Here, `h=6.63xx10^-34J`,`lamda=1.0nm=10^-9m`
a. The electron and the proton will possess the same momentum, which is given by
`p=(h)/(lamda)=(6.63xx10^-34)/(10^-9)=6.63xx10^-25kg ms^-1`
b. The energy of photon,
`E=(hc)/(lamda)=(6.63xx10^-34xx3xx10^8)/(10^-9)=1.99xx10^-16J`
c. The kinetic energy of electron,
`E=(p^2)/(2m)=((6.63xx10^-25)^2)/(2xx9.1xx10^-31)=2.415xx10^-19J`
`=(2.415xx10^-19)/(1.6xx10^-19)=1.51eV`