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$\text{Hint: Pitch } = v \cos \theta \times T$ $\text{Step 1: Find the velocity and radius of the proton.}$ $\text{The kinetic energy of the proton is given by:}$ $\frac{1}{2} mv^2 = 2 \text{ eV}$ $\Rightarrow v = \sqrt{\frac{4 \text{ eV}}{m}} = \sqrt{\frac{2K.E}{m}}$ $\text{The radius of the path of the proton is given by:}$ $\frac{mv^2}{R} = Bqv_\perp$ $R = \frac{mv \sin 60^\circ}{qB} \quad [v_\perp = v \sin 60^\circ]$ $\text{Step 2: Find the pitch of the path of the proton.}$ $\text{The time period of the path of the proton is given by:}$ $T = \frac{2\pi r}{v_\perp} = \frac{2\pi m}{Bq}$ $\text{The pitch of the path of the proton is given by:}$ $\text{Pitch} = v \cos 60^\circ \times \frac{T}{m}$ $\Rightarrow \text{Pitch} = \sqrt{\frac{2K.E}{m}} \times \cos 60^\circ \times \frac{2\pi m}{qB}$ $\Rightarrow \text{Pitch} = \sqrt{\frac{2 \times 2 \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27}}} \times \frac{1}{2} \times \frac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 1}$ $\Rightarrow \text{Pitch} = 2\pi \times \frac{1.67 \times 10^{-27}}{1.6 \times 10^{-19}}$ $\Rightarrow \text{Pitch} = 2 \times 3.14 \times 1.021 \times 10^{-4} = 6.415 \times 10^{-4} \text{ m}$ $\Rightarrow \text{Pitch} = 6.42 \times 10^{-4} \text{ m}$ $\text{Hence, option (2) is the correct answer.}$