An electron is accelerated from rest through a potential difference of 70kV in a vacuum. Calculate the maximum speed acquired by the electron (electronic charge = -1.6 x 10-19; mass of an electron = 9.1 x 10-31kg)
A. 3.00 x 108ms-1 B. 2.46 x 108ms-1 C. 1.57 x 108ms-1 D. 1.32 x 108ms-1 E. 1.11 x 108ms-1
Correct Answer: C
Explanation
ev = \(\frac{1}{2}\)mv2 v2 = \(\frac{ev}{\frac{1}{2}m}\) V = \(\sqrt{\frac{ev}{\frac{1}{2m}}} = \sqrt{\frac{1.6 \times 10^{-19} \times 70 \times 10^3}{\frac{1}{2} \times 9.1 \times 10^{-31}}}\) = \(\sqrt{246.2 \times 10^{14}}\) = 1.57 x 108
