An `alpha`-particle moves in a circular path of radius 0.83 cm in the presence of a magnetic field of `0.25 Wb//m^(2)`. What is the de Broglie wavelength associated with the particle?

Explanation

The radius of a charged particle rotating in a circular path in a magnetic field
`R = (mv)/(Bq) or, mv = RBq`
The de Broglie wavelength associated with the particle,
`lamda = (h)/(mv) = (h)/(RBq)`
Here, `R = 0.83 cm = 0.83 xx 10^(-2) m, B = 0.25 Wb.m^(-2)`,
`q = 2e = 2 xx 1.6 xx 10^(-19) C` [ `:' alpha`-particle]
`:. lamda = (6.6 xx 10^(-34))/(0.83 xx 10^(-2) xx 0.25 xx 2 xx 1.6 xx 10^(-19)) = 0.01 Å`