Question #168071

Calculate the de-Broglie wavelength of a proton whose kinetic energy is equal to the rest mass energy of an electron (mass of proton is 1836 times that of electron). 


1
Expert's answer
2021-03-02T18:04:55-0500

The rest mass energy of electron is E0=mec2.E_0 = m_e c^2. The de-Broglie wavelength is

λB=hpp.\lambda_B = \dfrac{h}{p_p} .

The momentum of proton is pp=mpv,p_p=m_p v , the kinetic energy is Ek=mpv22,E_k=\dfrac{m_pv^2}{2}, so mpv=mp2Ekmp=2mpEk=2mpmec2=2mpmec.m_pv = m_p \cdot \sqrt{\dfrac{2E_k}{m_p}} = \sqrt{2m_pE_k} = \sqrt{2m_p m_ec^2} = \sqrt{2m_p m_e}c . Therefore, the de-Broglie wavelength is λB=h2mpmec=6.631034Js21836(9.11031kg)23108m/s=41014m.\lambda_B = \dfrac{h}{\sqrt{2m_p m_e}c} = \dfrac{6.63\cdot10^{-34}\,\mathrm{J\cdot s}}{\sqrt{2\cdot 1836\cdot (9.1\cdot 10^{-31}\,\mathrm{kg})^2}\cdot 3\cdot10^8\,\mathrm{m/s}} = 4\cdot10^{-14}\,\mathrm{m}.


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