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A star ending its life with a mass of four to eight times the Sun's mass is expected to collapse and then undergo a a supernova event. In the remnant that is not carried away by the supernova explosion, protons and electrons combine to form a neutron star with approximately twice the mass of the Sun. Such a star can be thought of as a gigantic atomic nucleus. Assume $r=a A^{1 / 3}(\text { Eq. } 44.1) .$ If a star of mass $3.98 \times 10^{30} \mathrm{kg}$ is composed entirely of neutrons $\left(m_{n}=\right.$

$1.67 \times 10^{-27} \mathrm{kg} ),$ what would its radius be?

16.0 $\mathrm{km}$

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Cornell University

Rutgers, The State University of New Jersey

Simon Fraser University

McMaster University

So we want to find the radius of this star. We know the equation. Where onto uses r equals lower case, constant air times the mass number which is the number of neurons in this case to the one that power we know our constant aids will be 1.2 times 10 minutes of the negative 50 meters. So far, amassed number A or the number of neurons we know that's gonna be confined, that with this equation here, the total mass of the star you going So the massive wonder on times the number of neurons which is are a right there So we could rearrange this equation. So find the total number of neurons not in total massive. This stars 3.98 times 10 to the 30 kilograms over the massive wonder on which is 1.67 It sounds 10 to the negative 27 kilograms. Cover guns. Cantaloupe. Now leave us with a mass number or a number of neurons. Two point three a times. 10 to the 57. That's how many neurons we have. So now we just simply plug everything in himself for the radius. So are you cool? 1.2 times turns of the negative 15 meters times two point three a terms turn So the 57 so the 1/3 power, then we will get 16,000 on 29 meters for 16 kilometers.

University of Massachusetts Amherst