A 500kV is applied across an X-ray tube. Calculate the maximum velocity of the electrons produced [Me = 9.1 x 10\(^{-31}\)kg, e = 1.6 x 10\(^{-19}\)C]
A
4.2 x 108ms-1
B
1.9 x 108ms-1
C
4.2 x 105ms-1
D
1.8 x 105ms-1
Verified Correct Answer: A.
Physical Principles & Numerical Solution
500Kv = 5 x 10\(^5\)V
E = Qv = \(\frac{1}{2}\)mv\(^2\)
E = 1.6 x 10\(^{-19}\) x 9.1 x 10\(^{-31}\) = 8 x 10\(^{-14}\)
8 x 10\(^{-14}\) = \(\frac{1}{2}\)mv\(^2\)
8 x 10\(^{-14}\) = \(\frac{1}{2}\) x 9.1 x 10\(^{-31}\) x v\(^2\)
8 x 10\(^{-14}\) = 4.55 x 10\(^{-31}\) x v\(^2\)
v\(^2\) = \(\frac{8 \times 10^{-14}}{4.55 \times 10^{-31}}\) = 1.758 x 10\(^{17}\)
v = \(\sqrt{1.758 \times 10^{17}}\) = 4.2 x 10\(^8\)m/s
Source: Myschool Past Questions Archive (Record #34701)