In large CRT televisions, electrons are accelerated from rest by a potential difference of 23.88 kV and shot onto a phosphorescent screen to produce an image. What is the speed of the electrons when they reach the screen? ( qe = 1.602 x 10 19C ;mel = 9.11 x 10 31 kg) %3D Answer: x10' (express your answers in tenths place or one decimal digit only)

In large CRT televisions, electrons are accelerated from rest by a potential difference of 23.88 kV and shot onto a phosphorescent screen to produce an image. What is the speed of the electrons when they reach the screen? ( qe = 1.602 x 10 19C ;mel = 9.11 x 10 31 kg) %3D Answer: x10' (express your answers in tenths place or one decimal digit only)

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter28: Special Relativity

Section: Chapter Questions

Problem 65PE: A Van de Graaff accelerator utilizes a 50.0 MV potential difference to accelerate charged particles...

See similar textbooks

Similar questions

Unreasonable Results A proton has a mass of 1.671027 kg. A physicist measures the proton's total energy to be 50.0 MeV. (a) What is the proton's kinetic energy? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
A Van de Graaff accelerator utilizes a 50.0 MV potential difference to accelerate charged particles such as protons. (a) What is the velocity of a proton accelerated by such a potential? (b) An electron?
Alpha decay is nuclear decay in which a helium nucleus is emitted. If the helium nucleus has a mass of 6.801027 kg and is given 5.00 MeV of kinetic energy, what is its velocity?
Suppose an electron (q = -e = -1.6 x 10¬9 C,m=9.1 x 10¬3' kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for %3D the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K+U = 0 K = -U 1 Since K mv and using the formula for potential energy above, we arrive at an equation for speed: 2 v = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=
A proton is accelerated through a potential difference of 6 MV. (megavolts) from rest. Calc. the final velocity in m/s (a) 1.4 X 10' (b) 2.4 X 10' (c) 3.4 X 10' (d) 4.4 X 10'.
Suppose an electron (q= -e = -1.6 x 10^-19 C, m = 9.1 x 10^-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. Use the template in the attached photo to solve for the problem.
Problem 17: An evacuated tube uses a potential difference of AV= 0.56 kV to accelerate electrons, which then hit a copper plate and produce X-rays. . Part (a) Write an expression for the non-relativistic speed of these electrons v in terms of e, AV, and m, assuming the electrons start from rest. v = AV 7 9 HOME a b d. 4 5 e h 1 j k P END m S V VO BACKSPACE CLEAR DEL Submit Hint Feedback I give up! Hints: 0% deduction per hint. Hints remaining: 2 Feedback: 1% deduction per feedback. Part (b) Calculate the non-relativistic speed of these electrons v in m/s.
What is the speed of an electron after being accelerated from rest through a 2.5×107 V potential difference? Express your answer as a fraction of cc.
1. Find E at (0, 0, 5) m due to Q1 = 0.35 µ C at (0, 4, 0) m and Q2 = -0.55 µ C at (3, 0, 0) m (see the figure). E,E, 7. R, R, of (0, 4, 0) (3,0, 0)
In this problem, we will try to understand why chemical reactions cannot power the Sun, but nuclear reactions can. The energy scale of chemical reactions is a few eV, where eV is a unit of energy called an electron volt. 1 eV = 1.602 x 10-19 J. A typical chemical reaction involves an energy change of ~0.1 to 10 eV. In contrast, a nuclear reaction typically involves a change in energy of order a few MeV (mega electron volts; a factor of a million larger). Suppose that the Sun has a constant luminosity throughout its life, equal to its current luminosity of L⊙=3.827×1026J/s . Suppose also that the Sun is made entirely of hydrogen (or just protons, since the mass of the electron is about 2000 times smaller and is negligible in comparison). If every pair of two protons in the Sun undergo a one-time chemical reaction that nets ~1 eV of energy, how long would it take (in years) to expend all the available chemical energy?
A) Through what potential difference does an electron have to be accelerated, starting from rest, to achieve a speed of 0.965 cc? Express your answer in volts. B) What is the kinetic energy of the electron at this speed? Express your answer in joules. Express your answer in joules. C) What is the kinetic energy of the electron at this speed? Express your answer in electronvolts. Express your answer in electronvolts.
An electron is accelerated from rest through a potential difference of 181 V. What is the electron's final speed? (e = 1.60×10-¹9 C, Melectron 9.11x10-31 kg) =