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. `1Ã…`
B. `0.1Ã…`
C. `10Ã…`
D. `0.01Ã…`

Answer

D

Explanation

(d) : Radius of the circular path of a charged particle in a magnetic field is given by
`R=(mv)/(Bq)ormv=RBq`
Here, `R=0.83cm=0.83xx10^(-2)m`
`B=0.25"Wb m"^(-2)`
`q=2e=2xx1.6xx10^(-19)C`
`:.mv=(0.83xx10^(-2))(0.25)(2xx1.6xx10^(-19))`
de Broglie wavelength,
`lamda=(h)/(mv)=(6.6xx10^(-34))/(0.83xx10^(-2)xx0.25xx2xx1.6xx10^(-19))=0.01Ã…`