Question

Calculate the de Broglie wavelengths of an electron and a proton having same kinetic energy of $100eV$.

• A. ${\lambda }_e=123pm$; ${\lambda }_p=286pm$.
• B. ${\lambda }_e=286pm$; ${\lambda }_p=286pm$.
• C. ${\lambda }_e=286pm$; ${\lambda }_p=123pm$.
• D. ${\lambda }_e=123pm$; ${\lambda }_p=123pm$.

Answer 1

The correct option is A ${\lambda }_e=123pm$; ${\lambda }_p=286pm$.

$\lambda =\frac{h}{\sqrt{2mE}}$ where $E=$ Kinetic energy.

${\lambda }_e=\frac{h}{\sqrt{2m_{e}E}}=\frac{6.626\times {10}^{-33}}{\sqrt{2\times 9.1\times {10}^{-31}\times 100\times 1.6\times {10}^{-19}}}=123pm$

${\lambda }_p=\frac{h}{\sqrt{2m_{p}E}}=\frac{6.626\times {10}^{-33}}{\sqrt{2\times 1.6\times {10}^{-27}\times 100\times 1.6\times {10}^{-19}}}=286pm$