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. `1A`
B. `0.1A`
C. `10A`
D. `0.01A`

Answer

d

Explanation

Here, `r=0.83cm=0.83xx10^(-2)m`,
`B=0.25Wb//m^(2)`
`Bqv=mv^(2)//r` or `mv=Bqr`
`:. lambda=h/(mv) = h/(Bqr)`
`=(6.6xx10^(-34))/(0.25xx(2xx1.6xx10^(-19))xx(0.83xx10^(-2)))`
`~~0.1xx10^(-10)m=0.01Å`