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Problem 16 Easy Difficulty

Calculate the binding energy of the last neutron in the $^{43}_{20} \mathrm{Ca}$ nucleus. Hint: You should compare the mass of $^{43}_{20} \mathrm{Ca}$ with the mass of $^{43}_{20} \mathrm{Ca}$ plus the mass of a neutron. The mass of $_{20}^{12} \mathrm{Ca}=$ 41.958 622 u, whereas the mass of $\underset{20}{43} \mathrm{Ca}=42.958770 \mathrm{u}$

Answer

7.93 \mathrm{MeV}

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Top Physics 103 Educators
LB
Liev B.

Numerade Educator

Zachary M.

Hope College

Aspen F.

University of Sheffield

Jared E.

University of Winnipeg

Video Transcript

for number 16 were asked to find the binding energy of the last neutron in calcium 43. And we were given the massive calcium 42 calcium 43 Only only difference between them will be that last neutron, because this is one extra neutron. So to get that mask, I'm just going to subtract. I'm taking 41.95 862 two minus 42.958 770 And I get that difference. There's one point. 0148 You so its atomic mass units. So that's the mass of that extra bound neutron in calcium 43. I want to know what is the binding energy of that. So I I'm gonna subtract to figure out what the massive Just a free neutron Ruby. Um, the free neutron would be one point. Oh, eat 665 You. But this one's on Lee. 1.148 You, um, subtract those and then I'm gonna throw my conversion right in here. I'm going to convert that so Meg elect trundles. So these air in use. I'll put that in here. So these are measured in atomic mass units. And I'm just going to my conversion. I know that one. You is the same as, uh, 931.5 Mega electrons ALS. So my use cancel and I get That's a binding energy of that one. Neutron is 7.93 mega electron volts.

University of Virginia
Top Physics 103 Educators
LB
Liev B.

Numerade Educator

Zachary M.

Hope College

Aspen F.

University of Sheffield

Jared E.

University of Winnipeg