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Chapter 12: Q18E (page 556)

Approximately what energy would electrons need to have to be useful as probes that could reveal features as small as 10-18m,the approximate range of the weak force?

Short Answer

Expert verified

The energy would the electrons need to have to be useful as probes that could reveal features as small as10-18m, the approximate range of the weak force is 1.2TeV.

Step by step solution

01

Given data 

The size of an object, λ=10-18m.

02

Concept of Energy of electron

The energy of an electron is given as,

E2=m2c4+p2c2

E2≈p2c2E≈hλc

Where is the mass of an electron, 9.1×10-31kg,h is reduced Planck's constant, 6.63×10-34J·s, λis the size of an object to be examined and is the velocity of light, 3×108m/s.

03

Calculation of the energy for an electron 

Substitute the numerical values in equation (1) as:

E=6.63×10-34J·s3×108m/s10-18mE=1.989×10-7J=1.2TeV

The energy would the electrons need to have to be useful as probes that could reveal features as small as10-18m , the approximate range of the weak force is 1.2TeV.

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Most popular questions from this chapter

Equation (12-7) assumes a matter-dominated universe in which the energy density of radiation is insignificant. This situation prevails today and has to do with the different rates at which the densities of matter and radiation vary with the size of the universe. Matter density is simply inversely proportional to the volume, obeying ρmIR3 , whereρm is the matter density now. Radiation density, however, would be proportional to I/R4(Not only does the volume increase, but also all wavelengths are stretched in proportion to R. lowering the energy density by the extra factor.) This density drops faster As the universe grows, but it also grows more quickly in the backward time direction. In other words, long ago, the universe would have been radiation dominated. Show that if the function used for matter density in equation (12-7) is replaced by one appropriate to radiation, but retaining the assumption that K' andΩA are both 0, then the scale factor Rwould grow ast1/2

The electron mentioned in Section 12.3 for deep inelastic scattering experiments is20GeV , and the momentum is given as .20 GeV/c Why so simple a conversion?

In the following exercises, two protons are smashed together in an attempt to convert kinetic energy into mass and new particles. Indicate whether the proposed reaction is possible. If not, indicate which rules are violated. Consider only those for charge, angular momentum, and baryon number If the reaction is possible, calculate the minimum kinetic energy required of the colliding protons.

p+p→n+n¯+p+p¯

To produce new particle accelerators often smash two equal mass objects together proton and proton or electron and positron. The threshold energy is the kinetic energy before the collision needed simply to produce the final particle their mass thermal energy alone with no leftover kinetic energy. Consider a colliding beam accelerator in which two initial particles of mass m are moving at the same speed relative to the lab. Assume that the total mass of the stationary particles after the collision is M. Show that the threshold energy is M-2mc2.

Exercise 23 discusses the threshold energy for two particles of mass m in a colliding beam accelerator to produce a final stationary mass M. If the accelerator is instead a stationary target type, more initial kinetic energy is needed to produce the same final mass. Show the threshold energy is M22m-2mc2.
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