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

Explanation

We know that `lamda = h/P = (h)/(qBr) " " [ therefore (mv^2)/(r ) = q v B ]`
` therefore lamda = (6.62 xx 10^(-34) )/(2 xx 1.6 xx 10^(-19) xx 0.5 xx 83 xx 10^(-4) )`
`r lamda 0.01 Å`