A photoelectric plate is initially exposed to a spectrum of hydrogen gas excited to second energy level. Late when the same photoelectric plate is exposed to a spectrum of some unknown hydrogen like gas, excited to second energy level. It is found that the de-Broglie wavelength of the photoelectrons now ejected has increased `sqrt(6.1)` times. For this new gas difference of energies of first Lyman line and Balmer series limit is found to be two times the ionization energy of the hydrogen atom is ground state. Detect the atom and determine the work function of the photoelectric plate.

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(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.