Radiation from hydrogen gas excited to first excited state is used for illuminating certain photoelectric plate . When the radiation from same unknown hydrogen-like gas excited to the same level is used to expose the same plate , it is found that the de Broglie wavelength of the fastest photoelectron has decreased `2.3 time`gt It is given that the energy corresponding to the longest wavelength of the Lyman series of the known gas is `3` times the ionization energy of the hydrogen gas `(13.6 eV)`. Find the work function of photoelectric plate in `eV`. `[Take (2.3)^(2) = 5.25]`

Answer

3 eV

Explanation

We have
`10.2 = W + K_(max.1)` (i)
and `10.2 Z^(2) = W + K_(max.2)` (ii)
Also ` lambda _(de Broglie) = (h)/(p) = (h)/(sqrt(2 mK))`
`:. (lambda_(1))/(lambda_(2)) = sqrt((K_(2))/(K_(1))) = 2.3 implies 5.25K_(1)` (iii)
Also `10.2 Z^(2) =` energy corresponding to longest wavelength of the Lyman series `= 3 xx 13.6`
`implies Z = 2`
From Eqs, (i) , (ii) and (iii), `W = 3 e V`