Saved Bookmarks
| 1. |
Calculate the maximum kinetic energy and maximum velocity of the photoelectrons emitted when the stopping potential is 81V for the photoelectric emission experiment. |
|
Answer» Solution :MAXIMUM KINETIC energy of electron, `{:(K_(max),= eV_(0)""V_(0) = 81 V),(,= 1.6 xx 10^(-19) xx 81""e = 1.6 xx 10^(-19) C),(,=129.6 xx 10^(-19)""m = 9.1 xx 10^(-31) kg),(,=1.29 xx 10^(-17)),(K_(max),= 1.3 xx 10^(-17) J):}` Maximum velocity of photoelectron, `v_(max) = sqrt((2eV_(0))/(m))` `= sqrt((2 xx 1.6 xx 10^(-19) xx 81)/(9.1 xx 10^(-31))) = sqrt((259.2 xx 10^(-19))/(9.1 xx 10^(-31))) = sqrt(28.48 xx 10^(12))` `v_(max) = 5.3 xx 10^(6) ms^(-1)` |
|