Motion of a Proton in Electric and Magnetic Fields

[Variable_X]

Homework Statement

A proton is accelerated through a voltage of 248 kV.

a) Calculate the energy of the proton in electron volt and in Joule.

b) Assuming that the mass of the proton remains the same as its rest mass, calculate the speed of the photon.

c)The proton is curved by a magnetic field of 1.5 Telsa. Calculate the radius of the curvature of its path.

Relevant Equations

N/A

a) 248*10^3 eV for 248kV

Calculate the energy in J

K=248*10^3*1.6*10^-19

=396.8*10^-19 J

b)

K=(1/2)mv^2

v=sqrt(2k/m)

=sqrt((2*396.8*10^-19)/1.67*10^-27)

=218^10^3 m/s

c)
r=mv/qB

=1.67*10^-27*218*10^3/1.6*10^-19*1.5*10^-4

=15.17 mr=mv/qB

=1.67*10^-27*218*10^3/1.6*10^-19*1.5*10^-4

=15.17 m

 

[mfb]

You lost a factor 10³ in the kinetic energy calculation.
Your magnetic field has a strange factor 10^-4.

Generally:
Don't leave out units, they are important.
a/b*c could be read as $\frac a b \cdot c$. Put brackets around denominators: a/(b*c) is unambiguous.
Even better: Use LaTeX for formulas.

 

[HallsofIvy]

Also these are PROTON calculations not PHOTON calculations!

 

[Cutter Ketch]

You lost a factor 103 in the kinetic energy calculation.

Or actually earlier in the conversion to Joules.

 

[kuruman]

Or actually earlier in the conversion to Joules.

Which would have been much easier to pinpoint and troubleshoot if you solved the problem symbolically before substituting numerical values.

 

[PeroK]

Re-workings
a)

converting 248kV to eV gives me $$2.4 × 10^-13$$

What sort of calculator are you using?

You know that if you search online for "electron volts to joules" you can find something to do the calculation. You just type in the number.

 

[Variable_X]

What sort of calculator are you using?

You know that if you search online for "electron volts to joules" you can find something to do the calculation. You just type in the number.

I not changing eV to Joules at that point I'm changing kV to eV

 

[PeroK]

I not changing eV to Joules at that point I'm changing kV to eV

And how are you doing that?

 

[PeroK]

I guess I've done it wrong... E(eV) = V(V) × Q(e) so changing 248kV to V = 248000
248000x1.60x10^-19= 3.968x10^-14

An $eV$ is the energy an electron (or proton or anything else with the same charge) gains when accelerated through $1V$. I thought you got this right in your OP:

a) 248*10^3 eV for 248kV

 

[PeroK]

Ok so what about the rest?

I may be mistaken, but I think everything else is wrong.

 

[PeroK]

Let's see where we are. First, we can write down the kinetic energy in $eV$: $T = 248keV$.

Next, we use a calculator (or the Internet), to convert that to Joules: $T = 3.97 \times 10^{-14}J$.

Now, we need the speed of the proton. How would you do that?

 

[Variable_X]

Let's see where we are. First, we can write down the kinetic energy in $eV$: $T = 248keV$.

Next, we use a calculator (or the Internet), to convert that to Joules: $T = 3.97 \times 10^{-14}J$.

Now, we need the speed of the proton. How would you do that?

Yeah, I understand now, it's 6.9x10^20ms-1 (rouned)

 

[PeroK]

Yeah, I understand now, it's 6.9x10^20J (rouned)

What is that?

 

[PeroK]

Yeah, I understand now, it's 6.9x10^20ms-1 (rouned)

That's worse! What's the speed of light?

 

[Variable_X]

That's worse! What's the speed of light?

3.00x10^8 ms-1

 

[PeroK]

3.00x10^8 ms-1

And the speed of your proton?

 

[PeroK]

Yeah, I understand now, it's 6.9x10^20ms-1 (rouned)

I'm signing off now. But, really, you can't produce a nonsensical answer like this and not realize your calculation has gone awry.

