Question on: WAEC Physics - 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
Ask EduPadi AI for a detailed answer
Correct Option: C
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
Add your answer
Please share this, thanks!
No responses