Physics - Modern Physics Question with Solution | TestHub

When electron pass through the mercury vapor, it losses some of its energy. The loss in$KE$of electron$=\left(5\cdot 6-0.7\right)eV$$=4.9 eV$

$\therefore$energy of radiation emitted$= 4.9 eV$

$\therefore$wavelength of radiation,$\lambda =\frac{1.24\times {10}^4}{4.9}A\approx 250nm$