Answer:
2.07 pm
Explanation:
The problem given here is the very well known Compton effect which is expressed as
![\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)](https://tex.z-dn.net/?f=%5Clambda%5E%7B%27%7D-%5Clambda%3D%5Cfrac%7Bh%7D%7Bm_e%20c%7D%281-cos%5Ctheta%29)
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, ![\theta=147^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D147%5E%7B%5Ccirc%7D)
Putting the respective values, we get
![\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.](https://tex.z-dn.net/?f=%5Clambda%5E%7B%27%7D-%5Clambda%3D%5Cfrac%7B6.626%5Ctimes%2010%5E%7B-34%7D%20%7D%7B9.11%5Ctimes%2010%5E%7B-31%7D%5Ctimes%203%5Ctimes%2010%5E%7B8%7D%20%7D%20%281-cos147%5E%5Ccirc%20%29%20m%5C%5C%5Clambda%5E%7B%27%7D-%5Clambda%3D2.42%5Ctimes%2010%5E%7B-12%7D%20%281-cos147%5E%5Ccirc%20%29%20m.%5C%5C%5Clambda%5E%7B%27%7D-%5Clambda%3D2.42%281-cos147%5E%5Ccirc%20%29%20p.m.%5C%5C%5Clambda%5E%7B%27%7D-%5Clambda%3D4.45%20p.m.)
Here, the photon's incident wavelength is ![\lamda=2.78pm](https://tex.z-dn.net/?f=%5Clamda%3D2.78pm)
Therefore,
![\lambda^{'}=2.78+4.45=7.23 pm](https://tex.z-dn.net/?f=%5Clambda%5E%7B%27%7D%3D2.78%2B4.45%3D7.23%20pm)
From the conservation of momentum,
![\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}](https://tex.z-dn.net/?f=%5Cvec%7BP_%5Clambda%7D%3D%5Cvec%7BP_%7B%5Clambda%5E%7B%27%7D%7D%7D%2B%5Cvec%7BP_e%7D)
where,
is the initial photon momentum,
is the final photon momentum and
is the scattered electron momentum.
Expanding the vector sum, we get
![P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta](https://tex.z-dn.net/?f=P%5E2_%7Be%7D%3DP%5E2_%7B%5Clambda%7D%2BP%5E2_%7B%5Clambda%5E%7B%27%7D%7D-2P_%5Clambda%20P_%7B%5Clambda%5E%7B%27%7D%7Dcos%5Ctheta)
Now expressing the momentum in terms of De-Broglie wavelength
![P=h/\lambda,](https://tex.z-dn.net/?f=P%3Dh%2F%5Clambda%2C)
and putting it in the above equation we get,
![\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}](https://tex.z-dn.net/?f=%5Clambda_%7Be%7D%3D%5Cfrac%7B%5Clambda%20%5Clambda%5E%7B%27%7D%7D%7B%5Csqrt%7B%5Clambda%5E%7B2%7D%2B%5Clambda%5E%7B2%7D_%7B%27%7D-2%5Clambda%20%5Clambda%5E%7B%27%7D%20cos%5Ctheta%7D%7D)
Therefore,
![\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](https://tex.z-dn.net/?f=%5Clambda_%7Be%7D%3D%5Cfrac%7B2.78%5Ctimes%207.23%7D%7B%5Csqrt%7B2.78%5E%7B2%7D%2B7.23%5E%7B2%7D-2%5Ctimes%202.78%5Ctimes%207.23%5Ctimes%20cos147%5E%5Ccirc%20%7D%7D%20pm%5C%5C%5Clambda_%7Be%7D%3D%5Cfrac%7B20.0994%7D%7B9.68%7D%20%3D%202.07%20pm)
This is the de Broglie wavelength of the electron after scattering.