Question

A proton and an electron have same kinetic energy. Which one has greater de-Broglie wavelength and why?

Answer 1

Let $m_1,m_2$ be the masses and $v_1,v_2$ be the velocities for electron and proton respectively .

As the two particles have same kinetic energy , therefore

$1/2m_1{v}_1^2=1/2m_2{v}_2^2=E$

or $m_1{v}_1=\sqrt{2m_{1}E}$ and $m_2{v}_2=\sqrt{2m_{2}E}$

since $m_1<m_2$ or $m_1{v}_1<m_2{v}_2$ .

As de-Broglie wavelength , $\lambda =h/mv$ , it means that de-Broglie wavelength for electron is greater .