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adoni [48]
3 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?
Physics
1 answer:
yKpoI14uk [10]3 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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