuclear. So the number of nuclei therefore the number off nuclear that decay in the time period in the given interval off time defense upon half life period, then initial radio activity Radio at PVT off the the directive substances and time in true

To find he number off. You cry. Ah, beauty Fullback on using the activity since the activities related to the normal UK fire d decay constant. So over here we want to find What is he towards animals. In fact, that Kate, between some time interval, right, we didn't $10.12 hours. So all you need is t activity at $10 activity at Joe Flowers. And from there you can find what is the change de and failure. So right now, to fight audience activities, we will first need to use the decay law. Right? Reduced the activity to the initial activity. In fact, he not which is given this 40 micro curies. Let the backs would ensure negative. The t so lambda, in this case you take a constant is given a slot to off t half 64.8. So substituting that in Mississippi Pretty Michael. Curious what? Bye bye, Exponent. Show teeth on to 10/64. Point it. I saw the r U C unit step off in hours so they can sew it. Shut out! You got to do the same process for the Jeff. Always one. Okay, this is just 40. Michael. Curious Ensured. Already flown to show off 64 point it. Remember that this also equals two Lunda times. The number off do you cry? Shop hours this week goes to the I don't see number on your case. 10 of us. So in order to find the difference, right? The difference in the nobody, right? They will give us the noble. You got that, Kate, right? You just have to find we take each off o e 10 minus itself. You gotta buy Linda. Circular City number off the case that I could between these time, period. All right. And with this, in order to get the actual inability case, you have a convert. I'll listen to seconds. Michael Curie's in two back Gross. So 40 Michael Curie's You can call my debt by multiplying by 10 po off. Wait. Us six took over for Michael. Curious too Curious. In a multiplied by 3.7 times Sent Po of 10. It's number. Bekoe spoke Yuri for you. Always me much by that by treat About 600 seconds. Oh, if every Tatis expression final answer would be 9.47 posted above night


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