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

Explanation

We have , `10.2 =W + (K_("max"))_(1)" "…(i)`
and `10.2 Z^(2) = W+(K_("max"))_(2)" "….(ii)`
Also `lambda_("de -Brogile")=h/p =h/(sqrt(2mK))`
` :. (lambda_(1))/(lambda_(2)) = sqrt((K_(2))/(K_(1)))= 2.3 rArr K_(2)=5.25 K_(1)`
Also , `10.2 Z^(2)` = emergy corresponding to longest wavelength of the Lyman series = `3 xx 13.6`
`rArr Z=2`
From Eqs , (i) ,(ii) and (iii) , W= 3eV