An `alpha` -particle moves in a circular path of radius 1 cm in a uniform magnetic field of 0 .1 25 T . The de-Broglie wavelength associated with the `alpha `- particle is

Options

A. `1 . 65 xx 10^(-12) m `
B. `3 . 3 xx 10^(-12) m `
C. `4 . 95 xx 10^(-12) m `
D. `6 . 6 xx 10^(-12) m `

Answer

A

Explanation

Given magnetic field , B = 0 . 1 25 T, Radius of the circular path , R = 1 cm = `10^(-2)m`
We know radius of circular path in a uniform magnetic field
`R = (mv)/(qB) rArr mv = qBR`
and de-Broglie wavelenth `lambda = (h)/(mv) " " [:. q = q_(alpha , " For " alpha - " particle" ]]`
or `lambda = (h)/(q_(alpha BR)`
`lambda = (6 . 6 xx 10^(-34))/(2xx 1.6 xx 10^(-19) xx 0.125 xx 1 xx 10^(-2))" " [:. q_(alpha = + 2C = 2 xx 1.6 xx 10 ^(-19)]]`
`rArr lambda = 1.65 xx 10^(-12) m`
Hence, de- Broglie wavelength associated with `alpha `- particle is `1 . 65 xx 10^(12) m`