contestada

You will find in Chapter 39 that the electrons cannot move indefinite orbits within atoms, like the planets in our solar system.To see why, let us try to "observe" such an orbiting electron byusing a light microscope to measure the electron's presumed orbitalposition with a precision of, say, 8.2 pm(a typical atom has a radius of about 100 pm). The wavelength ofthe light used in the microscope must then be about 8.2 pm.
(a) What would be the photon energy of thislight?
1 keV

(b) How much energy would such a photon impart to an electron in ahead-on collision?
2 keV

(c) What do these results tell you about the possibility of"viewing" an atomic electron at two or more points along itspresumed orbital path? (Hint: The outer electrons of atoms arebound to the atom by energies of only a few electron-volts.)

Respuesta :

chamberlainuket

Answer:

1.. 151keV

2. 56.9keV

3. It isn't possible to view an electron with such a photon because of the high energy

Explanation:

See attached file for calculation

The energy of a photon is the product of the planks constant and frequency. The photon energy of this light is 151 keV.

The Energy of a photon:

It is the product of the planks constant and frequency.

[tex]E = \dfrac {hc}\lambda[/tex]

Where,

[tex]h[/tex] -Plank's constant  = [tex]\bold {6.63\times 10^{-34}\rm \ J/s}[/tex]

[tex]c[/tex]- speed of light = [tex]\bold { 3\times 10^8 \rm \ m/s }[/tex]

[tex]\lambda[/tex][tex]\labda[/tex]- wavelength = 8.2 pm = [tex]\bold {8.2 \times 10^{-12 }\rm \ m}[/tex]

Put the values in the equation,

[tex]E = \dfrac { 6.63\times 10^{-34}\rm \ J/s\times 3\times 10^8 \rm \ m/s }{ 8.2 \times 10^{-12 }\rm \ m}\\E = 151 \rm \ keV[/tex]

Therefore, the photon energy of this light is 151 keV.

Learn more about photon energy:

https://brainly.com/question/11733357

Otras preguntas