Question

An amoeba has 1.00×1016 protons and a net charge of 0.300 pC. (a) How many fewer electrons are there than protons? (b) If you paired them up, what fraction of the protons would have no electrons?

Answer 1

Explanation:

(a)   It is known that charge on a proton is equal to $1.6 \times 10^{-19} C$. And, net charge is given as $0.3 pC$ which is also equal to $0.3 \times 10^{-12}C$.

Therefore, we will calculate the number of electrons as follows.

           $\frac{0.3 \times 10^{-12}C}{1.6 \times 10^{-19}C}$

            = $1.87 \times 10^{6}$

Hence, there are $1.87 \times 10^{6}$ fewer electrons are there than protons.

(b)   Now, we will calculate the fraction of protons that would have no electrons as follows.

               $\frac{1.88 \times 10^{6}}{1 \times 10^{16}}$

               = $1.88 \times 10^{-10}$

Therefore, fraction of the protons that would have no electrons is $1.88 \times 10^{-10}$.