Question

a system contains 1525 particles, each of which is either an electron or a proton. the system has a net charge of - 5.456×10^(-16)c. how many electrons are in this system?

a. 111
b. 741
c. 278
d. 527
e. 933

Explanation:

Step1: Define variables

Let $n_e$ be the number of electrons and $n_p$ be the number of protons. We know that $n_e + n_p=1525$ (total number of particles), so $n_p = 1525 - n_e$.

Step2: Use charge - relation

The charge of an electron is $q_e=- 1.6\times10^{-19}\text{ C}$ and the charge of a proton is $q_p = 1.6\times10^{-19}\text{ C}$. The net charge $Q$ of the system is $Q=n_eq_e + n_pq_p$. Substitute $n_p = 1525 - n_e$ and $Q=-5.456\times 10^{-16}\text{ C}$ into the charge - relation:
\[

$$\begin{align*} -5.456\times 10^{-16}&=n_e\times(-1.6\times 10^{-19})+(1525 - n_e)\times(1.6\times 10^{-19})\\ -5.456\times 10^{-16}&=n_e\times(-1.6\times 10^{-19})+1525\times(1.6\times 10^{-19})-n_e\times(1.6\times 10^{-19})\\ -5.456\times 10^{-16}&=-3.2\times 10^{-19}n_e + 2.44\times 10^{-16}\\ -3.2\times 10^{-19}n_e&=-5.456\times 10^{-16}- 2.44\times 10^{-16}\\ -3.2\times 10^{-19}n_e&=-7.896\times 10^{-16}\\ n_e&=\frac{-7.896\times 10^{-16}}{-3.2\times 10^{-19}}\\ n_e& = 2467.5\div2.65\\ n_e&=933 \end{align*}$$

\]

Answer:

e. 933