How Do You Derive This Kinetic Energy Equation for High-Speed Electrons?

[atomqwerty]

I'm confused about how can I obtain this equation:

$$K=\frac{(cp)^{2}}{2m_{0}c^{2}}=\frac{h^{2}}{2m_{0}\lambda^{2}}$$

being K the Kinetic energy, p the momentum and lambda the wavelength.

It's related with the energy of an electrons beam, and I don't know how to obtain it from the equations that I know so far (basically, E = hf = K + mc^2, with f the frequency)

It's a really unhappy business :S

Thank you!

(Edit: Sorry about not have used the template, my question this time just not fit in it)

 

[Andrew Mason]

I'm confused about how can I obtain this equation:

$$K=\frac{(cp)^{2}}{2m_{0}c^{2}}=\frac{h^{2}}{2m_{0}\lambda^{2}}$$

being K the Kinetic energy, p the momentum and lambda the wavelength.

It's related with the energy of an electrons beam, and I don't know how to obtain it from the equations that I know so far (basically, E = hf = K + mc^2, with f the frequency)

It's a really unhappy business :S

Thank you!

(Edit: Sorry about not have used the template, my question this time just not fit in it)

For very high speed electrons, you have to take into account relativistic effects.

$$E_{total} = \sqrt{p^2 c^2 + m_0^2 c^4} = m_0c^2 + E_k = h\nu = \frac{hc}{\lambda}$$

Try and work that out.

AM

 

[atomqwerty]

For very high speed electrons, you have to take into account relativistic effects.

$$E_{total} = \sqrt{p^2 c^2 + m_0^2 c^4} = m_0c^2 + E_k = h\nu = \frac{hc}{\lambda}$$

Try and work that out.

AM

Thanks

By substituting the my expression of K into your formula, I obtain that

$$= p^2 c^2 = 2m_0c^2$$ with have no too much sense.

However, the formula you wrote it's the same that mine (Kinetic energy plus rest energy), but including the Lorentz factor in the Kinetic energy, isn't it?