Answer
$5.7\;keV$
Work Step by Step
Energy of the electron: $E=20\;keV=20000 \;eV$
Difference in wavelength:
$\Delta\lambda=\lambda_2-\lambda_1=130\;pm=0.13\;nm\\
\lambda_2=\Delta\lambda+\lambda_1$
$E=\frac{hc}{\lambda_1}+\frac{hc}{\lambda_2}$
$\frac{E}{hc}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}$
$\frac{1}{\lambda_0}=\frac{1}{\lambda_1}+\frac{1}{\lambda_2}$ [where, $\frac{hc}{E}=\lambda_0$] .....................(1)
Now,
$\lambda_0=\frac{hc}{E}$
or, $\lambda_0=\frac{6.626\times 10^{-34}\times3\times10^{8}}{20000\times1.6\times10^{-19}}\;m=6.21\times10^{-11}\;m=0.0621\;nm$
Now, eq. (1) becomes
$\frac{1}{\lambda_0}=\frac{1}{\lambda_1}+\frac{1}{\Delta\lambda+\lambda_1}$
or, $\lambda_1^2+\lambda_1(\Delta\lambda-2\lambda_0)-\lambda_0\Delta\lambda=0$
or, $\lambda_1^2+\lambda_1(0.13-2\times0.0621)-0.0621\times0.13=0$
or, $\lambda_1^2+0.0058\lambda_1-8.073\times10^{-3}=0$
The solution is:
$\lambda_1=\frac{-0.0058\pm\sqrt {0.0058^2+4\times1\times8.073\times10^{-3}}}{2}\;nm$
The valid solution is:
$\lambda_1=0.087\;nm$
Thus, the wavelength associated with the photon emitted in the First collision is: $\lambda_1=0.087\;nm$
Thus,
$\lambda_2=\Delta\lambda+\lambda_1=(0.13+0.087)\;nm=0.217\;nm$
Therefore, the wavelength associated with the photon emitted in the second collision is: $\lambda_2=0.217\;nm$
Thus the energy associated with the photon emitted in second photon is
$\frac{hc}{\lambda_2}=\frac{1240\;eV-nm}{0.217\;nm}=5714\;eV=5.7\;keV$
