An `alpha`-particle moves in curcular path of radius `0.83 cm` in the presence of a magnetic field of `0.25 Wb//m^(2)`. Find the De Brogile wavelength associated with the particle.

Options

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

Answer

D

Explanation

`lamda =h/P`
Since ,` r = (mv)/(qB)`
`implies mv = qrB = 2e xx0.83 xx10^(-2) xx1/4`
`lamda = (6.6 xx10^(-34) xx4)/(2 xx1.6 xx10^(-19)xx0.83 xx10^(-2))`
`implies lamda = 0.01 Å`