 

[Cutter Ketch]

I believe you know what you are doing. Your original post does the right things for the right reasons using the right equations. (Except you converted the magnetic field to Gauss for some reason. Tesla is already the SI unit). You just can’t seem to do the numerical calculations correctly. I don’t know if you type in things wrong or don’t use scientific notation correctly, but all your difficulties arise from your calculator skills.

You should try again. I think you’ve seen and understood that your energy in Joules was off by a factor of 1000. Since the speed is proportional to the square root of energy, the speed is off by a factor of about 33. Since you converted to Gauss when you should have used Tesla your radius is off by an additional factor of 10^4. So you should have a very good idea of what the correct answer is. If you get something wildly different, you are once again making a calculator error.

 

[Variable_X]

I believe you know what you are doing. Your original post does the right things for the right reasons using the right equations. (Except you converted the magnetic field to Gauss for some reason. Tesla is already the SI unit). You just can’t seem to do the numerical calculations correctly. I don’t know if you type in things wrong or don’t use scientific notation correctly, but all your difficulties arise from your calculator skills.

You should try again. I think you’ve seen and understood that your energy in Joules was off by a factor of 1000. Since the speed is proportional to the square root of energy, the speed is off by a factor of about 33. Since you converted to Gauss when you should have used Tesla your radius is off by an additional factor of 10^4. So you should have a very good idea of what the correct answer is. If you get something wildly different, you are once again making a calculator error.

Right so
a) is 248keV.
3.97×10−14JT=3.97×10−14J.

b) v = √(2K/m) should be √(2(3.97x10^-14)/1.67×10-31) = 6.89528x10^9 ms^-1
I'm assuming the question has typos and it's meant to be for the speed of the PROTON, not the PHOTON.

c) r = mv/(eB) r=1.67×10-31x6.9x10^9/(1.67×10-19x1.5) = 0.0046 nm

 

[mfb]

See #17 and #18 and consider if these values are realistic.

You seem to make many errors with the calculator, that makes it even more important to check if the answer is plausible.

 

[PeroK]

Right so

b) v = √(2K/m) should be √(2(3.97x10^-14)/1.67×10-31) = 6.89528x10^9 ms^-1
I'm assuming the question has typos and it's meant to be for the speed of the PROTON, not the PHOTON.

You still have $v > c$! Where did you get the mass of a proton?

 

[willem2]

b) v = √(2K/m) should be √(2(3.97x10^-14)/1.67×10-31) = 6.89528x10^9 ms^-1
I'm assuming the question has typos and it's meant to be for the speed of the PROTON, not the PHOTON.

c) r = mv/(eB) r=1.67×10-31x6.9x10^9/(1.67×10-19x1.5) = 0.0046 nm

you're using the mass of an electron and not of a proton at both b and c

 

[PeroK]

you're using the mass of an electron and not of a proton at both b and c

Actually, it's neither one thing nor the other:

Mass of proton: $1.67 \times 10^{-27}kg$

Mass of electron: $9.11 \times 10^{-31}kg$

Mass used in this problem: $1.67 \times 10^{-31}kg$

 

[PeroK]

My rest Mass of a proton is given as 1.67×10^-31 kg

Your protons must be different from all the others! That's less than an electron. Please check online.

 

[PeroK]

a) is 248keV.
3.97×10−14JT=3.97×10−14J.

b) v = √(2K/m) should be √(2x3.97x10^-14)/1.67×10^-19) = 1.68731x10^12 ms^-1

You appear to have used the charge on a proton instead of its mass.

 

[PeroK]

a) is 248keV.
3.97×10−14JT=3.97×10−14J.

b) v = √(2K/m) should be √(2x3.97x10^-14)/1.67×10^-27) = 1.68731x10^20 ms^-1

Don't post any more answers until you get a speed significantly less than the speed of light.

 

[mfb]

Please don't delete posts, especially not if people replied, this makes the thread difficult to follow.