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05-08-2019
Physics

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Determine the minimum amount of energy (in MeV) needed to remove a neutron from a 88Sr nucleus. (For all masses, keep six places beyond the decimal point when performing your calculations. Then round your final answer to at least three significant figures.)

Respuesta :

11-08-2019

Answer:

11.106 Mev.

Explanation:

Sr⁸⁸ → Sr ⁸⁷ + ₀n¹

87.905612 86.908877 1.008664 ( atomic masses in amu )

mass defect = ( 86.908877 + 1.008664) - 87.905612 = .011929 amu.

Equivalent energy in MeV = .011929 x 931 = 11.106 MeV

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