An \( \alpha \)-particle of mass \( 6.4 \times 10^{-27}\text{ kg} \) and charge \( 3.2 \times 10^{-19}\text{ C} \) is situated in a uniform electric field of \( 1.6 \times 10^5\text{ Vm}^{-1} \). The velocity of the particle at the end of \( 2 \times 10^{-2}\text{ m} \) path when it starts from rest is:
\( 2\sqrt{3} \times 10^5\text{ ms}^{-1} \)
\( 8 \times 10^5\text{ ms}^{-1} \)
\( 16 \times 10^5\text{ ms}^{-1} \)
\( 4\sqrt{2} \times 10^5\text{ ms}^{-1} \)
Solution:
Using the work-energy theorem, \( qEd = \frac{1}{2}mv^2 \). Rearranging gives \( v = \sqrt{\frac{2qEd}{m}} = \sqrt{\frac{2 \times 3.2 \times 10^{-19} \times 1.6 \times 10^5 \times 2 \times 10^{-2}}{6.4 \times 10^{-27}}} = 4\sqrt{2} \times 10^5\text{ m/s} \).
Leave a Reply