Saved Bookmarks
| 1. |
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. |
|
Answer» For minimum de-Broglie wavelength of electron, the momentum and hence `KE` of electron must be maximum. Case-I. When hydrogen spectra is used, for maximum energy of photoelectron `13.6[(1)/(1^(2))-(1)/(2^(2))]-phi=(1)/(2)mv_(1)^(2) rArr (1)/(2)mv_(1)^(2)=(3)/(4)xx13.6eV-phi` For unknown gas of atomic no.`z` `13.6z^(2)[(1)/(1^(2))-(1)/(2^(2))]-phi=(1)/(2)mv_(2)^(2) rArr (1)/(2)mv_(2)^(2)=z^(2)(3)/(4)xx13.6ev-phi` Comparing the de-Broglie wavelengths `(lambda_(1))/(lambda_(2))=(h//mv_(1))/(h//mv_(2))=((v_(2))/(v_(1)))` `rArr ((lambda_(1))/(lambda_(2)))^(2)=((3)/(4)xx13.6-phi)/((3)/(4)xx13.6z^(2)-phi)=(1)/(6.1)`....(`1`) For the unknown gas Energy of first lvman line-Energy of Balmer series limit `=2xx(13.6)eV` `(13.6)z^(2)((3)/(4))-13.6z^(2)((1)/(4))=2(13.6)` `z^(2)//2=2`, `z=2` `:. ` Unknown gas is helium Now from equation (`1`) `(10.2-phi)/(40.8-phi)=(1)/(6.1)` `rArr phi=4.2eV` |
|