A proton with an initial speed of 900,000ms is brought to rest by an electric...

Walter Clyburn

Walter Clyburn

Answered

2022-01-14

A proton with an initial speed of 900,000ms is brought to rest by an electric field.
What was the potential difference that stopped the proton?
Express your answer to two significant figures and include the appropriate units.
What was the initial kinetic energy of the proton, in electron volts?
Express your answer using two significant figures.

Answer & Explanation

Papilys3q

Papilys3q

Expert

2022-01-15Added 34 answers

Step 1
Solution:
a) 12mpv2=qVV=mpv2zq=1.67×1027×(9×105)22×1.6×1019v
v=4227×19v
b) Initial kinetic energy of proton
=4227×19eV
Serita Dewitt

Serita Dewitt

Expert

2022-01-16Added 41 answers

Step 1
Initial speed of proton vi=900,000ms
final speedf problem vf=0
Now,
vi=900,00ms
vf=0
Kinetic energy
KEi=12mpvi2
KEf=12mvf2=0
Potential Energy
PEi=0
PEf=?
Here, from Conservation of Energy. Kinetic Energy is Converted in to Potential Energy
12mpvi2=PEf
12×1.67×1027×(900000)2=PEf=677×1018
Tunles{kg(mls)2
Now, Potential difference Δv=ΔPEq
=PEfPEiq
Step 2
Potential difference to stop the proton given by
Δv=PEfq
(PEi=0)
=67741×1018Ton1.6×1019c
Δv=4228v
b) Now, initial kinetic energy KEi=12mpvi2
KEi=677.4×1018T
v=1.6×1019T
Kinetic energy is in ev is 4228 ev
i.e. 4228 ev

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