Question

In nuclear fission, a nucleus splits roughly in half.

A. what is the potential 4.00x10^-14m from a fragment that has 50 protons in it? (p=+1.602x10^-19C)

B. what is the potential energy of a similarly charged fragment at this distance?

Answer 1

A. Charge of the fragment:

$Q = 50 \;protons = 50 \times 1.6 \times 10^{-19} \;C$

Distance from the fragment:

$r = 4.00 \times 10^{-14} \;m$

Electric force constant:

$k = 9 \times 10^9 \;Nm^2/C^2$

The electric potential of a point charge:

$V = \frac{kQ}{r} \\ V = \frac{9 \times 10^9 \times 50 \times 1.6 \times 10^{-19}}{4.00 \times 10^{-14}} \\ = 1.80 \times 10^6 \;V$

B. The relationship between potential and electrical potential energy is:

$V = \frac{U}{q} \\ U = qV \\ U = 50 \times 1.80 \times 10^6 \\ = 9.0 \times 10^7 \;eV \\ = 90 \;MeV$