When light of wavelength 248nm falls on a metal of threshold energy 3.0 eV, the de-Broglie wavelength of emitted electronic is …….. A.  (Round off to the Nearest Integer). [Use: 3=1.73,h=6.63×10−34 ] Js ml=9.1×10−31kg;c=3.0×108ms;1eV=1.6×10−19J

Energy of incident radiation =hcλ6.63×10−34×3×108248×10−90.0802×10−178.02×10−19J(w) work function =3ev=3×1.602×10−19J =4.806×10−19JK.E of photo e−=I^inc−w=8.02×10−19−4.806×10−19=3.214×10−19Jλ=6.63×10−3458.49×10−50=6.637.64×10−9=0.86×10−9mλ=8.6A0