an electron moving with velocity c/2 having de brogle eqal to wavelength of photon.the ratio of their energy is????????????

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anonymous · Answer

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anonymous · Answer

Since the electron is moving at relativistic velocities, we must use the relativistic equation for the total energy of a particle.  This is$$E=\sqrt{p^2c^2+(mc^2)^2}$$where $p=mv$ is the momentum of the particle of mass $m$ and velocity $v$.  We are told that the velocity of the electron is $c/2$, where $c$ is teh speed of light.  Substituting in we get $$E=\sqrt{\left(\frac{mc}{2}\right)^2c^2+(mc^2)^2}$$which simplifies down to $$E=\sqrt{\left(\frac{m^2c^4}{4}\right)+(m^2c^4)}=\sqrt{\left(\frac{5}{4}\right)m^2c^4}=\frac{\sqrt{5}}{2}mc^2$$

Now a particle with deBroglie wavelength $\lambda$ will have an energy of $$E=\frac{hc}{\lambda}$$where $h$ is Planck's Constant.  We now equate the two expressions to give $$\frac{hc}{\lambda}=\frac{\sqrt{5}}{2}mc^2$$ Rearranging to make $\lambda$ the subject we get $$\lambda=\frac{2h}{\sqrt{5}mc}$$and hence if the mass of the electron is $m=9.109\times10^{-31}$ kg, then the deBroglie wavelength is thus $$\lambda=\frac{2(6.626\times10^{-34})}{\sqrt{5}(9.109\times10^{-31})(2.998\times10^8)}=2.170\times10^{-12}\rm{m}$$

What this means is that one can use relativistic electrons for imaging purposes  as if they had a photon wavelength of $2.170\times10^{-12}$ m, and so can resolve smaller features than is capable using visible wavelengths (around 500 nm).

anonymous · Answer

i guess johnyymca is correct