Question

What mass of electrons would be required to just neutralize the charge of 4.8 g of protons?

Answer 1

Let's calculate the total charge of M=4.8 g=0.0048 kg of protons.
Each proton has a charge of $q=1.6 \cdot 10^{-19} C$, and a mass of $m_p = 1.67 \cdot 10^{-27}kg$. So, the number of protons is
$N_p = \frac{M}{m_p}= \frac{0.0048 kg}{1.67 \cdot 10^{-27}kg}=2.87 \cdot 10^{24}$ And so the total charge of these protons is $Q_p = qN_p = (1.6 \cdot 10^{-19}C)(2.87 \cdot 10^{24})=4.6\cdot 10^5 C$

So, the neutralize this charge, we must have $N_e$ electrons such that their total charge is
$Q_e = -4.6 \cdot 10^5 C$
Since the charge of each electron is $q_e = -1.6 \cdot 10^{-19}C$, the number of electrons needed is
$N_e = \frac{Q_e}{q}= \frac{-4.6 \cdot 10^5 C}{-1.6 \cdot 10^{-19}C}=2.87 \cdot 10^{24}$
which is the same as the number of protons (because proton and electron have same charge magnitude). Since the mass of a single electron is $m_e=9.1 \cdot 10^{-31}kg$, the total mass of electrons should be
$M_e = N_e m_e = (2.87 \cdot 10^{24})(9.1 \cdot 10^{-31}kg)=2.6 \cdot 10^{-6}kg$