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.

unbalanced force

explanation:

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures