Respuesta :
Answer:
4.63 p.m.
Explanation:
The problem given here can be solved by the 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=157^{\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-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.[/tex]
Therfore,
[tex]\lambda^{'}-\lambda=4.64 p.m.[/tex]
Here, the photon's incident wavelength is [tex]\lamda=7.33pm[/tex]
So,
[tex]\lambda^{'}=7.33+4.64=11.97 p.m[/tex]
From the conservation of momentum,
[tex]\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}[/tex]
here, [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]
Therfore,
[tex]\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.[/tex]
This is the de Broglie wavelength of the electron after scattering.
