Question

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

• A. 5.47 ??/div>
• B. 10.9 ??/div>
• C. 2.7 ??/div>
• D. none of these

Answer 1

Answer: (A)

5.47 ??/div>

Answer 2

Wavelength associated with an electron $\lambda =\frac{h}{\sqrt{2 m E}}$ $=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 5 \times 1.6 \times 10^{-19}}}$ $=5.47 \times 10^{-10} m =5.47 ,??