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4 years ago

6

In nuclear fission, a nucleus splits roughly in half. (a) What is the electric potential 5.24 10-14 m from a fragment that has 3

6 protons in it? V (b) What is the electric potential energy in MeV of a similarly charged fragment at this distance?

1 answer:

yKpoI14uk [10]4 years ago

4 0

Answer:

Part A:

Electric potential = $V=9.8821*10^5 V$

Part B:

Electric Potential Energy=U= $35.57556 MeV$

Explanation:

Part A:

Charge on the fragment that has 36 Protons in it:

Charge on single roton= $1.6*10^{-19} C$

Charge on 36 protons= $36*1.6*10^{-19} C$

Charge on 36 protons= $5.76*10^{-18}\ C$

Distance=r= $5.24*10^{-14}\ m$

Formula For electric potential V:

$V=\frac{kQ}{r}$

Where

k is coulomb constant= $8.99*10^9\ N.m^2/C^2$

r is the distance

Q is the charge on 6 protons

$V=\frac{5.76*10^{-18}*8.99*10^9}{5.24*10^{-14}}\\V=988213.7405 V\\V=9.8821*10^5 V$

Electric potential = $V=9.8821*10^5 V$

Part B:

Electric Potential Energy=U

U=q*V

V is the electric potential

q is the number of protons

$U=36*9.8821*10^5 \\U=35575560 eV\\U=35.57556 MeV$

Electric Potential Energy=U= $35.57556 MeV$

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