Question 40.44: A 20 keV electron is brought to rest by colliding twice with......

(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 $E_2=\frac{h c}{\lambda_2}=\frac{1240 \,keV \cdot pm }{2.2 \times 10^2 \,pm }=5.7 \,keV$.