Question

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

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

Answer 1

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

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