Answer
$41.4pm; 4.80fJ$
Work Step by Step
We can find the required wavelength and energy as follows:
$\lambda_f=\lambda_i+10\% \space of \lambda_i$
$\lambda_f=\lambda_i+0.1\lambda_i=1.1\lambda_i$
Now $1.1\lambda_i-\lambda_i=\frac{h}{m_ec}(1-cos135^{\circ})$
We plug in the known values to obtain:
$0.1\lambda_1=2.43\times 10^{-12}m(1-cos135^{\circ})$
This simplifies to:
$\lambda_i=\frac{2.43\times 10^{-12}m(1-cos135^{\circ})}{0.1}$
$\lambda_i=41.4\times 10^{-12}m=41.4pm$
Now $E=\frac{hc}{\lambda_i}$
We plug in the known values to obtain:
$E=\frac{(6.63\times 10^{-34}J.s)(3\times 10^8m/s)}{41.8\times 10^{-12}m}$
$E=4.8\times 10^{-14}J=4.8fJ$
