Question

The kinetic energy of an electron is 5 eV. Calculate the de-Broglie wavelength associated with it $(h=6.6\times {10}^{-34}Js,me=9.1\times {10}^{-31}kg)$

• A. $5.47\dot{A}$
• B. $10.9\dot{A}$
• C. $2.7\dot{A}$
• D. None of these

Answer 1

The correct option is A $5.47\dot{A}$

$\lambda =\frac{h}{\sqrt{2mE}}=\frac{6.6\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 5\times 1.6\times {10}^{-19}}}=5.469\times {10}^{-10}m=5.47\dot{A}$