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The nucleus of $^{8} \mathrm{Be},$ which consists of 4 protons and 4 neutrons, is very unstable and spontaneously breaks into two alpha particles (helium nuclei, each consisting of 2 protons and 2 neutrons). (a) What is the force between the two alpha particles when they are $5.00 \times 10^{-15} \mathrm{m}$ apart, and (b) what is the initial magnitude of the acceleration of the alpha particles due to this force? Note that the mass of an alpha particle is 13 4.0026 $\mathrm{u} .$

a. 36.8 $\mathrm{N}$

b. $5.54 \times 10^{27} \mathrm{m} / \mathrm{s}^{2}$

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the force between a tool for particle is given by physical too. Okay, Constant times to e. Oh, this will be square divided by our square student values 8.99 times 10 to the power nine times are four times 1.6 times 10 to the power. Minus 19 is the charge on each alpha particle off square, divided by the distance that is a five times gentle power minus 15 meters square. Disc uses the force off a 36.8 Newton than the exploration produced by this force can be written is a selection physical force divided by the mass force. We just found his 36.8 Newton divided by Mass so converting the atomic mass unit into kilograms. That means been Hying four. When 00 to six times 1.66 times tent about 24 7 This whole factor divides. This uses that explanation off from ex elation of 5.54 times 10 to the power 27 meter for second square