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3 years ago

How does the number of each type of particle in a sodium ion compare with the number of each type of particle in a sodium atom?

Physics

1 answer:

anyanavicka [17]3 years ago

3 0

<h3>Answer and explanation;</h3>

A neutral sodium atom contains 11 protons, 12 neutrons and 11 electrons. By removing an electron from this atom we get a positively charged Na+ ion that has a net charge of +1. Thus; Sodium atom is neutral whereas sodium ion is a charged species with a charge of +1.
Sodium ion is positively charged. In sodium ion, there are 11 protons and 12 neutrons but 10 electrons, i.e., sodium ion contains lesser number of electrons. Additionally,the size of sodium ion is smaller than that of sodium atom.

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Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

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Explanation:

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$C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm$

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The electric field according to Gauss Law is given by:

$EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2} }=\frac{kQ}{r_{a}^{2}}\\ E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C$

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