Question 40.44: A 20 keV electron is brought to rest by colliding twice with......
A 20 keV electron is brought to rest by colliding twice with target nuclei as in Fig. 40-14. (Assume the nuclei remain stationary.) The wavelength associated with the photon emitted in the second collision is 130 pm greater than that associated with the photon emitted in the first collision. (a) What is the kinetic energy of the electron after the first collision? What are (b) the wavelength λ_1 and (c) the energy E_1 associated with the first photon? What are (d) λ_2 and (e) E_2 associated with the second photon?
(a) and (b) Let the wavelength of the two photons be λ_1 and \lambda_2=\lambda_1+\Delta \lambda . Then,
e V=\frac{h c}{\lambda_1}+\frac{h c}{\lambda_1+\Delta \lambda} \Rightarrow \lambda_1=\frac{-\left(\Delta \lambda / \lambda_0-2\right) \pm \sqrt{\left(\Delta \lambda / \lambda_0\right)^2+4}}{2 / \Delta \lambda} .
Here, Δλ = 130 pm and
\lambda_0=h c / e V=1240 \, keV \cdot pm / 20\, keV =62\, pm ,
where we have used hc = 1240 eV·nm = 1240 keV·pm. We choose the plus sign in the expression for λ_1 (since λ_1 > 0) and obtain
\lambda_1=\frac{-(130 \,pm / 62 \,pm -2)+\sqrt{(130 \,pm / 62\, pm )^2+4}}{2 / 62 \,pm }=87 \,pm .
The energy of the electron after its first deceleration is
K=K_i-\frac{h c}{\lambda_1}=20 \,keV -\frac{1240 \,keV \cdot pm }{87 \,pm }=5.7 \,keV .
(c) The energy of the first photon is E_1=\frac{h c}{\lambda_1}=\frac{1240 \,keV \cdot pm }{87 \,pm }=14 \,keV .
(d) The wavelength associated with the second photon is
\lambda_2=\lambda_1+\Delta \lambda=87 \,pm +130 \,pm =2.2 \times 10^2 \,pm .
(e) The energy of the second photon is E_2=\frac{h c}{\lambda_2}=\frac{1240 \,keV \cdot pm }{2.2 \times 10^2 \,pm }=5.7 \,keV .