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physicsboy321
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energy from electron passing between 2 electrodes, and calc. the freq of the elec.
Part 1:
The difference in potential between the cathode and anode of a spark plug is 17800 V.
What energy does the electron give up as it
passes between the electrodes?
Answer in units of J.
Part 2:
One fourth of the energy given up by an electron is converted into radiation in the form of
a single photon.
What is the frequency of the radiation?
Answer in units of Hz.
v=i*r?
Q = Δmc2
E=hf?
planck's constant=6.62607 × 10^-34 J · s
c² =931.5 MeV/u
1 eV=1.602×10−19 J
idk what else is needed sry
I'm not sure how to begin this problem, but my attempts are listed below
Pt.1: E= (1.602×10−19 J)(17800V)=2.85156 x 10^-15
Pt 2: E=hf
[(1/4)*(2.85156 x 10^-15)]/(6.62607 × 10^-34 J · s) =f <<<<-not sure
f=1.075886612 x 10^18 Hz??
Homework Statement
Part 1:
The difference in potential between the cathode and anode of a spark plug is 17800 V.
What energy does the electron give up as it
passes between the electrodes?
Answer in units of J.
Part 2:
One fourth of the energy given up by an electron is converted into radiation in the form of
a single photon.
What is the frequency of the radiation?
Answer in units of Hz.
Homework Equations
v=i*r?
Q = Δmc2
E=hf?
planck's constant=6.62607 × 10^-34 J · s
c² =931.5 MeV/u
1 eV=1.602×10−19 J
idk what else is needed sry
The Attempt at a Solution
I'm not sure how to begin this problem, but my attempts are listed below
Pt.1: E= (1.602×10−19 J)(17800V)=2.85156 x 10^-15
Pt 2: E=hf
[(1/4)*(2.85156 x 10^-15)]/(6.62607 × 10^-34 J · s) =f <<<<-not sure
f=1.075886612 x 10^18 Hz??
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