n1 = 2, n2 = 3
Energy possessed by Hα light
= 13.6 (1/n12 – 1/n22) = 13.6 x (1/4 – 1/9) = 1.89 eV.
For Hα light to be able to emit photoelectrons from a metal...
a) When λ = 400 nm
Energy of photon=hc/ λ=1240/400=3.1eV
This energy given to electron
But for the first collision energy lost = 3.1 ev x 10% = 0.31 ev for second collision energy lost = 3.1 ev...