InterviewSolution
Saved Bookmarks
| 1. |
A metal is irradiated with light of wavelength 660 nm. Given that the work that the work function of the metal is `1.0eV,` the de Broglie wavelength of the ejected electron is close to-A. `6.6xx10^(-7)m`B. `8.9xx10^(-11)m`C. `1.3xx10^(-9)m`D. `6.6xx10^(-13)m` |
|
Answer» Correct Answer - C `E_("Absorbed")=(hc)/(lamda)=(12400)/(600)=1.88eV` `E_("Absorbed")=E_(0)+KE` `KE=E_("absorbed")-E_(0)` `=(1.88-1)eV=0.88eV` de Broglie wavelength`lamda=(h)/(sqrt(2Em))=sqrt((150)/(KE(eV)))` `=sqrt((150)/(0.88))=sqrt(170.45)` `=13"Ã…"=1.3xx10^(-9)m` |
|