Question

X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{\circ}θ=30 ​∘ ​​ relative to the incident X-rays, what is the energy of the recoiling electron?

Answer 1

Answer:

37.91594 keV

Explanation:

$E_i$ = Incident energy = 400 keV

θ = 30°

h = Planck's constant = 4.135×10⁻¹⁵ eV s = 6.626×10⁻³⁴ J s

Incident photon wavelength

$\lambda_i=\frac{hc}{E_i}\\\Rightarrow \lambda_i=\frac{4.135\times 10^{-15}\times 3\times 10^8}{400\times 10^3}\\\Rightarrow \lambda_i=3.101\times 10^{-12}\ m$

Difference in wavelength

$\Delta \lambda=\frac{h}{m_ec}(1-cos\theta)\\\Rightarrow \Delta \lambda=\frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8}(1-cos30)\\\Rightarrow \Delta \lambda=3.248\times 10^{-13}\ m$

$\lambda_f=\lambda_i+\Delta \lambda\\\Rightarrow \lambda_f=3.101\times 10^{-12}+3.248\times 10^{-13}\\\Rightarrow \lambda_f=3.426\times 10^{-12}$

Final photon wavelength

$\lambda_f=\frac{hc}{\lambda_f}\\\Rightarrow E_f=\frac{4.135\times 10^{-15}\times 3\times 10^8}{3.426\times 10^{-12}}\\\Rightarrow E_f=362084.06\ eV = 362.08406\ keV$

Energy of the recoiling electron

$\Delta E=E_i-E_f\\\Rightarrow \Delta E=400-362.08406=37.91594\ keV$

Energy of the recoiling electron is 37.91594 keV

Answer 2

Answer:

The energy of recoiling electron=192.44 keV

Explanation:

Energy of x-ray$E_o=400 keV$

Web know that compton shift is

$\Delta \lambda =\dfrac{h}{m_eC(1-cos\theta)}$

$m_e$ is the mass of electron and C is the velocity of sound.

Given that θ=30°

Now by putting the values

$\Delta \lambda =\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8(1-cos30)}$

$\Delta \lambda =2.8\times 10^{-3}nm$

$\Delta \lambda_o=\dfrac{hc}{E_o}$

By putting the values

$\Delta \lambda_o=\dfrac{6.63\times 10^{-34}3\times 10^8}{400\times 1.602\times 10^{-16}}$

$\Delta \lambda_o=3.31times 10^{-3}nm$

$\lambda =\Delta \lambda +\lambda _o$

$\lambda=2.8\times 10^{-3}+3.31\times 10^{-3}nm$

$\lambda=6.11`\times 10^{-3}nm$

Energy $E=\dfrac{hC}{\lambda }$

So $E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{6.11\times 10^{-12}}$

E=207.55 keV

The energy of recoiling electron=400-207.55 keV

The energy of recoiling electron=192.44 keV