Answer
change in wavelength will be
$\Delta \lambda=0.3256nm$
Work Step by Step
given that scattering angle $\theta=30^{0}$
Wavelength shift or Compton shift is given by
$\Delta \lambda= \lambda- \lambda_{0}=\lambda_{C} (1-cos\theta)$
$\Delta \lambda= \lambda- \lambda_{0}=\lambda_{C} (1-cos30^{0})$
$\Delta \lambda=\lambda_{C} (1-0.8660)=0.134\lambda_{C}$
for scattering by electron $\lambda_{C}=\frac{h}{m_{e}c}=2.43\times10^{-3}nm$
change in wavelength will be
$\Delta \lambda=0.134\times2.43\times10^{-3}nm=0.3256nm$