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]`
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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(2mK))`, `therefore (lambda _(1))/(lambda_(2))=sqrt((K_(2))/(K_(1)))=2.3 implies K_(2)=5.25 K_(1)` …..(iii) Also `10.2 Z^(2)= energy corresponding to longest wavelength of the Lyman series =3`xx 13.6 implies Z=2. therefore From equations (i),(ii), and (iii) W=3eV=3.
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