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.

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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.

Papilys3q · Accepted Answer

Step 1  Solution:  a) 12mpv2=qV⇒V=mpv2zq=1.67×10−27×(9×105)22×1.6×10−19v  ⇒v=4227×19v  b) Initial kinetic energy of proton  =4227×19eV

Serita Dewitt · Accepted Answer

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=677×10−18  Tunles{kg(mls)2  Now, Potential difference Δv=ΔPEq  =PEf−PEiq  Step 2  Potential difference to stop the proton given by  Δv=PEfq  (∵PEi=0)  =677−41×10−18Ton≤1.6×10−19c  Δv=4228v  b) Now, initial kinetic energy KEi=12mpvi2  KEi=677.4×10−18T  ≤v=1.6×10−19T  ∴ Kinetic energy is in ev is 4228 ev  i.e. 4228 ev

A proton with an initial speed of 900,000 m/s is brought to rest

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