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Problem 38 Hard Difficulty

One method of producing neutrons for experimental use is to bombard $_{3}^{7}$Li with protons. The neutrons are emitted
$$_{1}^{1} \mathrm{H}+_{3}^{7} \mathrm{Li} \rightarrow_{4}^{7} \mathrm{Be}+_{0}^{1} \mathrm{n}$$
(a) Calculate the mass in atomic mass units of the particles on the left side of the equation. (b) Calculate the mass (in atomic mass units) of the particles on the right side of the equation.
(c) Subtract the answer for part (b) from that for part (a) and convert the result to mega electron volts, obtaining the Q value for this reaction. (d) Assuming lithium is initially at rest, the proton is moving at velocity v, and the resulting beryllium and neutron are both moving at velocity V after the collision, write an expression describing conservation of momentum for this reaction in terms of the masses $m_{p}, m_{B}, m_{n},$ and the velocities. (e) Write an expression relating the kinetic energies of particles before and after together with Q. (f ) What minimum kinetic energy must the incident proton have if this reaction is to occur?

Answer

a. 8.023829 u
b. 8.025594 u
c. -1.64 M e V
d. m_{p} v=\left(m_{B c}+m_{n}\right) V
e. (1 / 2) m_{p} v^{2}=(1 / 2)\left(m_{B e}+m_{n}\right) V^{2}-Q
f. 1.88 \mathrm{McV}

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Video Transcript

from number 38 were given this reaction and first were asked, what is the sum of the masses of the left side of the equation, and then this? Some of the masses on the right side equation. So in appendix B in the back, I looked at the masses of these. So this was a massive one point. Oh, 78 to 5 was looking was a massive seven 0.0, 16 004 This beryllium had a massive 7.1 six, 9 to 9 and the single neutron had a massive 1.0, eight, 665 So apart, eh? And ask We're the sums on the left side. So that was just these two added together, and I got 8.0, 23 8 to 9. And then it s what's someone the right side of the equation report Be so that was just adding these two. And I got 8.0, 25594 Ellen part See said the subtract. So we're taking just the right. I mean, the left minus the right side. So left minus right. We turned out to be negative 0.0 176 life And then we're as toe change that to energy. So these are all measured and use atomic mass units. And now this one I'm converting to energy. So I know there's 9 31.5 in Mega Electron volts is the same as one U. So my you canceled. And here my answer part. See, In May, electron volts will be negative 1.64 for mega electron volts For port de the give me expressions for velocities, uh, initially was lifting was at rest. This hydrogen has a velocity of a lower case of E. And then after the collision of these, both have a velocity of a capital V. I'm just putting me up there to remember, and we're to write an expression in port in Port de were to write an expression for conservation momentum. Well, this was moving. This wasn't so totem in him before will be just the momentum of this. And where's called this? The massive proton rats post with this number now. So the mass of the proton tempts the velocity approach on which was his lower case vey. That's the total momentum Before that would equal the total momentum after through the momentum of this, plus the momentum of this. Um so this we're going the mass of beryllium terms of velocity, which was this Capital V plus the mass of the neutron times the velocity, this capital V So that's my expression for conservation momentum, Total men. And before it was equal to total momentum after. And then we're to do the same thing with kinetic energy for Port E. So now I'm just using the kinetic energy equation 1/2 empty squared instead of the momentum equation M v squared. So this will be 1/2 miss the Proton Times lower case v squared who equal 1/2 the mess the beryllium times capital V squared plus momentum. Other neutron bumps. 1/2 Remember the mass of the neutron times capital of these squared. I could I could just pull the V's out of these, you know, to distribute. I mean, I could factor of Viet of both of those and V squared, but it's is the same mathematical. And then in port f, I'm to find, um, an expression for the kinetic energy, Um, the minimum kinetic energy that this Proton would have when it comes in because you until you have this negative, I'm so isn't it? It's not just that, because it was also a mass defect. You, um So I'm using just the equation in the book where it's the I'll Change the Blue here, where the minimum kinetic energy a CZ one plus the myths of the incident particle over the mess of the target particle times the absolute value Q. And what we got right here this was Q. That's Q. So now when I go in this equation, I'm trying to find the mineral kinetic. My incident particle. Is this her origin? So that's this massive 1.78 25 It's a plus. My target particle is the lithium. So that's this mass appear 7.16 04 times the absolute value of this Q. Absolutely negative. 1.644 I do that when I get 1.8. That's mega Electron rules

University of Virginia
Top Physics 103 Educators
LB
Liev B.

Numerade Educator

Marshall S.

University of Washington

Farnaz M.

Other Schools

Aspen F.

University of Sheffield