Question

An electron has kinetic energy $2.8\times {10}^{-23}J$. De-Broglie wavelength will be nearly

$(me=9.1\times {10}^{-31}kg)$

• A. 9.28 × 10^-4 m
• B. 9.28 × 10^-7 m
• C. 9.28 × 10^-8 m
• D. 9.28 × 10^-10 m

Answer 1

The correct option is C

9.28 × 10^-8 m

From De-Broglie wave equation, we get

$\lambda$ = $\frac{h}{p}$ $\Rightarrow$ $\lambda$ = $\frac{h}{mv}$

$\&K.E=\frac{1}{2}m{v}^2\Rightarrow$ v = $\sqrt{\frac{2KE}{m}}$

$Now,\lambda$ = $\frac{h}{\lambda }$

$\&v=\sqrt{\frac{2KE}{m}}$

$\Rightarrow$ $\lambda$ = $\frac{h}{\sqrt{2mKE}}$

$\Rightarrow$ $\lambda$ =$\frac{6.62\times {10}^{-24}}{\sqrt{2\times 9.1\times {10}^{-31}\times 2.8\times {10}^{-23}}}$

$\Rightarrow$ $\lambda =9.28$ $\times$ ${10}^{-8}$m