Question

Find the de Broglie wavelength of the following. (a) a 4-MeV proton 14.3 Correct: Your answer is correct. fm (b) a 40-GeV electron 6.13623 Incorrect: Your answer is incorrect. fm (c) an electron moving at v

Answer 1

a) de Broglie wavelength of a 4-MeV proton: 14.3 fm

b) de Broglie wavelength of a 40-GeV electron: 0.031 fm

Explanation:

a)

The de Broglie wavelength of an object is given by

$\lambda=\frac{h}{p}$ (1)

where

h is the Planck constant

p is the momentum of the particle

Here we want to find the de Broglie wavelength of a 4-MeV proton. The rest of mass of the proton in MeV is

$m_0 = 938 MeV$

And since $4MeV$, this means that the proton is non-relativistic. So its kinetic energy is related to its momentum by

$E=\frac{p^2}{2m}$

which means

$p=\sqrt{2Em}$

where

$E=4 MeV \cdot 10^6 eV/MeV \cdot 1.6\cdot 10^{-19] J/eV=6.4\cdot 10^{-13} J$ is the kinetic energy

$m=1.67\cdot 10^{-27} kg$ is the proton mass

Substituting, we find

$\lambda=\frac{h}{\sqrt{2Em}}=\frac{6.63\cdot 10^{-34}}{\sqrt{2(6.4\cdot 10^{-13})(1.67\cdot 10^{-27})}}=14.3\cdot 10^{-15} m = 14.3 fm$

b)

In this case, the electron has kinetic energy of 40 GeV, while the rest mass of an electron is

$m_0 = 0.511 MeV$

Since $40 GeV >> 0.511 MeV$, the electron is ultra-relativistic: so we can rewrite its energy as

$E = pc$

The equation (1) can also be rewritten as

$\lambda = \frac{hc}{pc}$

where c is the speed of light. The quantity at the denominator is the energy, so

$\lambda=\frac{hc}{E}$

where:

$E=40 GeV = 40\cdot 10^9 eV \cdot (1.6\cdot 10^{-19})=6.4\cdot 10^{-9} J$ is the energy of the electron

And substituting, we find:

$\lambda=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{6.4\cdot 10^{-9}}=3.1 \cdot 10^{-17} m = 0.031 fm$

Learn more about de Broglie wavelength:

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