Respuesta :
Answer:
2.07 pm
Explanation:
The problem given here is the very well known Compton effect which is expressed as
[tex]\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)[/tex]
here, [tex]\lambda[/tex] is the initial photon wavelength, [tex]\lambda^{'}[/tex] is the scattered photon wavelength, h is he Planck's constant, [tex]m_e[/tex] is the free electron mass, c is the velocity of light, [tex]\theta[/tex] is the angle of scattering.
Given that, the scattering angle is, [tex]\theta=147^{\circ}[/tex]
Putting the respective values, we get
[tex]\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.[/tex]
Here, the photon's incident wavelength is [tex]\lamda=2.78pm[/tex]
Therefore,
[tex]\lambda^{'}=2.78+4.45=7.23 pm[/tex]
From the conservation of momentum,
[tex]\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}[/tex]
where,[tex]\vec{P_\lambda}[/tex] is the initial photon momentum, [tex]\vec{P_{\lambda^{'}}}[/tex] is the final photon momentum and [tex]\vec{P_e}[/tex] is the scattered electron momentum.
Expanding the vector sum, we get
[tex]P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta[/tex]
Now expressing the momentum in terms of De-Broglie wavelength
[tex]P=h/\lambda,[/tex]
and putting it in the above equation we get,
[tex]\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}[/tex]
Therefore,
[tex]\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm[/tex]
This is the de Broglie wavelength of the electron after scattering.
