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Problem 41 Hard Difficulty

Natural gold has only one isotope, $^{197}_{79}$ Au. If gold is bombarded with slow neutrons, $e^{-}$ particles are emitted. (a) Write the appropriate reaction equation. (b) Calculate the maximum energy of the emitted beta particles. The mass of $_{80}^{198} \mathrm{Hg}$ is 197.966 75 u.

Answer

a. _{79}^{197} A u+_{0}^{1} n \rightarrow_{80}^{198} H g+_{-1}^{0} e+v_{c}
b. 7.89 M c V
c.

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Video Transcript

for number 41. We're told that the golden Adam is bombarded with neutrons, slow neutrons and gives off electrons and would have right the chemical, the nuclear reaction for that. Okay, so I started my gold that I was given this isotope. It was hit with the neutron. So a new trial as a massive one, but no charge. And for neutron And I know it gives off electrons. So that's what this is beta decay. So it's going to give open electron. Um, usually put that next. So I'm gonna leave a space here for what the one product is. I know the other product is going to be electron. An electron has no mass and charge a bit negative one. And whenever you have a native baited decay, you also give often Aunt Cindy tree No. So I'm gonna put that here too. So the barber top making that an answering neutrino. So now I can figure out by balance since equation what? My other product is here. My other my daughter. Um, so mass on this. That is 1 97 plus one more 1 98 and these don't have mess. So my mass must be 1 98 and my heart tones or my charge on the bottom Here at 79 on this side, Uh, minus one on that side. Because remember this neutron when in nucleus. So it's one more proton and give off this electron. So now So there's 80 and then I look on the productive when I see that number 80 is lucky. So here's my equation. This is my three products, and I'm to find what the kinetic energy of this. So I'm just gonna figure out how much you know Mass is missing here, and that is the kinetic energy of this electron. So just look up in the back and appendix B A mess of gold 1 96 quaint 966 552 a neutron 1.8 665 I was actually given in the proud problem. This is when they seven point nine 6675 and that doesn't have a mass. And that doesn't have mess. So if I'm gonna figure out how much mass is missing, gonna add this side subtracted that side and I get the missing mass is quaint. 00! 84 67 I remember this would be in you. These role in these messes were measured in you atomic mass units. I'm just gonna convert that, though to its energy equivalents. I know that one. You is the same as oh, 9 31.5 Mega electron volts. So you canceled and multiply. And the answer is seven 0.89 Mega the electron volts.

University of Virginia
Top Physics 103 Educators
LB
Liev B.

Numerade Educator

Farnaz M.

Other Schools

Zachary M.

Hope College

Aspen F.

University of Sheffield