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 = (M)/(m_p)= (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 = (Q_e)/(q)= (-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`