Alexandra Richardson

2022-07-22

f a photon of wavelength 0.04250 nm strikes a free electron and is scattered at an angle of ${35.0}^{\circ}$ from its original direction, find the change in the wavelength of this photon

Selden1f

Beginner2022-07-23Added 14 answers

Here, change in wavelength is determined by the relation is shown below,

$\mathrm{\u25b3}\lambda =\frac{h}{mc}(1-\mathrm{cos}\varphi )$

Here,

c = velocity of light

m = mass

h= Planck's constant

By utilizing the change in wavelength and original wavelength determine the wavelength of scattered light. The difference in the energies of the incident photons and the scattered photons is equal to the change in energy of the photon. The loss in the energy of a photon is equal to the energy gained by an electron.

$\mathrm{\u25b3}\lambda =\frac{(6.626\times {10}^{-34}\text{}J\cdot s)}{(9.109\times {10}^{-31}\text{}kg)(3\times {10}^{8})}(1-\mathrm{cos}{35}^{\circ})\phantom{\rule{0ex}{0ex}}=0.43844\times {10}^{-12}\text{}m(\frac{{10}^{9}\text{}nm}{1\text{}m})\phantom{\rule{0ex}{0ex}}=4.384\times {10}^{-4}\text{}nm$

Hence, the value of change in wavelength is $4.384\times {10}^{-4}\text{}nm$

$\mathrm{\u25b3}\lambda =\frac{h}{mc}(1-\mathrm{cos}\varphi )$

Here,

c = velocity of light

m = mass

h= Planck's constant

By utilizing the change in wavelength and original wavelength determine the wavelength of scattered light. The difference in the energies of the incident photons and the scattered photons is equal to the change in energy of the photon. The loss in the energy of a photon is equal to the energy gained by an electron.

$\mathrm{\u25b3}\lambda =\frac{(6.626\times {10}^{-34}\text{}J\cdot s)}{(9.109\times {10}^{-31}\text{}kg)(3\times {10}^{8})}(1-\mathrm{cos}{35}^{\circ})\phantom{\rule{0ex}{0ex}}=0.43844\times {10}^{-12}\text{}m(\frac{{10}^{9}\text{}nm}{1\text{}m})\phantom{\rule{0ex}{0ex}}=4.384\times {10}^{-4}\text{}nm$

Hence, the value of change in wavelength is $4.384\times {10}^{-4}\text{}nm$

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