Question

The energy of an $\alpha$-particle is $6.8\times {10}^{-18}J$. What will be the wavelength associated with it?

Answer 1

Using the De-Broglie's equation

$E=6.8\times {10}^{-18}J\lambda =\frac{h}{mv}\Rightarrow {\lambda }^2=\frac{h^2}{(2\times \frac{1}{2}{m}^2{v}^2)}\Rightarrow \lambda =\frac{h}{\sqrt{2Em}}\lambda =\frac{6.626\times {10}^{-34}}{\sqrt{2\times 6.8\times {10}^{-18}\times 4\times 1.66\times {10}^{-27}}}\lambda =2.2\times {10}^{-12}m$