Questions: An archaeon has 9.8 × 10^16 protons and a net charge of 6.6 pC. Part (a) How many fewer electrons are there than protons? Np-Ne=

Solution Steps

The net charge is positive, indicating there are more protons than electrons. The net charge is given by $Q = (N_p - N_e)e$, where $N_p$ is the number of protons, $N_e$ is the number of electrons, and $e$ is the elementary charge, $e = 1.602 \times 10^{-19}$ C.

We are given that the net charge is $Q = 6.6 \text{ pC} = 6.6 \times 10^{-12}$ C. We are also given that $N_p = 9.8 \times 10^{16}$. We need to find $N_p - N_e$, which represents the number of excess protons (or fewer electrons).

From the equation $Q = (N_p - N_e)e$, we can solve for $N_p - N_e$: $N_p - N_e = \frac{Q}{e}$

Substituting the given values, we get: $N_p - N_e = \frac{6.6 \times 10^{-12} \text{ C}}{1.602 \times 10^{-19} \text{ C}}$ $N_p - N_e = 4.11985 \times 10^7$ $N_p - N_e \approx 4.1 \times 10^7$

Final Answer