Question

Two ions carrying equal charges repel
with a force of 1.48 x 10^-⁸ N when they
are separated by a distance of 5 x 10^-¹⁰ m.
How many electrons have been removed
from each ion.
​

Answer 1

Answer :

The number of electrons has been removed from each ion is 2.

Explanation :

The force of repulsion between two equal charges is $1.48 * 10^{-8} N$

Distance of separation (d) = $5 * 10^{-10} m$

We need to calculate the number of electrons that have been removed from each ion,

Let $q_{1} =q_{2} =q$

Using Coulomb's Law, we have

$F =\frac{1}{ 9 *10^{9} } * \frac{q_{1}q_{2} }{D^{2} }\\ \\F = \frac{1}{ 9 *10^{9} } *\frac{q^{2} }{d^{2} } \\\\q^{2} = \frac{1}{ 9 *10^{9} } *1.48 * 10^{-8} * (5 * 10^{-10} )^{2} \\\\q = 4.1 *10^{-19} C$

Now, q = ne

$n =\frac{q}{e}$

$n = \frac{4.1 *10^{-19} }{1.6 *10^{-19} } \\\\n = 2.5625$

So, electrons have been removed from each ion is 2.

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Answer 2

Answer:

4

Explanation: