A proton with a kinetic energy of 2 . 0eV moves into a region of...

The given data is

$K . E = 2eV$

$B = \dfrac{\pi}{2} \times {10}^{- 3}T$

$\theta = {60}^{o}$

The pitch of the proton is given by

$P = \dfrac{2\pim}{qB} \times vcos\theta . . . ( i )$

By using

$K = \dfrac{m{v}^{2}}{2} ⇒ v = \sqrt{\dfrac{2K}{m}}$

Substituting the values in equation (i)

$= \dfrac{2\pi \times \sqrt{2mKE} \times \dfrac{1}{2} \times 2}{1 . 6 \times {10}^{- 19} \times \pi \times {10}^{- 3}}$

$= \dfrac{2 \times \sqrt{2 \times 1 . 6 \times {10}^{- 27} \times 2 \times 1 . 6 \times {10}^{- 19}} \times {10}^{3}}{1 . 6 \times {10}^{- 19}}$

$= 2 \times 2 \times {10}^{– 1} = 0 . 4m$