Energy of emitted photon is \\\\(E=h\
u\\\\) For a photon its momentum is given as \\\\(\\dfrac{Energy}{velocity \\ of \\ light}\\\\) which is nothing but \\\\(\\dfrac{E}{c}\\\\) or \\\\(\\dfrac{h\
u}{c}\\\\) According to momentum conservation value of the momentum of recoiled nucleus should also be \\\\(\\dfrac{h\
u}{c}\\\\) but in opposite direction to make net momentum zero before and after emission.Now recoil energy \\\\(E_2\\\\) = \\\\(\\dfrac { p^2}{2M}\\\\) where \\\\(p\\\\) is recoil momentum putting \\\\(p\\\\) \\\\(=\\\\) \\\\(\\dfrac{h\
u}{c}\\\\) we get recoil energy as \\\\(\\dfrac{h^{2} \
u^{2}}{2Mc^2}\\\\) i.e., option C