Physics
WAEC 1998
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
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