Answer to Question #168071 in Quantum Mechanics for dheeraj

Question

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). 

Answer 1

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

$\lambda_B = \dfrac{h}{p_p} .$

The momentum of proton is $p_p=m_p v ,$ the kinetic energy is $E_k=\dfrac{m_pv^2}{2},$ so $m_pv = m_p \cdot \sqrt{\dfrac{2E_k}{m_p}} = \sqrt{2m_pE_k} = \sqrt{2m_p m_ec^2} = \sqrt{2m_p m_e}c .$ Therefore, the de-Broglie wavelength is $\lambda_B = \dfrac{h}{\sqrt{2m_p m_e}c} = \dfrac{6.63\cdot10^{-34}\,\mathrm{J\cdot s}}{\sqrt{2\cdot 1836\cdot (9.1\cdot 10^{-31}\,\mathrm{kg})^2}\cdot 3\cdot10^8\,\mathrm{m/s}} = 4\cdot10^{-14}\,\mathrm{m}.$