# Chapter 2 - Sections 2.1-2.9 - Exercises - Problems by Topic: 50

$7.33\times10^{3}$ electrons

#### Work Step by Step

Mass of 1 proton = 1.00727 amu Mass of 1 neutron = 1.00866 amu Total mass of Helium nucleus = $2\times(1.00727) + 2\times(1.00866)$ = 4.03186 amu Mass of 1 electron = 0.00055 amu. Nmber of electrons to be equal to mass of nucleus = $\frac{4.03186}{0.00055}$ = $7.33\times10^{3}$ electrons.

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