A `alpha` -parhticle moves in a circular path of radius `0.83 cm` in the presence of a magnetic field of `0.25 Wb//m^(2)`. The de-Broglie wavelength assocaiated with the particle will be

Options

A. 10 Å
B. 0.01 Å
C. 1 Å
D. 0.1 Å

Answer

B

Explanation

radius `R=(m v)/(q B)` and de-Broglie wavelength `lamda=(h)/(m v)` on solving we get `lamda=0.01` Å