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From equations (3-4) to (3-6) obtain equations (3.8). It is easiest to start by eliminating φ between equations (3-4) and (3-5) using cos2φ+²õ¾±²Ô2φ = 1 The electron speed u may then be eliminated between the remaining equations.

Short Answer

Expert verified

Using the given equations, the following can be obtained is λ'-λ=hmec1-cosθ.

Step by step solution

01

Given data

hλ=hλ'cosθ+γumeucosϕ…..(1)0=hλ'sinθ-γumeusinϕ…..(2)hcλ-mec2=hcλ'+γumec2…..(3)

02

Concept  used

Einstein's mass-energy equivalence relation can be expressed as,

E = mc2

03

Use the equations and solve

Rearranging the above equations (1 and 2) will give:

γumeucosϕ=hλ-hλ'cosθγumeusinϕ=hλ'sinθ

Squaring both will give:

γumeu2cos2ϕ=hλ-hλ'cosθ2γumeu2sin2ϕ=hλ'sinθ2

Adding both equations will lead to:

γumeu2=hλ2-2h2λλ'cosθ+hλ'2…..(4)

The Square of equation 3 will give

γumeu2=hλ2+hλ2-2h2λλ'+2mechλ-hλ'+me2c2…..(5)

04

Subtract equation 5 from equation 4

Simplify further,

me2γu2c2-γu2c2=-2h2λλ'1-cosθ+2mechλ'-hλ+mec22h2λλ'1-cosθ=2mechλ'-hλhλλ'1-cosθ=mec1λ'-1λλ-λ'=hmec1-cosθ

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Most popular questions from this chapter

A photon and an object of mass m have the same momentum p.

  1. Assuming that the massive object is moving slowly, so that non-relativistic formulas are valid, find in terms of m , p and c the ratio of the massive object’s kinetic energy, and argue that it is small.
  2. Find the ratio found in part (a), but using relativistically correct fomulas for the massive object. (Note: E2=p2c2+m2c4may be helpful.)
  3. Show that the low-speed limit of the ratio of part (b) agrees with part (a) and that the high-speed limit is 1.
  4. Show that at very high speed, the kinetic energy of a massive object approaches .

The electromagnetic intensity of all wavelengths thermally radiated by a body of temperature T is given by

I=σT4whereσ=5.66×10-8W·m2·K4

This is the Stefan-Boltzmann Law. To derive it. show that the total energy of the radiation in a volume V attemperature T is U=8Ï€5kB4VT4/15h3c3 by integrating Planck's spectral energy density over all frequencies. Note that

∫0∞x3ex-1dx=π415

Intensity, or power per unit area, is then the product of energy per unit volume and distance per unit time. But because the intensity is a flow in a given direction away from the blackbody, c is not the correct speed. For radiation moving uniformly in all directions, the average component of velocity in a given direction is14c .

A low-intensity beam of light is sent toward a narrow single slit. On the far side, individual flashes are seen sporadically at detectors over a broad area that is orders of magnitude wider than the slit width. What aspects of the experiment suggest a wave nature for light, and what aspects suggest a particle nature?

A beam of 500 nm light strikes a barrier in which there is a narrow single slit. At the very center of a screen beyond the single slit, 1012photons are detected per square millimeter per second.

(a) What is the intensity of the light at the center of the screen?

(b) A secood slit is now added very close to the first. How many photons will be detected per square millineter per sec and at the center of the screen now?

Equation (3-1) expresses Planck's spectral energy density as an energy per range df of frequencies. Quite of ten, it is more convenient to express it as an energy per range of wavelengths, By differentiatingf=C/λ we find thatdf=-C/λ2dλ . Ignoring the minus sign (we are interested only in relating the magnitudes of the ranges df and dλ). show that, in terms of wavelength. Planck's formula is

dUdλ=8πVhcehc/λkBT-1×1λ5

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