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

Answer

A

Explanation

`lamda=h/(mv)`
`qvB=(mv^2)/R`
`qB=(mv)/R`
`mv=qBR`
`lamda=h/(qBR)=(6.63xx10^(-34))/(2xx1.6xx10^(-19)xx0.25xx0.83xx10^(-2))`
`" "[:.q=2e]`
`=9.8xx10^(-13)=0.0098xx10^(-10)=0.01Å`