Saved Bookmarks
| 1. |
Photoelectric work- function of a metal is `1 eV.` Light of wavelength `lambda = 3000 Å` falls on it. The photoelectrons come out with maximum velocityA. `10ms^(-1)`B. `10^(3)ms^(-1)`C. `10^(4)ms^(-1)`D. `10^(6)ms^(-1)` |
|
Answer» Correct Answer - D `(1)/(2)mv^_(max)^(2)=(hc)/(lamda)-phi_(0)` `(1)/(2)mv_(max)^(2)=(12375eV)/(3000)-1eV` `=3.125xx1.6xx10^(-19)` `v_(max)=sqrt((2xx3.125xx1.6xx10^(-19))/(9.1xx10^(-31)))approx10^(6)ms^(-1)` |
|