A student wants to make a very small particle accelerator using a 9.0 V battery. What speed will
(b) an electron have after being accelerated from rest through the 9.0 V potential difference?
Verified step by step guidance
1
Identify the charge of an electron, which is approximately -1.6 \times 10^{-19} coulombs.
Recognize that the kinetic energy gained by the electron when accelerated through a potential difference V is given by the equation KE = qV, where q is the charge of the electron and V is the potential difference.
Substitute the values into the equation: KE = (-1.6 \times 10^{-19} C) \times (9.0 V). Note that the negative sign indicates the direction of the charge and does not affect the calculation of energy magnitude.
Use the relationship between kinetic energy and velocity, KE = \frac{1}{2}mv^2, where m is the mass of the electron (approximately 9.11 \times 10^{-31} kg) and v is the velocity. Solve for v.
Rearrange the equation to solve for v: v = \sqrt{\frac{2 \times KE}{m}}. Substitute the kinetic energy calculated and the mass of the electron to find the velocity.