Calculating the speed of a β-particle?

[Mikry]

I'm busy preparing for my upcoming physics final by going through some old papers. I came across this question and I honestly have no idea what to do. Here's the question:

A β-particle was emitted with an energy of 5.7 x 10³ eV.
Calculate the speed of the β-particle.


What I was thinking was that I could use E_k=(0.5)mv²

Thus I would get 5.7 x 10³ = (0.5)(9.11 x 10^-31)(v²)
Which would give me the value of v = 1.1186 ms^-1 but I don't know if that is correct.

I think the biggest problem though is that I'm not actually sure what eV is...I'm just assuming it's equivalent to joules.

Thanks in advance!β

 

[vanhees71]

One electron volt, 1 eV, is the energy an electron gains when it is running through an electric potential difference of 1 V. To evaluate the speed, you don't need to transform to Joule first. Just use that the mass of the electron is equivalent to the rest energy m c^2=511 \; \mathrm{keV}.

 

[CAF123]

1 eV is the amount of energy an electron would gain if it is accelerated by a potential difference of 1V. You know W= qΔV = q (1V) = 1 eV => q = 1 eV /1V. The charge on an electron is ≈ 1.6 x 10^-19 C so then 1 eV $\equiv$ 1.6 x 10^-19J since [J/V] = [C].

This is the conversion you should use if you wish to convert everything to joules.

 

[DeltaFunction]

Thus I would get 5.7 x 10³ = (0.5)(9.11 x 10^-31)(v²)
Which would give me the value of v = 1.1186 ms^-1 but I don't know if that is correct.

when you do this calculation you actually get 1.1186 x 10^17. Many orders of magnitude faster than the speed of light. As the others have said you've just got to make sure your units are consistent.

 

[Curious3141]

I'm busy preparing for my upcoming physics final by going through some old papers. I came across this question and I honestly have no idea what to do. Here's the question:

A β-particle was emitted with an energy of 5.7 x 10³ eV.
Calculate the speed of the β-particle.What I was thinking was that I could use E_k=(0.5)mv²

Thus I would get 5.7 x 10³ = (0.5)(9.11 x 10^-31)(v²)
Which would give me the value of v = 1.1186 ms^-1 but I don't know if that is correct.

I think the biggest problem though is that I'm not actually sure what eV is...I'm just assuming it's equivalent to joules.

Thanks in advance!β

You don't *have* to convert everything to J. But if you're curious, $1eV \approx 1.6 \times 10^{-19}J$ (the number is (not coincidentally) exactly equal to the electron charge in coulomb).

Are you expected to know and use Special Relativity Theory? The difference from the classical answer is only about 1%, but if you're expected to use Special Relativity and you use Newtonian mechanics in your working, you might be penalised. So you need to clarify this.

 

[Emilyjoint]

One electron volt, 1 eV, is the energy an electron gains when it is running through an electric potential difference of 1 V. To evaluate the speed, you don't need to transform to Joule first. Just use that the mass of the electron is equivalent to the rest energy m c^2=511 \; \mathrm{keV}.

I don't think you can simply replace the mass with the rest energy in eV to calculate the speed.
The most sensible thing to do is convert the energy in eV into Joules...this works for any case when the energy of a particle (electron or otherwise) is given in eV

 

[janhaa]

I'm busy preparing for my upcoming physics final by going through some old papers. I came across this question and I honestly have no idea what to do. Here's the question:
Thus I would get 5.7 x 10³ = (0.5)(9.11 x 10^-31)(v²)
Which would give me the value of v = 1.1186 ms^-1 but I don't know if that is correct.
I think the biggest problem though is that I'm not actually sure what eV is...I'm just assuming it's equivalent to joules.Thanks in advance!β

this equation will get the correct answer

(5.7 x 10³)x (1.6 x 10^-19)= 0.5x(9.11 x 10^-31)x(v²)