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Question 13: Answer the following questions regarding nuelear chemistry Write nuclear equation when germanium-75 undergoes beta decay Determine the nuclear binding ...

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Question 13: Answer the following questions regarding nuelear chemistry Write nuclear equation when germanium-75 undergoes beta decay Determine the nuclear binding energy of the daughter nucleus formed in Question A. The daughter nucleus from that reaction has mass of 74.9625 amu Convert the following nuclear reaction to nuclear transmutation notation '93Rh Eca 6n '3Tb Calculate AE of the process in Question € given the following information below. Assume mole of rhodium-[04 and cale

Question 13: Answer the following questions regarding nuelear chemistry Write nuclear equation when germanium-75 undergoes beta decay Determine the nuclear binding energy of the daughter nucleus formed in Question A. The daughter nucleus from that reaction has mass of 74.9625 amu Convert the following nuclear reaction to nuclear transmutation notation '93Rh Eca 6n '3Tb Calculate AE of the process in Question € given the following information below. Assume mole of rhodium-[04 and caleium-40 Mass mol IRh 103.9067 WCa 39,9626 00s66 14"16 138.948} Throium-234 undergoes alpha decay. followed by beta decuy; and two more alphu decays Write nuelear equations for the entire decay process



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Questions 1–3 (A) Alpha decay (B) Beta decay (C) Gamma ray (D) Fission (E) Fusion Select the term from above that identifies the nuclear reactions described in questions 1 to 3. An isotope of bismuth, $^{214}_{83}$ $\mathrm{Bi}$ , undergoes a transmutation into an isotope of polonium, $^{214}_{84} \mathrm{Po}$

In alpha decay. The radium are a are a, um to 26 to 26 radiate, um 1 38 and decays into, uh, irony are in 222 86 then 1 36 boss. Helium nuclear. So hee four two and to hear, end off the problem. Thank you for watching.

So in this problem we're seeing breaking, breaking off on a town with large number off protons, which is our uranium here. So let me just Gee, I don't know atomic number and mass numbers. So uranium here is breaking down into strong Siham, and they're non along with the release of three neutrons and a large amount of magic. So basically, this process is breaking apart off an item with large number off Britain's and nutrients into two smaller atoms. That is this John C, um, and Zeno. Okay, no, through the process, free nutrients, as you can see at the least, along that tremendous amount of energy and hence tree Nugent's on a large amount of energy is released on in order to in order for the reaction to occur the, uh, infighting a new trick so that this atom larger down will break and this process got responds to efficient reaction in the car. Hence, the correct answer will be deep

All right there. Two reactions taking place of the strong Schum decay. Ah, beta decay of strontium. Strontium 90 38 decays to That's a nine there decays to your atrium. 90 39 There's one electron, uh, with the negative charge there. And tiny tree No, uh, and then the atrium 90 39 uh, decays you treatem 90 39 decays to, uh, zirconium 90 40 again baited a case. So there's an electron there. Okay, so, uh, what we want to do, what we want to use is the radioactive decay equation and equals and not terms exponents of minus lambda t So land us is equal to the log. A rhythm of an over and not over tea. Um, and therefore that's equal to log, too. Over 1/2 life. That's the definition of love. Lamba. Be arranging this. We find that t equals ah half life over love two times log of n over and not half life is 29 years. Divide that by locked, too. And then multiply by the by log rhythm off 0.1 Because we want the level of which, the time of which it falls to 1%. of its initial levels. Point pull one is that and so therefore you have the time taken would be, Ah, 192.7 years. So approximately 1 93 years.

Each of the following isotopes undergoes radioactive decay my production of a beta particle and we have to you right, the balance to formula for that radioactive decay. Ding. So remember that a beta particle can be represented as zero minus one you. So for each of these reactions, one of the products is going to be that beta particle. Remember that on both the left and the right D mass number and atomic number, you have to be the same. So for party, our second product, he's gonna have to have atomic number of seven seven minus one will give us six. Like on the left, you'll have mass number of 14. So from the periodic table will find that the element with atomic number of seven is nitrogen. For part B, the mass number for a missing particle is gonna be 140 and the atomic number is gonna have to be 56 from the periodic table will find that the element with atomic number of 56 is very, um for part C, the mass number will be 234 in the atomic number, it's gonna have to be 91 from the periodic table will find that the element with atomic number 91 is protecting you


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