Aphoton of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. what is the de broglie wavelength of the electron after the photon has been scattered?

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

here,  is the initial photon wavelength, is the scattered photon wavelength, h is he Planck's constant, is the free electron mass, c is the velocity of light,  is the angle of scattering.

Given that, the scattering angle is,

Putting the respective values, we get

Therfore,

Here, the photon's incident wavelength is

So,

From the conservation of momentum,

here, is the initial photon momentum, is the final photon momentum and is the scattered electron momentum.

Expanding the vector sum, we get

Now expressing the momentum in terms of De-Broglie wavelength

and putting it in the above equation we get,

Therfore,

This is the de Broglie wavelength of the electron after scattering.

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

hope this ! this is not copied! search it up, it's free of plagiarism!

c) power lifting, will be your answer.

,

deku ❤