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De Broglie postulated that the relationship $\lambda=h / p$ is valid for relativistic particles. What is the de Broglie wavelength for a (relativistic) electron having a kinetic energy of 3.00 $\mathrm{MeV}$ ?

3.5810^{-13} m

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here for the solution, given that spirit that the sea off light is equal to three multiplied by 10 to the power 8 m per second kind. Eric Energy, That's K off Electron is 3.0 m e v rest mass, that is M E off electron will be 9.11 multiplied by 10 to depart minus 31. Casey, now rest Energy. It is e r is equal to m C square off Clacton equal to 0.51 m E V plank's constant magical toe 6.6 to multiply by 10 to department is 34. They s now the moment um, let s p for the relate do for the relativistic political will be p equal to under oath. He square minus e are square divide by See the total energy that is equal to kinetic energy. Okay, plus e. R. So this will be 3.51 MTV now the d broccoli wavelength off the Britain will be lambda called toe Catch by p. So here, by substituting the value, it will be at sea by under route E square minus e square are. And here we substituted the numerical value off all valuable or variables. On calculating this, we get the final answer 3.58 Multiply by 10 to the power minus 13 m. So this is the solution. Step by step, please go through.

DR APJ ABDUL KALAM TECHNICAL UNIVERSITY