Answer
$60^{\circ}$
Work Step by Step
We can find the required scattering angle as follows:
$\Delta \lambda=\frac{h}{m_e c}(1-cos\theta)$
We plug in the known values to obtain:
$\Delta \lambda=\frac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^8}(1-(180^{\circ}))$
$\Delta \lambda=4.8\times 10^{-12}m$
Set the wavelength shift to one fourth the maximum:
$\Delta \lambda=\frac{1}{4}(4.8\times 10^{-12}m)=1.215\times 10^{-12}m$
$\Delta \lambda=\frac{h}{m_e c}(1-cos\theta)$
We plug in the known values to obtain:
$1.215\times 10^{-12}=\frac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^8}(1-cos\theta)$
$\implies 0.5=1-cos\theta$
$\implies \theta=60^{\circ}$
