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. `1Å`
B. `0.1Å`
C. `10Å`
D. `0.01Å`

Answer

D

Explanation

Radius of the circular path of a charged particle in a magnetic field is given by
`R = (mv)/(Bq) " or " mv = RBq`
Here, `R = 0.83 cm = 0.83 xx 10^(-2)m`
`B = 0.25 Wbm^(-2)`
`q = 2e = 2 xx 1.6 xx 10^(-19)C`
`:. Mv = (0.83 xx 10^(-2)) (0.25)(2 xx 1.6 xx 10^(-19))`
de Broglie wavelength,
`lamda=h/(mv) =(6.6xx10^(-34))/(0.83 xx10^(-2) xx 0.25 xx 2xx1.6 xx 10^(-19))=0.01 Å`