Answer
$\lambda = 2.485\times 10^{-11}~m$
Work Step by Step
We can find the energy of the scattered photon:
$E_f = E_0-K$
$E_f = (240.0~keV)-(190.0~keV)$
$E_f = 50.0~keV$
We can find the wavelength of the scattered photon:
$E_f = \frac{hc}{\lambda}$
$\lambda = \frac{hc}{E_f}$
$\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(50.0\times 10^3~eV)(1.6\times 10^{-19}~J/eV)}$
$\lambda = 2.485\times 10^{-11}~m$
