Problem 21.10
A proton with an initial speed of 800000 m/s is brought to rest by an electric field.
Part B -
What was the potential difference that stopped the proton?
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Part C -
What was the initial kinetic energy of the proton, in electron volts?
To solve Part B, we need to find the potential difference that stopped the proton. We can use the equation:
ΔV = KE/q
where ΔV is the potential difference, KE is the kinetic energy, and q is the charge of the proton.
Since the proton is brought to rest, its final kinetic energy is zero. Therefore, the potential difference ΔV is equal to the initial kinetic energy of the proton.
To solve Part C, we can use the equation:
KE = (1/2)mv^2
where KE is the kinetic energy, m is the mass of the proton, and v is its initial speed.
The mass of a proton is approximately 1.67 x 10^-27 kg.
To convert the kinetic energy from joules to electron volts (eV), we can use the conversion factor: 1 eV = 1.6 x 10^-19 J.
Now let's calculate the answers step by step:
Part B:
ΔV = KE/q
Since the proton is brought to rest, its final kinetic energy is 0 J.
Therefore, the potential difference ΔV is also 0 V.
Answer for Part B: 0 V
Part C:
KE = (1/2)mv^2
m = 1.67 x 10^-27 kg
v = 800,000 m/s
KE = (1/2)(1.67 x 10^-27 kg)(800,000 m/s)^2
Now we calculate the value of KE (in Joules):
KE = (1/2)(1.67 x 10^-27 kg)(640,000,000,000 J)
KE ≈ 5.34 x 10^-14 J
Now we convert the result to electron volts (eV):
KE (in eV) = (5.34 x 10^-14 J)/(1.6 x 10^-19 J/eV)
KE ≈ 3.34 x 10^5 eV
Answer for Part C: approximately 3.34 x 10^5 eV.
So the potential difference that stopped the proton is 0 V, and the initial kinetic energy of the proton is approximately 3.34 x 10^5 eV.