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: Option 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