Question

The de Broglie wavelength of an electron (mass $= 1 × 10^{-30}$ kg, charge = $1.6 × 10^{ -19}$ C) with a kinetic energy of $200\, eV$ is (Planck’s constant = $6.6 \times 10^{-34}$ J s)

• A. $9.60 × 10^{-11} \,m$
• B. $8.25 × 10^{-11}\, m$
• C. $6.25 × 10^{-11}\, m$
• D. $5.00 × 10^{-11}\, m$

Answer 1

Answer: (B)

The de-Broglie wavelength, $\lambda=\frac{h}{\sqrt{2 m k}}$
Given, $h =6.6 \times 10^{-34} J - s $
$m =1 \times 10^{-30}\, kg$
$K=200 \,eV =200 \,\times 1.6 \times 10^{-19} \,J$
Substituting all these values
$\lambda =\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 1 \times 10^{-30} \times 200 \times 1.6 \times 10^{-19}}}$
$=0.825 \times 10^{-10}=8.25 \times 10^{-11} \,m$