Answer
(a) $933$
(b) $9.9\times 10^{-25}Kg$
Work Step by Step
(a) We know that
$N_p+N_e=1525$
$\implies N_p=1525-N_e$....eq(1)
We also know that
$e(N_p-N_e)=Q$
Substituting values of $N_p$ from eq(1) into the above equation, we obtain:
$e[(1525-N_e)-N_e]=Q$
This simplifies to:
$N_e=\frac{1}{2}(1525-\frac{Q}{e})$
We plug in the known values to obtain:
$N_e=\frac{1}{2}[1525-\frac{-5.456\times 10^{-17}}{1.6\times 10^{-19}}]$
$N_e=933$
(b) We can find the mass of the system as
$N_p=1525-N_e=1525-933=529$
Now $M_{total}=N_e m_e+N_pm_p$
We plug in the known values to obtain:
$M_{total}=(933)(9.109\times 10^{-31})+(529)(1.673\times 10^{-27})$
$M_{total}=9.9\times 10^{-25}Kg$
