A yellow coloured spectral line was observed at $5890{\mathrm{A}}^{\mathrm{o}}$ 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

$\lambda =5890\times {10}^{-10}\mathrm{m};c=3.0\times {10}^8{\mathrm{ms}}^{-}$

We know that $E=hv;E=\frac{hc}{\lambda }$

$E=\frac{6.626\times {10}^{-34}\mathrm{J}.\mathrm{s}\times 3.0\times {10}^8{\mathrm{ms}}^{-}}{5890\times {10}^{-10}\mathrm{m}}$

$=3.37\times {10}^{-19}\mathrm{J}$

But, $1\mathrm{eV}=1.6\times {10}^{-19}\mathrm{J}$

Hence, $E=3.37\times 10-19\mathrm{J}\times \frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}$

$=2.1\mathrm{eV}$

= 2.1 V x charge on one electron

$\therefore$ 2.1 Vis needed to accelerate the electron away from the atom.