Modern Physics
3rd Edition
ISBN: 9781111794378
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
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Chapter 5, Problem 29P
To determine
The ratio of the probability of an electron arriving at an maximum interference and to the minimum interference.
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a) In a double-slit experiment, a stream of electrons with speed v = 10 m/s is aimed at two parallel
slits. On the screen one meter behind the slits an interference pattern is observed in the number of
electrons arriving at different points on the screen. The distance between the first and the second
intensity maxima is 1 cm. Find the distance between the slits.
A linear particle accelerator using beta particles collides electrons with
their anti-matter counterparts, positrons. The accelerated electron hits the
stationary positron with a velocity of 77 x 10° m/s, causing the two
particles to annihilate.
If two gamma photons are created as a result, calculate the energy of
each of these two photons, giving your answer in MeV (mega electron
volts), accurate to 1 decimal place. Take the mass of the electron to be
5.486 x 104 u, or 9.109 x 1031 kg.
Note: Assume that the kinetic energy is also converted into the gamma
rays, and is included in the two photons.
A beam of electrons, each with the same ki- netic energy, illuminates a pair of slits sepa rated by a distance of 63 nm. The beam forms bright and dark fringes on a screen located a distance 1.1 m beyond the two slits. The arrangement is otherwise identical to that used in the optical two-slit interference experiment. The bright fringes are found to be separated by a distance of 0.1 mm.
What is the kinetic energy of the elec- trons in the beam? Planck’s constant is 6.63 × 10^−34 J · s.
Answer in units of keV.
Chapter 5 Solutions
Modern Physics
Ch. 5.1 - A 0.20-kg ball is thrown upward. How much work is...Ch. 5.5 - Prob. 5ECh. 5 - Prob. 1QCh. 5 - Prob. 2QCh. 5 - Prob. 3QCh. 5 - Prob. 4QCh. 5 - Prob. 5QCh. 5 - Prob. 7QCh. 5 - Prob. 8QCh. 5 - Prob. 9Q
Ch. 5 - Prob. 10QCh. 5 - Prob. 11QCh. 5 - Prob. 1PCh. 5 - Prob. 2PCh. 5 - Prob. 3PCh. 5 - Prob. 4PCh. 5 - Prob. 5PCh. 5 - Prob. 6PCh. 5 - Prob. 7PCh. 5 - Prob. 8PCh. 5 - Prob. 9PCh. 5 - Prob. 10PCh. 5 - Prob. 11PCh. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Show that the group velocity for a nonrelativistic...Ch. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Prob. 19PCh. 5 - Prob. 20PCh. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 26PCh. 5 - Prob. 27PCh. 5 - Prob. 28PCh. 5 - Prob. 29PCh. 5 - Prob. 30PCh. 5 - Prob. 31PCh. 5 - Prob. 32PCh. 5 - Prob. 33PCh. 5 - Prob. 34PCh. 5 - Prob. 35PCh. 5 - Prob. 36PCh. 5 - Prob. 37P
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- The width of the central bright fringe in a diffraction pattern on a screen is identical when either electrons or optical light pass through a single slit. The distance between the screen and the slit is the same in each case and is large compared to the slit width. If the wavelength of the optical light is 552 nm, how fast are the electrons moving?Give your answer in units of m/s, accurate to 2 decimal placesarrow_forwardThe table gives relative values for three situations for the barrier tunneling experiment of the figures. Electron Energy Barrier Height Barrier Thickness (a) 5E L (b) 17E L/2 (c) 2E 2L Energy V-0 V<0 V-0 Electron * 0 x l. Rank the situations according to the probability of the electron tunneling through the barrier. If multiple situations rank equally, use the same rank for each, then exclude the intermediate ranking (i.e. if objects A, B, and C must be ranked, and A and B must both be ranked first, the rạnking would be A:Greatest, B:Greatest, C:Third greatest). If all situations rank equally, rank each as 'Greatest'. (a) (b) (c)arrow_forwardFor the double slit experiment with electrons, which one of the following statements is true according to the standard (Copenhagen) interpretation of quantum mechanics? It is possible to measure which slit an electron went through, but if you make this measurement, the beam of electrons will no longer form the multiple fringe pattern on the screen. It is, in principle, possible to measure which slit an electron went through and still see a multiple fringe pattern, if the technology is sophisticated enough. Each electron must have gone through one slit or the other, but it is impossible to measure which slit any one particular electron went through. No answer text provided.arrow_forward
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