Question

An a-particle moves in a circular path of radius 0.83cm in the presence of a magnetic field of 0.25wb/m^2. The de-broglie wavelength associated with the particle will be

1-10A

2-0.01A

3-1A

4-0.1A

Answer 1

$$\begin{gathered} \mathrm{De}\mathrm{broglie}\mathrm{wavelength}\mathrm{is}\mathrm{given}\mathrm{by}, \\ \mathrm{\lambda }=\frac{\mathrm{h}}{\mathrm{p}} \\ \mathrm{and}\mathrm{for}\mathrm{alpha}\mathrm{particle},\mathrm{p}=2\mathrm{eBr} \\ \mathrm{So}, \\ \mathrm{\lambda }=\frac{\mathrm{h}}{2\mathrm{eBr}} \\ \mathrm{Putting}\mathrm{the}\mathrm{values}, \\ \mathrm{\lambda }=\frac{6.6\times {10}^{-34}}{2\times 1.6\times {10}^{-19}\times 0.25\times 0.83\times {10}^{-2}} \\ =9.94\times {10}^{-13} \\ \mathrm{or}0.01\stackrel{0}{\mathrm{A}}\left(\mathrm{Approx}\right) \end{gathered}$$