A yellow coloured spectral line was observed at 5890Ao in the spectrum of a sodium vapour lamp. Calculate the minimum accelerating potential that is required to excite this line in an electron tube containing sodium vapour

λ=5890×10−10m;c=3.0×108ms−We know that E=hv;E=hcλE=6.626×10−34J.s×3.0×108ms−5890×10−10m=3.37×10−19JBut, 1eV=1.6×10−19JHence, E=3.37×10−19J×1eV1.6×10−19J=2.1eV= 2.1 V x charge on one electron∴ 2.1 Vis needed to accelerate the electron away from the atom.